A mixture of right angled triangle questions.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], [], []], "questions": [{"name": "15.a Calculating angles using cos", "extensions": ["eukleides"], "custom_part_types": [], "resources": ["question-resources/Picture1_caMIdF1.png", "question-resources/Picture2_6KE4ZpW.png"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$Opposite = \\\\var{bc}$
$Hyptonuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $arcsin$ or notated $sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = arcsin(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse cos\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arccos(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$Opposite = \\\\var{bc}$
$Adjacent = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $arctan$ or notated $tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = arctan(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = x$
$Hypotenuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Hypotenuse = \\\\var{ab}$
$Adjacent = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = \\\\var{bc}$
$Adjacent = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "If(SCT='c',arccos(ac/ab),if(SCT = 's',arcsin(bc/ab),arctan(bc/ac)))", "description": "", "templateType": "anything", "can_override": false}, "gen_ac": {"name": "gen_ac", "group": "Unnamed group", "definition": "random(3 .. 12#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gen_bc": {"name": "gen_bc", "group": "Unnamed group", "definition": "random(5 .. 15#0.1)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "300"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "SCT", "AngORside", "answer", "angle", "gen_ac", "gen_bc"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\n
Give your answer in degrees (make sure you calculator is in the right mode), correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arcsin(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$Adjacent = \\\\var{ac}$
$Hyptonuse = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse cos\\' button on the calculator (also called $arccos$ or notated $cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = arccos(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$Opposite = \\\\var{bc}$
$Adjacent = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $arctan$ or notated $tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = arctan(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = x$
$Hypotenuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Hypotenuse = \\\\var{ab}$
$Adjacent = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = \\\\var{bc}$
$Adjacent = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "If(SCT='c',arccos(ac/ab),if(SCT = 's',arcsin(bc/ab),arctan(bc/ac)))", "description": "", "templateType": "anything", "can_override": false}, "gen_ac": {"name": "gen_ac", "group": "Unnamed group", "definition": "random(3 .. 12#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gen_bc": {"name": "gen_bc", "group": "Unnamed group", "definition": "random(5 .. 15#0.1)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "300"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "SCT", "AngORside", "answer", "angle", "gen_ac", "gen_bc"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\n
Give your answer in degrees (make sure you calculator is in the right mode), correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$Opposite = \\\\var{bc}$
$Hyptonuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $arcsin$ or notated $sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = arcsin(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$Adjacent = \\\\var{ac}$
$Hyptonuse = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse cos\\' button on the calculator (also called $arccos$ or notated $cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = arccos(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse tan\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arctan(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = x$
$Hypotenuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Hypotenuse = \\\\var{ab}$
$Adjacent = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = \\\\var{bc}$
$Adjacent = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "If(SCT='c',arccos(ac/ab),if(SCT = 's',arcsin(bc/ab),arctan(bc/ac)))", "description": "", "templateType": "anything", "can_override": false}, "gen_ac": {"name": "gen_ac", "group": "Unnamed group", "definition": "random(3 .. 12#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gen_bc": {"name": "gen_bc", "group": "Unnamed group", "definition": "random(5 .. 15#0.1)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "300"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "SCT", "AngORside", "answer", "angle", "gen_ac", "gen_bc"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\n
Give your answer in degrees (make sure you calculator is in the right mode), correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\nAngles are given in degrees (make sure you calculator is in the right mode)
Give your answer correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotonuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arcsin(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse cos\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arccos(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse tan\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arctan(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = x$
$Hypotenuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Hypotenuse = \\\\var{ab}$
$Adjacent = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = \\\\var{bc}$
$Adjacent = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "If(SCT='c',arccos(ac/ab),if(SCT = 's',arcsin(bc/ab),arctan(bc/ac)))", "description": "", "templateType": "anything", "can_override": false}, "gen_ac": {"name": "gen_ac", "group": "Unnamed group", "definition": "random(3 .. 12#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gen_bc": {"name": "gen_bc", "group": "Unnamed group", "definition": "random(5 .. 15#0.1)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "300"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "SCT", "AngORside", "answer", "angle", "gen_ac", "gen_bc"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\n
Give your answer in degrees (make sure you calculator is in the right mode), correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\nAngles are given in degrees (make sure you calculator is in the right mode)
Give your answer correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\nAngles are given in degrees (make sure you calculator is in the right mode)
Give your answer correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\nAngles are given in degrees (make sure you calculator is in the right mode)
Give your answer correct to 2 decimal place.
Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{max_height(25,diagram)}
", "advice": "Avoid using rounded values in calculations and just round for the final answer.
{advice}
In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hyptonuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given a right angled triangle as shown calculate the value of x.
\nAngles are given in degrees (make sure you calculator is in the right mode)
Give your answer correct to 2 decimal place.
Draws a triangle based on 3 side lengths and randomises asking for hypotenuse or not.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{statement}
\nFind $x$.
", "advice": "Only round your final answer to 1 decimal place.
\n{advice}
\nUse this link to find some resources to help you revise how to use pythagoras' theorem.
", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"setup": {"name": "setup", "group": "Varying q and advice", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "answerside": {"name": "answerside", "group": "Varying q and advice", "definition": "sh1", "description": "", "templateType": "anything", "can_override": false}, "answerhyp": {"name": "answerhyp", "group": "Varying q and advice", "definition": "hyp", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Varying q and advice", "definition": "if(setup=1,answerside,answerhyp)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "Varying q and advice", "definition": "if(setup=1,advice1,advice2)", "description": "", "templateType": "anything", "can_override": false}, "advice2": {"name": "advice2", "group": "Varying q and advice", "definition": "\"Avoid using rounded values in calculations and just round for the final answer.
Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:
\\\\[a^2 + b^2 = c^2\\\\]
\\nLet\\'s call the unknown value $x$, therefore we can write:
\\n$a = \\\\var{sh1}$, $b =\\\\var{sh2}$ and $c = x$
\\nSo
\\\\[\\\\var{sh1}^2 + \\\\var{sh2}^2 = x^2\\\\]
and therefore
\\n\\\\[x^2 = \\\\var{sh1^2} + \\\\var{sh2^2}\\\\]
\\\\[x = \\\\sqrt{\\\\var{sh1^2} + \\\\var{sh2^2}}\\\\]
\\\\[x = \\\\sqrt{\\\\var{sh1^2+sh2^2}}\\\\]
$x = \\\\var{hyp}$ to 1 d.p.
Avoid using rounded values in calculations and just round for the final answer.
Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:
\\\\[a^2 + b^2 = c^2\\\\]
\\nLet\\'s call the unknown value $x$, therefore we can write:
\\n$a = x$, $b =\\\\var{sh2}$ and $c = \\\\var{hyp}$
\\nSo
\\n\\\\[x^2 + \\\\var{sh2}^2 = \\\\var{hyp}^2\\\\]
\\nand therefore
\\n\\\\[x^2 = \\\\var{hyp^2} - \\\\var{sh2^2}\\\\]
\\n\\\\[x = \\\\sqrt{\\\\var{hyp^2-sh2^2}}\\\\]
$x = \\\\var{sh1}$ to 1 d.p.
one of two shortest sides for calculations.
", "templateType": "randrange", "can_override": false}, "sh1": {"name": "sh1", "group": "Unnamed group", "definition": "precround(sh1_gen,1)", "description": "", "templateType": "anything", "can_override": false}, "sh2_gen": {"name": "sh2_gen", "group": "Unnamed group", "definition": "random(4 .. 9#0.1)", "description": "", "templateType": "randrange", "can_override": true}, "sh2": {"name": "sh2", "group": "Unnamed group", "definition": "precround(sh2_gen,1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "10"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["hyp", "sh1_gen", "sh1", "sh2_gen", "sh2"]}, {"name": "Varying q and advice", "variables": ["setup", "answerside", "answerhyp", "ans", "advice", "advice2", "advice1", "statement1", "statement2", "statement"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "$x=$[[0]] to 1 d.p.
", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "ans", "maxValue": "ans", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "1", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "15.j Calculating angles of right angled from description", "extensions": ["eukleides"], "custom_part_types": [], "resources": ["question-resources/Picture1_caMIdF1.png", "question-resources/Picture2_6KE4ZpW.png"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "Draws a triangle based on 3 side lengths. Randomises asking angle or side.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "{Question}
", "advice": "Let $x$ be the angle between the ladder and the wall. If we draw the triangle given by the ladder, the wall, and the ground we get the following diagram:
\n{max_height(25,diagram)}
\n{advice}
\n\n", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"ab": {"name": "ab", "group": "Unnamed group", "definition": "precround(sqrt(ac^2+bc^2),1)", "description": "", "templateType": "anything", "can_override": false}, "ac": {"name": "ac", "group": "Unnamed group", "definition": "precround(gen_ac,1)", "description": "", "templateType": "anything", "can_override": false}, "bc": {"name": "bc", "group": "Unnamed group", "definition": "precround(gen_bc,1)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_1": {"name": "d_t_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label(bc)\n , a..c label('x')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_1": {"name": "d_c_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , a..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "diagram": {"name": "diagram", "group": "Unnamed group", "definition": "if(SCT='s',\n if(AngORside='ang',\n random(d_s_a_1),\n random(d_s_s_1,d_s_s_2)),\n if(SCT='t',\n if(AngORside='ang',\n random(d_t_a_1),\n random(d_t_s_1,d_t_s_2)),\n if(SCT='c',\n if(AngORside='ang',\n random(d_c_a_1),\n random(d_c_s_1,d_c_s_2)),'X')))\n ", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_1": {"name": "d_s_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_1": {"name": "d_c_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , a..c label(ac)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_1": {"name": "d_s_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_1": {"name": "d_t_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..c label(ac)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_2": {"name": "d_t_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle)\n , b..c label(bc)\n , a..c label('x')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_2": {"name": "d_c_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x')\n , a..c label(ac)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "SCT": {"name": "SCT", "group": "Unnamed group", "definition": "random('s','c','t')", "description": "", "templateType": "anything", "can_override": false}, "AngORside": {"name": "AngORside", "group": "Unnamed group", "definition": "'ang'", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_2": {"name": "d_c_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle)\n , a..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_2": {"name": "d_s_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_2": {"name": "d_s_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_2": {"name": "d_t_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0), \n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x')\n , b..c label(bc)\n , a..c label(ac)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "answer": {"name": "answer", "group": "Unnamed group", "definition": "precround(angle*180/pi,2)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "advice", "definition": "if(SCT='s',\n if(AngORside='ang',\n {sin_a},\n {sin_bc}),\n if(SCT='c',\n if(AngORside='ang',\n {cos_a},\n {cos_ac}),\n if(AngORside='ang',\n {tan_a},{tan_ac})))", "description": "", "templateType": "anything", "can_override": false}, "sin_a": {"name": "sin_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:
\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arcsin(\\\\var{bc}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hypotenuse} = \\\\var{ab}$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:
\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]
We need to use the \\'inverse cos\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arccos(\\\\var{ac}/\\\\var{ab})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:
\\n$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:
\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]
We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\arctan(\\\\var{bc}/\\\\var{ac})\\\\]
\\n\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]
\\nRound as required:
\\n\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = x$
$Hypotenuse = \\\\var{ab}$
We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $sin$ formula:
\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Hypotenuse = \\\\var{ab}$
$Adjacent = x$
We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:
\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]
and rearrange to give:
\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypothenuse\\' in relation to the angle we know:
\\n$Opposite = \\\\var{bc}$
$Adjacent = x$
We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $tan$ formula:
\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]
\\nNow we subsitute the values we have in this particular question
\\n\\\\[ tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]
and rearrange to give:
\\\\[ x = \\\\frac{\\\\var{bc}}{tan(\\\\var{angle})} \\\\]
Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!
\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]
\\nRound as required:
\\n\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]
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