// Numbas version: exam_results_page_options {"name": "Right angled triangles: SOHCAHTOA 2 - find side", "extensions": ["eukleides"], "custom_part_types": [], "resources": [["question-resources/Picture1_caMIdF1.png", "/srv/numbas/media/question-resources/Picture1_caMIdF1.png"], ["question-resources/Picture2_6KE4ZpW.png", "/srv/numbas/media/question-resources/Picture2_6KE4ZpW.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Right angled triangles: SOHCAHTOA 2 - find side", "tags": [], "metadata": {"description": "

Draws a triangle based on 3 side lengths.  Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{max_height(25,diagram)}

", "advice": "

Avoid using rounded values in calculations and just round for the final answer.

{advice}

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"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": "'side'", "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": 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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": "if(SCT='s',\n if(AngORside='ang',\n angle,\n bc),\n if(SCT='t',\n if(AngORside='ang',\n angle,\n ac),\n if(AngORside='ang',\n angle,ac)))", "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 \\'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:

\\n

\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]

\\n

Now 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!

\\n

\\\\[ x = arcsin(\\\\var{bc}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]

\\n

Round 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:

\\n

\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]

\\n

Now 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!

\\n

\\\\[ x = arccos(\\\\var{ac}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]

\\n

Round 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:

\\n

\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]

\\n

Now 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!

\\n

\\\\[ x = arctan(\\\\var{bc}/\\\\var{ac})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]

\\n

Round 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:

\\n

\\\\[ sin(Angle) = \\\\frac{Opposite}{Hypotenuse}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ 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!

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]

\\n

Round 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:

\\n

\\n

\\\\[ cos(Angle) = \\\\frac{Adjacent}{Hypotenuse}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ 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!

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]

\\n

Round 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:

\\n

\\\\[ tan(Angle) = \\\\frac{Opposite}{Adjacent}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]

and rearrange to give:

\\n

\\\\[ 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!

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]

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Given a right angled triangle as shown calculate the value of x.

\n

Angles are given in degrees (make sure you calculator is in the right mode)

Give your answer correct to 2 decimal place.

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