// Numbas version: exam_results_page_options {"name": "GB6 Trigonometry - non-right angled trig", "extensions": ["geogebra"], "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": "GB6 Trigonometry - non-right angled trig", "tags": [], "metadata": {"description": "

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

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{diagram}

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Find x.

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{Advice}

\n

Use this link to find some resources which will help you revise this topic.

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First recognise that the diagram is a non-right angled triangle and that there are the lengths of two sides given and the angle specifically between those two sides. Further to this, the instruction is to find the other missing side. These are the conditions for when to use the $\\\\textit{cosine rule}$.

\\n

The formula for a missing side using the cosine rule is:

\\n

\\\\[ a^2 = b^2 + c^2 - 2bc \\\\cos(A)\\\\]

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The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the angle given and it\\'s opposite side.

\\n

In this case:

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\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A=\\\\var{Cang},\\\\]

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where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

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So, we now have:

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\\\\[x^2 = \\\\var{a}^2 +\\\\var{b}^2-2\\\\times\\\\var{a}\\\\times\\\\var{b}\\\\times\\\\cos{(\\\\var{Cang})},\\\\]

\\n

hence,

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\\\\[x=\\\\sqrt{\\\\var{a^2 +b^2-2*a*b*(cos(Cang))}}\\\\]

\\n

\\\\[x=\\\\var{c}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

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case 1: missing side in the cosine rule.

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First recognise that the diagram is a non-right angled triangle and that there are the lengths of all three sides given. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{cosine rule}$ but in its rearranged form to find an angle. You need to identify which side is \\\"$a$\\\" as being the one opposite the angle you are asked to find.

\\n

The formula for a missing angle using the cosine rule is:

\\n

\\\\[ A = \\\\arccos\\\\left(\\\\frac{b^2+c^2-a^2}{2bc}\\\\right)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the side opposite the angle that is asked for and the angle that is asked for.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A= x,\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x = \\\\arccos\\\\left(\\\\frac{\\\\var{a}^2+\\\\var{b}^2-\\\\var{c_round}^2}{2\\\\times\\\\var{a}\\\\times\\\\var{b}}\\\\right),\\\\]

\\n

hence,

\\n

\\\\[x=\\\\var{(180/pi)*arccos((a^2 +b^2-c_round^2)/(2*a*b))}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

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First recognise that the diagram is a non-right angled triangle and that a single length is provided, along with two angles, crucially including the angle opposite the given side. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding a side using the sine rule can be written as:

\\n

\\\\[ \\\\frac{a}{\\\\sin(A)}=\\\\frac{b}{\\\\sin(B)}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the side being asked for is in the above notation $a$.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad A=\\\\var{Cang}, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{x}{\\\\sin{(\\\\var{Cang})}}=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}}\\\\times\\\\sin{(\\\\var{Cang})},\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 3

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First recognise that the diagram is a non-right angled triangle and that two lengths are provided, along with an angle, crucially including an angle opposite a given side. Further to this, the instruction is to find the a missing side. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding an angle using the sine rule can be written as:

\\n

\\\\[ \\\\frac{\\\\sin(A)}{a}=\\\\frac{\\\\sin(B)}{b}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the angle being asked for is in the above notation $A$.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad A= x, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{\\\\sin{(x)}}{\\\\var{c_round}}=\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\arcsin\\\\left(\\\\var{c_round}\\\\times\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}}\\\\right),\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

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case 4

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side length a

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side length b

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C angle in degrees

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angle A in degrees

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Calculated answer for c from rounded values - as these will be seen information by student.

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$x =$[[0]]

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