// Numbas version: exam_results_page_options {"name": "Maria's copy of Sine, cosine, and area rules", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": [], "variablesTest": {"condition": "angleA<>90", "maxRuns": 100}, "parts": [{"extendBaseMarkingAlgorithm": true, "correctAnswerStyle": "plain", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "marks": 1, "unitTests": [], "maxValue": "angleB", "precisionMessage": "You have not given your answer to the correct precision.", "variableReplacements": [], "minValue": "angleB", "mustBeReducedPC": 0, "strictPrecision": true, "type": "numberentry", "scripts": {}, "precisionType": "dp", "allowFractions": false, "showPrecisionHint": true, "precision": "1", "showFeedbackIcon": true, "notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "correctAnswerFraction": false, "prompt": "

What is the approximate value of angle $B$ in degrees?

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What is the approximate area of the triangle $ABC$?

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What is the approximate length of the side $AB$?

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Draws a triangle based on 2 angles and a side length.

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The following questions refer to this triangle diagram.

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{plotgraph(a,b,c,angleA,angleB,angleC)}

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180-(angleB+angleC)

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anglerandom(30..60)

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random(30..60 except angleB)

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For part (a), the sine rule tells us that $\\frac{\\sin(A)}{a} = \\frac{\\sin(B)}{b}$,

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so here $B = \\sin^{-1}\\left(\\frac{\\var{b}\\times\\sin(\\var{angleA}^{\\circ})}{\\var{a}}\\right)$.

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For part (b), the angle sum of a triangle tells us that angle $C$ is given by $180^{\\circ}-A-B = 180^{\\circ} - \\var{angleA}^{\\circ} - \\var{precround(angleB,1)}^{\\circ}$.

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Then we can use the area formula $|ABC| = \\frac{1}{2} a b \\sin(C)$, which in this case is $\\frac{1}{2} \\times \\var{a} \\times \\var{b} \\times \\sin(\\var{precround(angleC,1)}^{\\circ})$.

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For part (c), the cosine rule tells us that $c^2 = a^2 + b^2 - 2ab\\cos(C)$,

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so here $c = \\sqrt{\\var{a}^2+\\var{b}^2 - 2 \\times \\var{a} \\times \\var{b} \\times \\cos(\\var{precround(angleC,1)}^{\\circ})}$.

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It's also possible to use the sine rule or area rule to find $AB$ here.