// Numbas version: finer_feedback_settings
{"name": "Apply the cosine rule (3 sides given, degrees)", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Triangle_700_JKWvhba.gif", "/srv/numbas/media/question-resources/Triangle_700_JKWvhba.gif"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Apply the cosine rule (3 sides given, degrees)", "tags": [], "metadata": {"description": "<p>A question testing the application of the Cosine Rule when given three side lengths. In this question, the triangle is always acute. A secondary application is finding the area of a triangle.</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>Suppose that $\\Delta ABC$ is a triangle with all interior angles less than $90^{\\circ}$.&nbsp;Sides and angles are labelled&nbsp;as shown in the diagram below (not to scale).</p>\n<p style=\"margin: 0px;\">Given the following three side lengths, determine the three angles using the Cosine Rule. Give your answers correct to the nearest degree.</p>\n<p style=\"margin: 0px;\"></p>\n<p style=\"margin: 0px;\"><img src=\"resources/question-resources/Triangle_700_JKWvhba.gif\"/></p>", "advice": "<p>Use the Cosine Rule to find $\\cos A$: $\\cos A =\\dfrac{b^2+c^2-a^2}{2bc}$. Therefore</p>\n<p>\\[\\cos A =\\dfrac{\\var{b0}^2+\\var{c0}^2-\\var{a0}^2}{2 \\times \\var{b0} \\times \\var{c0}}=\\dfrac{\\var{b0^2+c0^2-a0^2}}{\\var{2 *b0*c0}}\\]</p>\n<p>\\[=\\var{(b0^2+c0^2-a0^2)/(2 *b0*c0)}\\]</p>\n<p>and so $A=\\cos^{-1}(\\var{(b0^2+c0^2-a0^2)/(2 *b0*c0)})=\\var{aa0}$.</p>\n<p></p>\n<p>Similarly $\\cos B =\\dfrac{a^2+c^2-b^2}{2ac}$ and $\\cos C =\\dfrac{a^2+b^2-c^2}{2ab}$. So</p>\n<p>\\[\\cos B =\\dfrac{\\var{a0}^2+\\var{c0}^2-\\var{b0}^2}{2 \\times \\var{a0} \\times \\var{c0}}=\\var{(a0^2+c0^2-b0^2)/(2 *a0*c0)}\\]</p>\n<p>and so $B=\\cos^{-1}(\\var{(a0^2+c0^2-b0^2)/(2 *a0*c0)})=\\var{bb0}$.</p>\n<p>\\[\\cos C =\\dfrac{\\var{a0}^2+\\var{b0}^2-\\var{c0}^2}{2 \\times \\var{a0} \\times \\var{b0}}=\\var{(a0^2+b0^2-c0^2)/(2 *a0*b0)}\\]</p>\n<p>and so $C=\\cos^{-1}(\\var{(a0^2+b0^2-c0^2)/(2 *a0*b0)})=\\var{cc0}$.</p>\n<p></p>", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "extensions": [], "variables": {"cc4": {"name": "cc4", "group": "Ungrouped variables", "definition": "pi-AA3-BB3", "description": "", "templateType": "anything"}, "cc1": {"name": "cc1", "group": "Ungrouped variables", "definition": "pi-aa0-bb0", "description": "", "templateType": "anything"}, "b3": {"name": "b3", "group": "Ungrouped variables", "definition": "random(7..20)", "description": "", "templateType": "anything"}, "s5": {"name": "s5", "group": 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