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{"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(arccos(q3),4)", "description": "", "name": "bb3"}, "p3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "(c3^2+b3^2-a3^2)/(2*c3*b3)", "description": "", "name": "p3"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sin(aa2)", "description": "", "name": "s2"}, "s3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sin(AA3)", "description": "", "name": "s3"}, "c02": {"templateType": "anything", "group": "Ungrouped variables", "definition": "floor(max(a0+0.9*b0,b0+0.9*a0))", "description": "", "name": "c02"}, "q0": {"templateType": "anything", "group": "Ungrouped variables", "definition": "(a0^2+c0^2-b0^2)/(2*a0*c0)", "description": "", "name": "q0"}, "u3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sin(CC3)", "description": "", "name": "u3"}, "cc5": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(CC4,3)", "description": "", "name": "cc5"}, "bb0": {"templateType": "anything", "group": "Ungrouped variables", "definition": "\n //The angle B\n precround(arccos(q0),4)\n \n ", "description": "", "name": "bb0"}, "c0": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(c01..c02)", "description": "", "name": "c0"}, "t5": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sin(BB5)", "description": "", "name": "t5"}, "a0": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(10..25)", "description": "", "name": "a0"}, "s0": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sin(aa0)", "description": "", "name": "s0"}, "cc2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(cc1,3)", "description": "", "name": "cc2"}, "b0": {"templateType": "anything", "group": "Ungrouped variables", 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"maxValue": "{aa0}+0.0001", "precision": 4, "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": true, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "showCorrectAnswer": true, "marks": 2}, {"precisionPartialCredit": 0, "allowFractions": false, "correctAnswerFraction": false, "minValue": "{bb0}-0.0001", "maxValue": "{bb0}+0.0001", "precision": 4, "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": true, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "showCorrectAnswer": true, "marks": 2}, {"precisionPartialCredit": 0, "allowFractions": false, "correctAnswerFraction": false, "minValue": "{cc0}-0.0001", "maxValue": "{cc0}+0.0001", "precision": 4, "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": true, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "showCorrectAnswer": true, "marks": 1}], "type": "gapfill", "prompt": "
$a=\\var{a0}$, $b=\\var{b0}$, $c=\\var{c0}$
\nAngle $A=$ [[0]]
\nAngle $B=$ [[1]]
\nAngle $C=$ [[2]]
", "steps": [{"type": "information", "prompt": "Use the Cosine Rule to find $\\cos A$: $\\cos A =\\dfrac{b^2+c^2-a^2}{2bc}$. Then use $\\cos^{-1}$ to find $A$. Apply similar rules to find $B$ and $C$. Take care over $C$: $\\cos C <0$, which means that $C$ lies between $\\dfrac{\\pi}{2}$ and $\\pi$.
", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"precisionPartialCredit": 0, "allowFractions": false, "correctAnswerFraction": false, "minValue": "{Area}-0.001", "maxValue": "{Area}+0.001", "precision": 3, "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": true, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "showCorrectAnswer": true, "marks": 2}], "type": "gapfill", "prompt": "The area is [[0]]
", "showCorrectAnswer": true, "marks": 0}], "statement": "Suppose that $\\Delta ABC$ is a triangle with $C> \\dfrac{\\pi}{2}$ (so it is an obtuse triangle). Here all angles are expressed in radians. Suppose also that standard naming conventions are used as indicated in the picture below (not necessarily an accurate picture of $\\Delta ABC$).
\nGiven the following three side lengths, determine the three angles using the Cosine Rule. Write down the angles (in radians) as decimals to 4dp. [Before submitting answers, you can check that the sum of the three angles is $\\pi$.]
\n \nAlso calculate the area of the triangle, giving your answer as a decimal to 3dp.
\n \n ", "tags": ["Area of a triangle", "checked2015", "cosine rule", "Cosine Rule", "SFY0001", "Solving triangles", "Three side lengths", "Triangle"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t \t\tI want an obtuse triangle with side lengths $a,b,c$. I need $a^2+b^2<c^2<(a+b)^2$. I start with $c_1=ceil(\\sqrt{a^2+b^2})+1$, $c_2=\\max\\{b+0.9 a, a + 0.9 b\\}$ to establish a range of values for $c$ so that the triangle is neither too flat nor too close to a right-angled triangle. The upper limit ensures that $-\\cos C \\leq 0.9$ and so $\\sin C \\geq 0.435$. Specifying that $a \\leq 11b, b \\leq 11a$ ensures that $\\sin A, \\sin B$ are not too small and thereby ensures that percentage errors are below 0.5%. This last figure points to $a,b \\leq 100$ and there are benefits in $a,b \\geq 10$.
\n \t\t \t\t \n \t\t \n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "A question testing the application of the Cosine Rule when given three side lengths. In this question, the triangle is always obtuse. A secondary application is finding the area of a triangle.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "(a) Use the Cosine Rule to find $\\cos A$: $\\cos A =\\dfrac{b^2+c^2-a^2}{2bc}$. Therefore
\n\\[\\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}}\\]
\n\\[=\\var{(b0^2+c0^2-a0^2)/(2 *b0*c0)}\\]
\nand so $A=\\cos^{-1}(\\var{(b0^2+c0^2-a0^2)/(2 *b0*c0)})=\\var{aa0}$.
\n \nSimilarly $\\cos B =\\dfrac{a^2+c^2-b^2}{2ac}$ and $\\cos C =\\dfrac{a^2+b^2-c^2}{2ab}$. So
\n\\[\\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)}\\]
\nand so $B=\\cos^{-1}(\\var{(a0^2+c0^2-b0^2)/(2 *a0*c0)})=\\var{bb0}$.
\n\\[\\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)}\\]
\nand so $C=\\cos^{-1}(\\var{(a0^2+b0^2-c0^2)/(2 *a0*b0)})=\\var{cc0}$.
\n(b) We can use any of the formulae $\\dfrac{1}{2}ac \\sin B$, $\\dfrac{1}{2}bc \\sin A$ or $\\dfrac{1}{2}ab \\sin C$ for the area. For example
\n\\[\\dfrac{1}{2}bc \\sin A = \\dfrac{1}{2} \\times \\var{b0} \\times \\var{c0} \\times \\sin \\var{aa0}\\]
\n\\[=\\dfrac{1}{2} \\times \\var{b0 * c0} \\times \\var{sin(aa0)}=\\var{Area}\\]
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}