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$a=\\var{a0}$, $b=\\var{b0}$, $c=\\var{c0}$

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Angle $A=$ [[0]]

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Angle $B=$ [[1]]

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Angle $C=$ [[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 $\\pi$.

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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 $\\Delta ABC$).

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Given 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 $\\pi$.] 

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Also calculate the area of the triangle, giving your answer as a decimal to 3dp.

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", "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\t

I 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.

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(a) Use the Cosine Rule to find $\\cos A$: $\\cos A =\\dfrac{b^2+c^2-a^2}{2bc}$. Therefore

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\\[\\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}}\\]

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\\[=\\var{(b0^2+c0^2-a0^2)/(2 *b0*c0)}\\]

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and so $A=\\cos^{-1}(\\var{(b0^2+c0^2-a0^2)/(2 *b0*c0)})=\\var{aa0}$.

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Similarly $\\cos B =\\dfrac{a^2+c^2-b^2}{2ac}$ and $\\cos C =\\dfrac{a^2+b^2-c^2}{2ab}$. So

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\\[\\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)}\\]

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and so $B=\\cos^{-1}(\\var{(a0^2+c0^2-b0^2)/(2 *a0*c0)})=\\var{bb0}$.

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\\[\\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)}\\]

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and so $C=\\cos^{-1}(\\var{(a0^2+b0^2-c0^2)/(2 *a0*b0)})=\\var{cc0}$.

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(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 

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\\[\\dfrac{1}{2}bc \\sin A = \\dfrac{1}{2} \\times \\var{b0} \\times \\var{c0} \\times \\sin \\var{aa0}\\]

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\\[=\\dfrac{1}{2} \\times \\var{b0 * c0} \\times \\var{sin(aa0)}=\\var{Area}\\]

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