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$A=\\var{AA0}$, $a=\\var{a0}$, $c=\\var{c0}$

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Angle $C=$ [[0]]

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Angle $B=$ [[1]]

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Side length $b=$ [[2]]

", "steps": [{"type": "information", "prompt": "

Use the Sine Rule to find $\\sin C$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$, and then find $C$. Note that $C$ lies between $\\dfrac{\\pi}{2}$ and $\\pi$, whereas $\\sin^{-1}$ returns a value between $0$ and $\\dfrac{\\pi}{2}$; this means that $C$ will be $\\pi -$ (value calculated using $\\sin^{-1}$). Remember that $A+B+C=\\pi$. Use the Sine Rule to find $b$: $\\dfrac{a}{\\sin A}=\\dfrac{b}{\\sin B}$.

", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\n

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 two angles and a side length, determine the other two side lengths and the angle. Write down the side lengths as whole numbers and the angle (in radians) as a decimal to 3dp.

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\n ", "tags": ["checked2015", "SFY0001", "sine rule", "Sine Rule", "Solving triangles", "Triangle", "Two sides and an angle"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \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 $a,b \\geq 10$. 

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\n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

A question testing the application of the Sine Rule when given two sides and an angle.  In this question the triangle is obtuse and the first angle to be found is obtuse.

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We use the Sine Rule to find $C$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$. Thus $\\sin C=\\dfrac{c \\sin A}{a}=\\dfrac{\\var{c0}* \\var{s0}}{\\var{a0}}=\\var{c0*s0/a0}$. To find $C$ we need to calculate $\\sin^{-1} (\\var{c0*s0/a0})$, calculating the angle between $\\dfrac{\\pi}{2}$ and $\\pi$, so $C=\\var{cc01}$ (to 3 decimal places). [Using a calculator, $\\sin^{-1} (\\var{c0*s0/a0})$ returns $\\var{arcsin(u1)}$, which lies between $0$ and $\\dfrac{\\pi}{2}$, so the value we need is $\\pi-\\var{arcsin(u1)}=\\var{pi-arcsin(u1)}$.]

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Since $A+B+C=\\pi$, we calculate $B=\\pi-A-C=\\var{bb11}$. To 3dp, this gives $\\var{bb21}$.

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We use the Sine Rule to find $b$: $\\dfrac{a}{\\sin A}=\\dfrac{b}{\\sin B}$. Thus $b=\\dfrac{a \\sin B}{\\sin A}=\\dfrac{\\var{a0}* \\var{t21}}{\\var{s0}}=\\var{a0*t21/s0}$. The closest integer is then $\\var{b0}$. Note that this solution uses the 3dp value of $B$; the answer using $\\var{bb11}$ would give a slightly different long decimal value of $b$, but the integer value would be the same.

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