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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.

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Suppose that $\\Delta ABC$ is a triangle with all interior angles less than $90^{\\circ}$. Sides and angles are labelled as shown in the diagram below (not to scale).

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Given the following three side lengths, determine the three angles using the Cosine Rule. Give your answers correct to the nearest degree.

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

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Angle $A=$ [[0]]

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Angle $B=$ [[1]]

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Angle $C=$ [[2]]

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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 $C$.

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