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a) Use the Cosine Rule to find $a$: $a^2=b^2+c^2-2bc \\cos A$.
\n\\[a^2=\\var{b0}^2+\\var{c0}^2-2 \\times \\var{b0}\\times\\var{c0} \\times \\cos (\\var{aa0})=\\var{b0^2}+\\var{c0^2}-\\var{2*b0*c0} \\times \\var{cos (aa0)}\\]
\n\\[=\\var{b0^2+c0^2-2*b0*c0* cos (aa0)}.\\]
\nHence $a=\\sqrt{\\var{b0^2+c0^2-2*b0*c0* cos (aa0)}}=\\var{sqrt(b0^2+c0^2-2*b0*c0* cos (aa0))}$. To the nearest integer, this is $\\var{a0}$.
\nb) Use the Cosine Rule to find $b$: $b^2=a^2+c^2-2ac \\cos B$.
\n\\[b^2=\\var{a3}^2+\\var{c3}^2-2 \\times \\var{a3}\\times\\var{c3} \\times \\cos (\\var{bb3})=\\var{a3^2}+\\var{c3^2}-\\var{2*a3*c3} \\times \\var{cos (bb3)}\\]
\n\\[=\\var{a3^2+c3^2-2*a3*c3* cos (bb3)}.\\]
\nHence $b=\\sqrt{\\var{a3^2+c3^2-2*a3*c3* cos (bb3)}}=\\var{sqrt(a3^2+c3^2-2*a3*c3* cos (bb3))}$. To the nearest integer, this is $\\var{b3}$.
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In these questions, the triangle is always acute and both of the given side lengths are adjacent to the given angle.
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