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Question
The sides of a triangle are a = 4, b = 6 and c = 8. Show that \[8 \cos A + 16 \cos B + 4 \cos C = 17\]
Solution
Given: \[a = 4, b = 6 \text{ and } c = 8 .\]
Then,
\[\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{16 + 64 - 36}{2 \times 4 \times 8} = \frac{11}{16}\]
\[\cos A= \frac{b^2 + c^2 - a^2}{2bc} =\frac{36 + 64 - 16}{2 \times 6 \times 8}=\frac{7}{8}\]
\[\cos C= \frac{b^2 + a^2 - c^2}{2ab} =\frac{16 + 36 - 64}{2 \times 4 \times 6}=\frac{- 1}{4}\]
\[Now, \]
\[ 8\cos A+16\cos B+4\cos C=8 \times\frac{7}{8}+16 \times\frac{11}{16}-4 \times\frac{1}{4}\]
\[ \Rightarrow 8\cos A + 16\cos B + 4\cos C=7+11-1 = 17\]
Hence proved.
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