Question

The sides of a triangle are a = 4, b = 6 and c = 8. Show that $8\cos A+16\cos B+4\cos C=17$.

Answer 1

Given: $a=4,b=6\text{and}c=8.$
Then,
$$\begin{gathered} \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{16+64-36}{2\times 4\times 8}=\frac{11}{16} \\ \cos A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}=\frac{36+64-16}{2\times 6\times 8}=\frac{7}{8} \\ \cos C={\displaystyle \frac{b^2+a^2-c^2}{2ab}}=\frac{16+36-64}{2\times 4\times 6}=\frac{-1}{4} \\ \mathrm{Now}, \\ 8\cos A+16\cos B+4\cos C=8\times \frac{7}{8}+16\times \frac{11}{16}-4\times \frac{1}{4} \\ \Rightarrow 8\cos A+16\cos B+4\cos C=7+11-1=17 \end{gathered}$$

Hence proved.