Question

In Triangle ABC, if a = 25, b = 52 and c = 63, find cos A and sin A.

Answer 1

Cos A = \(\frac{b^2 + C^2 -a^2}{2bc} = \frac{52^2 + 63^2 - 25^2}{2 \times 52 \times 63}\)

\(\frac{2704 + 3969-625}{6552}\) = \(\frac{6048}{6552}\) = \( \frac{12}{13}\)

sin A = \(\sqrt{1-cos^2x}\) 

= \(\sqrt{1 - \frac{144}{169}}\) = \(\frac{5}{13}\)