Q4. Given 15 cot A = 8, find sin A and sec A.

Given,
$$15 cot A \ = \ 8$$
=> $$cotA \ = \ \frac{8}{15} \ = \ \frac{B}{P} \ = \ \frac{AB}{BC}$$

Let AB = 8k and BC = 15k.

Then, by Pythagoras theorem

$$AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{(8k)^2 + (15k)^2} \ = \ \sqrt{ 64k^2 + 225k^2} \ = \ \sqrt{289k^2} \ = \ 17k$$

∴ $$sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{15k}{17k} \ = \ \frac{15}{17}$$

and, $$secA \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \ = \ \frac{17k}{8k} \ = \ \frac{17}{8}$$