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

Given,

⇒ 15 cot A = 8

⇒ cot A = $\dfrac{8}{15}$

Let us draw a right angle triangle ABC.

We know that,

cot A = $\dfrac{\text{Side adjacent to ∠A}}{\text{Side opposite to ∠A}}$

Substituting values we get,

$\dfrac{8}{15} = \dfrac{AB}{BC}$

Let AB = 8k and BC = 15k.

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = (8k)² + (15k)²

⇒ AC² = 64k² + 225k²

⇒ AC² = 289k²

⇒ AC = $\sqrt{289k^2}$ = 17k.

We know that,

sin A = $\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{15k}{17k} = \dfrac{15}{17}$.

sec A = $\dfrac{\text{Hypotenuse}}{\text{Side adjacent to ∠A}} = \dfrac{AC}{AB} = \dfrac{17k}{8k} = \dfrac{17}{8}$.

Hence, $\text{sin A} = \dfrac{15}{17} \text{ and sec A} = \dfrac{17}{8}$.