Question

Given 15 cot A = 8. Find sin A and sec A

Answer 1

Consider a right-angled triangle, right-angled at B.

cot A =`(AB)/(BC)`

`= (AB)/(BC)`

it is given that

cot A = 8/15

`(AB)/(BC)=8/15`

Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

sin A `= (15k)/(17k) = 15/17`

sec A ` = (AC)/(AB) = 17/8`