Question

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

Answer 1

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

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`