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

Given that: 15 cot A = 8
⇒ cot A = 8/15
Let cot A = 8k/15k, where k is a real number.
In ∆ ABC, by Pythagoras theorem, we have
AC² = AB² + BC²
= (8k)²+(15k)²
= 64k²+225k²
= 289k²
⇒ AC = √(289k²) = 17k
Hence, sin A = BC/AC = 15k/17k = 15/17 and sec A = AC/AB = 17k/8k = 17/8

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