Question

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

Answer 1

SOLUTION

Let ΔABC be a right-angled triangle, right-angled at B. We know that cot A = AB/BC = 8/15   (Given) Let AB be 8k and BC will be 15k where k is a positive real number. By Pythagoras theorem we get, AC2 = AB2 + BC2 AC2 = (8k)2 + (15k)2 AC2 = 64k2 + 225k2 AC2 = 289k2 AC = 17 k sin A = BC/AC = 15k/17k = 15/17 sec A = AC/AB = 17k/8 k = 17/8

Answer 2

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Let ΔABC be a right-angled triangle, right-angled at B.
We know that cot A = AB/BC = 8/15 (Given)
Let AB be 8k and BC will be 15k where k is a positive real number.
By Pythagoras theorem we get, AC2 = AB2 + BC2 AC2 = (8k)2 + (15k)2 AC2 = 64k2 + 225k2 AC2 = 289k2 AC = 17 k sin A = BC/AC = 15k/17k = 15/17 sec A = AC/AB = 17k/8 k = 17/8