Question

GIVEN 15 COT A =8, find sin A and sec A

Answer 1

$\blue{HELLO\: MATE}$
see in attachment

Answer 2

Answer:

In a triangle right angled at B

Given, 15cotA =8

=>cot A=8/15

=>AB/BC =8/15

Therefore, AB =8 and BC =15

By using Pythagoras Theorem we get _

$ac^{2} = ab ^{2} + bc^{2}$

$ac = \sqrt{8^{2} + 15^{2} }$

$\sqrt{289}$

Therefore AC=17

Therefore Sin A = BC/AC =15/17

And, sec A =AB/AC =8/17

MAY THIS HELP YOU