Question

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

Answer 1

SOLUTION :

Given:  15 cot A = 8

cot A = 8/15

Cot A = base / Perpendicular  = 8/15

Base = 8  

Perpendicular side = 15  

We draw a ∆ ABC right angled at B.  

In ΔABC,  

Let BC = Perpendicular = 15  , AB = Base = 8  

Hypotenuse=  (AC)

AC² = AB² + BC²

[by using Pythagoras theorem]

AC² = 8² + 15²

AC² = 64 + 225

AC² = 289

AC = √289

AC = 17

Therefore, hypotenuse (AC) =17

sin A= Perpendicular / Hypotenuse

sin A = BC/AC

sin A=15/17

sec A= Hypotenuse / Base

sec A= AC / AB  

sec A = 17/8

Hence, sin A = 15/17, sec A = 17/8

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

cot A = cos A / sin A = 8/15
cot A = Base / Perpendicular = 8/15
Base = 8k
Perpendicular = 15k
Hypotenuse²
= (15k)² + (8k)²
= 225k²+64k²
= 289k²
Hypotenuse = 17k
sin A = perpendicular/Hypotenuse
= 15k/17k = 15/17
sec A = Hypotenuse/Base = 17k/8k = 17/8