Question

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

Answer 1

Let us draw a right triangle ABC, right angled at B.
It is given that: 15 cot A = 8
$cotA=\frac{8}{15}=\frac{AB}{BC}$



Let, AB = 8K, BC = 15K
Using Pythagoras theorem, we have
$A{C}^2=A{B}^2+B{C}^2=(8K{)}^2+(15K{)}^2=64K^2+225K^2=289K^2\therefore AC=17K$
Now,
AC = 17K

$\therefore sinA=\frac{BC}{AC}=\frac{15K}{17K}=\frac{15}{17}$, and
$secA=\frac{AC}{AB}=\frac{17K}{8K}=\frac{17}{8}$