Question

Question 4
Given 15 cot A = 8, find sin A and sec A.

Answer 1

Let $\Delta$ABC be a right-angled triangle, right-angled at B.
We know that cot A = $\frac{AB}{BC}$ = $\frac{8}{15}$
Let AB be 8k and BC will be 15k where k is a positive real number.

By Pythagoras theorem; we get,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(8k{)}^2+(15k{)}^2$
$A{C}^2=64k^2+225k^2$
$A{C}^2=289k^2$
AC = 17k
sin A = $\frac{BC}{AC}$ = $\frac{15k}{17k}$ = $\frac{15}{17}$
sec A = $\frac{AC}{AB}$ = $\frac{17k}{8k}$ = $\frac{17}{8}$