Question

Given 15 cot A = 8, find sin A.

• A. $\frac{15}{17}$
  
• B. $\frac{8}{17}$
  
• C. $\frac{17}{15}$
  
• D. $\frac{15}{8}$
  

Answer 1

The correct option is A $\frac{15}{17}$


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 = $\frac{BC}{AC}$ = $\frac{15k}{17k}$ = $\frac{15}{17}$