Question

If triangle ABC is right angled at B and
$sinA=\frac{3}{5}$, then find the value of cos A.

• A. $\frac{5}{4}$
• B. $\frac{3}{5}$
• C. $\frac{5}{3}$
• D. $\frac{4}{5}$

Answer 1

The correct option is D $\frac{4}{5}$

Given, triangle ABC is right angled at B.
Since,
$sinA=\frac{3}{5}=\frac{BC}{AC}$

Let BC = 3k and AC = 5k.

Applying Pythagoras theorem,
$A{B}^2+B{C}^2=A{C}^2$

$A{B}^2=(5k{)}^2-(3k{)}^2=16k^2$

$AB=4k$

$cosA=\frac{AB}{AC}=\frac{4}{5}$