Question

If cos B = $\frac{3}{5}$ and (A + B) = ${90}^o$, find the value of sin A.

Answer 1

Given, $A+B={90}^o$

$\Rightarrow B={90}^o-A$

Now, $cosB=\frac{3}{5}$

$\Rightarrow cos({90}^o-A)=\frac{3}{5}$

$\Rightarrow sinA=\frac{3}{5}$