Question:
If $A+B=90^{\circ}$ and $\cos B=\frac{3}{5}$, what is the value of $\sin A$ ?
Solution:
We have:
$A+B=90^{\circ}$
$\cos B=\frac{3}{5}$
$A+B=90^{\circ}$
$\Rightarrow A=90^{\circ}-B$
$\Rightarrow \sin A=\sin \left(90^{\circ}-B\right)$
$\Rightarrow \sin A=\cos B$
$\Rightarrow \sin A=\frac{3}{5}\left[\sin \left(90^{\circ}-B\right)=\cos B\right]$
Hence the value of $\sin A$ is $\frac{3}{5}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.