**Question:**

If $\sin A=\frac{9}{41}$, compute $\cos A$ and $\tan A$.

**Solution:**

Given: $\sin A=\frac{9}{41}$...(1)

To Find: $\cos A, \tan A$

By definition,

$\sin A=\frac{\text { Perpendicular side opposite to } \angle \mathrm{A}}{\text { Hypotenuse }}$…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 9 and

Hypotenuse = 41

Now using the perpendicular side and hypotenuse we can construct $\triangle A B C$ as shown below

Length of side $\mathrm{AB}$ is unknown in right angled $\triangle A B C$,

To find length of side AB, we use Pythagoras theorem.

Therefore, by applying Pythagoras theorem in $\triangle A B C$,

We get,

$A C^{2}=A B^{2}+B C^{2}$

$41^{2}=A B^{2}+9^{2}$

$A B^{2}=41^{2}-9^{2}$

$A B^{2}=1681-81$

$A B^{2}=1600$

$A B=\sqrt{1600}$

$A B=40$

Hence, length of side AB = 40

Now,

By definition,

$\cos A=\frac{\text { Base side adjacent to } \angle \mathrm{A}}{\text { Hypotenuse }}$

$\cos A=\frac{A B}{\mathrm{AC}}$

$\cos A=\frac{40}{41}$

Now,

By definition,

$\tan A=\frac{\text { Perpendicular side opposite to } \angle \mathrm{A}}{\text { Base side adjacent to } \angle \mathrm{A}}$

$\tan A=\frac{B C}{\mathrm{AB}}$

$\tan A=\frac{9}{40}$

Answer: $\cos A=\frac{40}{41}$ and $\tan A=\frac{9}{40}$