If $sin A Question: If$\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}=1681-81A B^{2}=1600A 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}\$