Question

$\Delta ABC$ is right angled at C with sides AB = 10 cm and AC = 6 cm. Find cos B.

• A. $\frac{3}{5}$
• B. $\frac{8}{6}$
• C. $\frac{6}{10}$
• D. $\frac{4}{5}$

Answer 1

The correct option is D

$\frac{4}{5}$

In $\Delta ABC,$ AB = 10 cm and AC=6 cm.
Applying Pythagoras' theorem to $\Delta ABC,$ we have

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

$6^2+B{C}^2={10}^2$

$⟹B{C}^2=100-36$

$⟹BC=\sqrt{64}=8cm$

The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.

$\therefore \text{cos}B=\frac{BC}{AB}=\frac{8}{10}=\frac{4}{5}$