Question

A right angled triangle $\Delta ABC$ is right angled at C with sides AB = 10 cm and AC = 6 cm. Find cosine of $\angle$B.

• A. $\frac{7}{10}$
• B. $\frac{6}{10}$
• C. $\frac{4}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is C $\frac{4}{5}$


$In\Delta ABC,AB=10cm,AC=6cm$

Applying Pythagoras theorem in $\Delta ABC$, we get,

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

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

$⟹B{C}^2=100-36$

$⟹BC=\sqrt{64}=8cm$

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

$\therefore cos\left(\angle B\right)=\frac{BC}{AB}=\frac{8}{10}=\frac{4}{5}.$