Question

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

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

Answer 1

The correct option is A $\frac{3}{4}$


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

Applying Pythagoras' theorem we get,
$A{C}^2+B{C}^2=A{B}^2$

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

$⟹B{C}^2=100-36=64$

$⟹BC=\sqrt{64}=8cm$

We know that, $tan\theta =\frac{OppositeSide}{AdjacentSide}$

$\therefore tanB=\frac{AC}{BC}=\frac{6}{8}=\frac{3}{4}$