# $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$. If $2 \angle \mathrm{P}=3 \angle \mathrm{Q}$ and $\angle C=100^{\circ}$, find $\angle B$.

Given:

$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.

$2 \angle \mathrm{P}=3 \angle \mathrm{Q}$ and $\angle C=100^{\circ}$.

To do:

We have to find $\angle B$.

Solution:

$△ABC \sim △PQR$

When two triangles are similar their corresponding angles are equal and corresponding angles are equal and corresponding sides are in proportion.

Therefore,

$\angle A=\angle P$

$\angle B=\angle Q$

$\angle C=\angle R=100^o$

$2\angle P=3\angle Q$

$\angle P=\frac{3}{2}\angle Q$

This implies,

$\angle A=\angle P=\frac{3}{2}\angle Q$

$=\frac{3}{2}\angle B$

Sum of the angles in a triangle is $180^o$.

Therefore,

$\angle A+\angle B+\angle C=180^o$

$\frac{3}{2}\angle B+\angle B+100^o=180^o$

$\frac{3+2}{2}\angle B=180^o-100^o$

$\frac{5}{2}\angle B=80^o$

$\angle B=\frac{2}{5}\times80^o$

$=32^o$

Hence, $\angle B=32^o$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

33 Views

##### Kickstart Your Career

Get certified by completing the course