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