$ \triangle \mathrm{ABC} \sim \triangle \mathrm{QPR} . $ If $ \angle \mathrm{A}+\angle \mathrm{B}=130^{\circ} $ and $ \angle B+\angle C=125^{\circ} $, find $ \angle Q $.

Given:

\( \triangle \mathrm{ABC} \sim \triangle \mathrm{QPR} . \) If \( \angle \mathrm{A}+\angle \mathrm{B}=130^{\circ} \) and \( \angle B+\angle C=125^{\circ} \).

To do:

We have to find \( \angle Q \).

Solution:

\( \triangle \mathrm{ABC} \sim \triangle \mathrm{QPR} \)

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

Therefore,

$\angle A=\angle Q$, $\angle B=\angle P$ and $\angle C=\angle R$

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

This implies,

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

$\angle A+125^o=180^o$

$\angle A=180^o-125^o=55^o$

$\angle A+\angle B=130^o$

$55^o+\angle B=130^o$

$\angle B=130^o-55^o=75^o$

$\angle B+\angle C=125^o$

$75^o+\angle C=125^o$

$\angle C=125^o-75^o=50^o$

$\Rightarrow \angle Q=\angle A=55^o$

Hence, \( \angle Q=55^o \).