# $\triangle \mathrm{ABC} \sim \triangle \mathrm{QPR} .$ 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, $\angle Q=55^o$.

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Updated on: 10-Oct-2022

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