In Fig. 6.39, sides $ \mathrm{QP} $ and $ \mathrm{RQ} $ of $ \triangle \mathrm{PQR} $ are produced to points $ \mathrm{S} $ and T respectively. If $ \angle \mathrm{SPR}=135^{\circ} $ and $ \angle \mathrm{PQT}=110^{\circ} $, find $ \angle \mathrm{PRQ} $. "
Given:
Sides $QP$ and $RQ$ of $\triangle PQR$ are produced to points $S$ and $T$ respectively.
$\angle SPR=135^o$ and $\angle PQT=110^o$.
To do:
We have to find $\angle PRQ$.
Solution:
We know that,
The sum of the measures of the angles in linear pairs is always $180^o$.
This implies,
$\angle TQP+\angle PQR=180^O$
By substituting the value of $\angle TQP$ we get,
$110^o+\angle PQR=180^o$
This implies,
$\angle PQR=180^o-110^o$
$\angle PQR= 70^o$
We also know that,
The sum of the interior angles is equal to the exterior angle.
From $\triangle PQR$ we get,
$\angle PQR+\angle PRQ=135^o$
By substituting the value of $\angle PQR$ we get,
$\angle PRQ=135^o-70^o$
This implies,
$\angle PRQ=65^o$
Hence, $\angle PRQ=65^o$.
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