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# 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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