In the figure,sides QP and RQ of Delta PQR ar | vedclass

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(A) Since $TQR$ is a straight line,$	herefore \angle TQP + \angle PQR = 180^o$ [Linear pair axiom]$\Rightarrow 110^o + \angle PQR = 180^o$$\Rightarro

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(A) Since $TQR$ is a straight line,$	herefore \angle TQP + \angle PQR = 180^o$ [Linear pair axiom]$\Rightarrow 110^o + \angle PQR = 180^o$$\Rightarrow \angle PQR = 180^o - 110^o = 70^o$Now,for $\Delta PQR$,the side $QP$ is produced to $S$.Therefore,$\angle SPR$ is an exterior angle of $\Delta PQR$.By the exterior angle property,the exterior angle is equal to the sum of the two interior opposite angles.$	herefore \angle SPR = \angle PQR + \angle PRQ$$135^o = 70^o + \angle PRQ$$\Rightarrow \angle PRQ = 135^o - 70^o$$\Rightarrow \angle PRQ = 65^o$

In the figure,sides $QP$ and $RQ$ of $\Delta PQR$ are produced to points $S$ and $T$ respectively. If $\angle SPR = 135^o$ and $\angle PQT = 110^o$,find $\angle PRQ$. (in $^o$)

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