In the figure,if lines PQ and RS intersect at | vedclass

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(D) In $\Delta PRT$,$\angle P + \angle R + \angle PTR = 180^\circ$ [By the angle sum property of a triangle]$\Rightarrow 95^\circ + 40^\circ + \angle

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$60$

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(D) In $\Delta PRT$,$\angle P + \angle R + \angle PTR = 180^\circ$ [By the angle sum property of a triangle]$\Rightarrow 95^\circ + 40^\circ + \angle PTR = 180^\circ$$\Rightarrow \angle PTR = 180^\circ - 135^\circ = 45^\circ$Since lines $PQ$ and $RS$ intersect at $T$,the vertically opposite angles are equal.$	herefore \angle QTS = \angle PTR = 45^\circ$Now,in $\Delta TQS$,$\angle TSQ + \angle STQ + \angle SQT = 180^\circ$ [By the angle sum property of a triangle]$\Rightarrow 75^\circ + 45^\circ + \angle SQT = 180^\circ$$\Rightarrow 120^\circ + \angle SQT = 180^\circ$$\Rightarrow \angle SQT = 180^\circ - 120^\circ = 60^\circ$Thus,$\angle SQT = 60^\circ$.

In the figure,if lines $PQ$ and $RS$ intersect at point $T$,such that $\angle PRT = 40^\circ$,$\angle RPT = 95^\circ$ and $\angle TSQ = 75^\circ$,find $\angle SQT$. (in $^\circ$)

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