Side QR of Delta PQR is extended on both the | vedclass

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Question

(A) Since $S-Q-R-T$ is a straight line,$\angle PQS$ and $\angle PQR$ form a linear pair. Therefore,$\angle PQR = 180^{\circ} - 100^{\circ} = 80^{\circ

Vedclass · Accepted Answer

$\angle P = 50^{\circ}, \angle Q = 80^{\circ}, \angle R = 50^{\circ}$

Vedclass · Answer

(A) Since $S-Q-R-T$ is a straight line,$\angle PQS$ and $\angle PQR$ form a linear pair. Therefore,$\angle PQR = 180^{\circ} - 100^{\circ} = 80^{\circ}$.Similarly,$\angle PRT$ and $\angle PRQ$ form a linear pair. Therefore,$\angle PRQ = 180^{\circ} - 130^{\circ} = 50^{\circ}$.In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.$\angle P + 80^{\circ} + 50^{\circ} = 180^{\circ}$.$\angle P + 130^{\circ} = 180^{\circ}$.$\angle P = 180^{\circ} - 130^{\circ} = 50^{\circ}$.Hence,the angles are $\angle P = 50^{\circ}, \angle Q = 80^{\circ}, \angle R = 50^{\circ}$.

Side $QR$ of $\Delta PQR$ is extended on both the sides so that exterior angles $\angle PQS$ and $\angle PRT$ are formed. If $\angle PQS = 100^{\circ}$ and $\angle PRT = 130^{\circ}$,find the measure of each angle of $\Delta PQR$.

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