Question : $\triangle PQR$ and $\triangle SQR$ are both isosceles triangles on a common base $QR$ such that $P$ and $S$ lie on the same side of $QR$. If $\angle QSR=60^{\circ}$ and $\angle QPR=100^{\circ}$, then find $\angle SRP$.
Option 1: $80^{\circ}$
Option 2: $60^{\circ}$
Option 3: $100^{\circ}$
Option 4: $20^{\circ}$
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Correct Answer: $20^{\circ}$
Solution :
Given: That $\triangle PQR$ and $\triangle SQR$ are isosceles triangles, $\angle QPR =100^{\circ}$ and $\angle QSR = 60^{\circ}$.$\triangle PQR$ and $\triangle SQR$ are isosceles triangles, such that $PQ=PR$ and $QS=SR$.
$PQ=PR$
$\therefore \angle PQR=\angle PRQ$
$180^{\circ}=100^{\circ}+2 \angle PQR$
⇒ $\angle PQR=\angle PRQ=40^{\circ}$...(i)
⇒ $QS=RS$
$\therefore \angle QRS=\angle RQS$
⇒ $180^{\circ}=60^{\circ}+2 \angle QRS$
⇒ $\angle QRS=\angle SQR=60^{\circ}$...(ii)
$\therefore\angle SRP=\angle QRS-\angle QRP$
From equations (i) and (ii),
$\angle SRP=60^{\circ}-40^{\circ}=20^{\circ}$
Hence, the correct answer is $20^{\circ}$.
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