Question : PQR is a triangle. The bisectors of the internal angle $\angle Q$ and external angle $\angle R$ intersect at S. If $\angle QSR=40^{\circ}$, then $\angle P$ is:
Option 1: $40^{\circ}$
Option 2: $60^{\circ}$
Option 3: $80^{\circ}$
Option 4: $30^{\circ}$
Correct Answer: $80^{\circ}$
Solution :
$\angle$QSR = 40$^\circ$
The exterior angle property of a triangle i.e. the exterior angle of a triangle is equal to the sum of the interior opposite angle of a triangle.
In $\triangle$PQR
$\angle$PRT = $\angle$PQR + $\angle$QPR (exterior angle property of a triangle)
⇒ 2b = 2a + $\angle$QPR --------------------- (1)
In $\triangle$SQR
$\angle$SRT = $\angle$SQR + $\angle$QSR (exterior angle property)
⇒ b = a + 40 --------------------------- (2)
Put the value of b in equation (1)
2(a + 40) = 2a + $\angle$QPR
⇒ 2a + 80$^\circ$ = 2a + $\angle$QPR
⇒ 80$^\circ$ = $\angle$QPR
⇒ $\angle$QPR = 80$^\circ$
Hence, the correct answer is 80$^\circ$.
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