Question : In $\triangle XYZ$, points P, Q, and R are points on the sides XY, YZ, and XZ respectively $\angle YXZ = 50^\circ$, XR = PR, ZR = QR, and $\angle RQZ = 80^\circ$. What is the value of $\angle PRQ$?
Option 1: $80^\circ$
Option 2: $90^\circ$
Option 3: $75^\circ$
Option 4: $60^\circ$
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Correct Answer: $80^\circ$
Solution :
In $\triangle XPR$
$\angle PXR = 50^\circ$
Since XR = PR, the triangle is isosceles.
$\angle$XPR = $\angle$PXR = 50$^\circ$
⇒ $\angle XRP = 180^\circ-50^\circ-50^\circ= 80^\circ$
Since ZR = QR, the triangle is isosceles.
Similarly,
$\angle RQZ = \angle RZQ = 80^\circ$
⇒ $\angle QRZ = 180^\circ- 80^\circ- 80^\circ = 20^\circ$
Now, $\angle PRQ = 180^\circ - \angle RQZ - \angle XPR$
$⇒\angle PRQ = 180^\circ-20^\circ- 80^\circ=80^\circ$
Hence, the correct answer is $80^\circ$.
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