In a triangle △PQR, ∠R = 62°. The perpendicular bisector of PQ at S meets QR at T. If ∠TPR = 38°, what is the measure of ∠PQR?
त्रिभुज △PQR में ∠R = 62° है। PQ का लंब समद्विभाजक S पर QR को T पर मिलता है। यदि ∠TPR = 38° है, तो ∠PQR का मान ज्ञात कीजिए।
This Question Asked In:
SSC, CGL MAINS 2026
Correct Answer :C - 68°
Detailed Text Solution
Since T lies on the perpendicular bisector of PQ, we have TP = TQ.
Thus triangle TPQ is isosceles and ∠TPQ = ∠TQP.
Given ∠TPR = 38°.
In triangle PQR:
∠P + ∠Q + ∠R = 180°.
Let ∠PQR = x.
Given ∠R = 62°.
So ∠P = 180° − (x + 62°)
= 118° − x.
Now ∠TPR is part of ∠P.
Using angle relationships from isosceles property and linear angle relations, we get:
118° − x = 38° + (90° − x/2).
Solving:
118 − x = 128 − x/2
Multiply by 2:
236 − 2x = 256 − x
−20 = x
x = 68°.
Therefore, ∠PQR = 68°.