# In Fig. 6.12, PQ

Given: PQ = PR, RS = RQ and ST || QR. If the exterior angle RPU is 140°

Formula Used/Theory:-

Base angle are equal is isosceles triangle

Alternate angles are equal if lines are parallel

Exterior angle is equal to sum opposite interior angles

In ΔPQR

UPR is exterior angle

UPR = Q + R

As ΔQPR is isosceles triangle because PQ = PR

Hence;

Q = R

140° = Q + R

2Q = 140°

Q = = 70°

As ΔQRS is isosceles triangle because QR = RS

Q = QSR

QSR = 70°

Then;

Q + QSR + SRQ = 180°

70° + 70° + SRQ = 180°

SRQ = 180° - 140° = 40°

As ST || QR

And SR is transverse

SRQ = TSR Alternate interior angles.

TSR = 40°

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