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

2∠Q = 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°