In Fig., PQ = PR, RS = RQ and ST || QR. If the exterior angle RPU is 1

1.	In Fig., PQ = PR, RS = RQ and ST \|\| QR. If the exterior angle RPU is 140°, then the measure of angle TSR is (a) 55° (b) 40° (c) 50° (d) 45°
Answer» (b) 40^o Consider the ΔPQR. From the exterior angle property = ∠RPU – ∠PRQ + ∠PQR ∠RPU = ∠PRQ + ∠PQR 140^O = 2 ∠PQR … [given PQ = PR] ∠PQR = 140/2 ∠PQR = 70^o Given, ST \|\| QR and QS is transversal. From the property of corresponding angles, ∠PST = ∠PQR = 70^o Now, consider the ΔQSR RS = RQ … [from the question] So, ∠SQR – ∠RSQ = 70^O Then, PQ is a straight line. ∠PST + ∠TSR + ∠RSQ = 180^o 70^o + ∠TSR + 70^o = 180^o 140^o + ∠TSR = 180^o ∠TSR = 180^o – 140^o ∠TSR = 40^o

In Fig., PQ = PR, RS = RQ and ST || QR. If the exterior angle RPU is 140°, then the measure of angle TSR is (a) 55° (b) 40° (c) 50° (d) 45°

Answer»

(b) 40^o

Consider the ΔPQR.

From the exterior angle property = ∠RPU – ∠PRQ + ∠PQR

∠RPU = ∠PRQ + ∠PQR

140^O = 2 ∠PQR … [given PQ = PR]

∠PQR = 140/2

∠PQR = 70^o

Given, ST || QR and QS is transversal.

From the property of corresponding angles, ∠PST = ∠PQR = 70^o

Now, consider the ΔQSR

RS = RQ … [from the question]

So, ∠SQR – ∠RSQ = 70^O

Then, PQ is a straight line.

∠PST + ∠TSR + ∠RSQ = 180^o

70^o + ∠TSR + 70^o = 180^o

140^o + ∠TSR = 180^o

∠TSR = 180^o – 140^o

∠TSR = 40^o