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