Question

In fig 6.12 PQ=PR,RS=RQ and ST||QR.If the exterior angle RPU is 140°the the measurement of TSR is??

Answer 1

triangle PQR is an isosceles triangle.....PQ=PR
therefore, angle PRQ = angle PQR
angle RPU = 140.....given
angle RPU = angle PRQ+angle PQR
angle PQR = 70

triangle SQR is an isosceles triangle........RS=RQ
therefore angle SQR=angle QSR=70....angle SQR=angle PQR

ST parallel to QR and SQ is transversal
angle TSQ+angle SQR=180......interior angles on same side of transversal
Angle TSQ=180-70=110

Angle TSQ= angle TSR+angle QSR
110 =angle TSR+70
angle TSR = 40

Answer 2

Answer:

Step-by-step explanation:

Your answer is: 40 degree

Consider the ΔPQR.

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

∠RPU = ∠PRQ + ∠PQR

140 degree= 2 ∠PQR … [given PQ = PR]

∠PQR = 140/2

∠PQR = 70 degree

Given, ST || QR and QS is transversal.

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

Now, consider the ΔQSR

RS = RQ … [from the question]

So, ∠SQR – ∠RSQ = 70 degree

Then, PQ is a straight line.

∠PST + ∠TSR + ∠RSQ = 180 degree

70o + ∠TSR + 70o = 180 degree

140o + ∠TSR = 180 degree

∠TSR = 180o – 140

∠TSR = 40 degree