Question

In ΔPQR, QR is produced to S so that ∠PRS = 130°. T is a point on QR so that PT = QT = TR. Find ∠QPR.

Answer 1

In ΔPQR, the side QR is produced to S and the exterior angle ∠PRS = 130°.

The angle ∠PRQ is supplementary to ∠PRS.

So, ∠PRQ = 180° – 130°

= 50°

We are given that T is a point on QR such that PT = QT = TR. 

Consider ΔPTR.

Since PT = TR, it is an isosceles triangle. 

The angles opposite the equal sides are equal:

∠TPR = ∠PRT = ∠PRQ = 50°

The sum of angles in ΔPTR is 180°.

So, ∠PTR = 180° – (∠TPR + ∠PRT)

= 180° – (50° + 50°)

= 80°

Consider ΔPTQ.

Since PT = QT, it is an isosceles triangle. 

The angles opposite the equal sides are equal:

∠TPQ = ∠PQT = ∠PQR

The angle ∠PTR is the exterior angle for ΔPTQ at vertex T.

The exterior angle is the sum of the two opposite interior angles:

∠PTR = ∠TPQ + ∠PQT

80° = 2 × ∠PQT

∠PQT = 40°

And also, ∠TPQ = 40°.

The angle ∠QPR is the sum of ∠TPQ and ∠TPR:

∠QPR = ∠TPQ + ∠TPR

= 40° + 50°

= 90°