Question

In the semicircle with centre O, chords PQ = QR = RS. Find ∠QOR, ∠OPR, ∠OPQ and ∠QPR.

Answer 1

Step 1:

Since PQ = QR = RS, the angles subtended at the center are equal: ∠POQ = ∠QOR = ∠ROS.

The angles along the diameter form a straight line: ∠POQ + ∠QOR + ∠ROS = 180°

Let ∠QOR = x

Then x + x + x = 180°

3x = 180°

`x = 180^circ/3 = 60^circ`

Thus, ∠QOR = 60

Step 2:

In ΔOPQ, OP = OQ (radii), so it’s an isosceles triangle.

Angles opposite equal sides are equal: ∠OPQ = ∠OQP

The sum of angles in ΔOPQ is 180°: ∠OPQ + ∠OQP + ∠POQ = 180°

Since ∠POQ = 60°, we have 2∠OPQ + 60° = 180°

2∠OPQ = 120°

`∠OPQ = 120^circ/2 = 60^circ`

Step 3:

In ΔOQR, OQ = OR (radii), so it’s an isosceles triangle.

Angles opposite equal sides are equal: ∠OQR = ∠ORQ

The sum of angles in ΔOQR is 180°: ∠OQR + ∠ORQ + ∠QOR = 180°

Since ∠QOR = 60°, we have 2∠OQR + 60° = 180°

2∠OQR = 120°

`∠OQR = 120^circ/2 = 60^circ`

In ΔOPR, OP = OR (radii), so it’s an isosceles triangle.

∠OPR = ∠ORP

∠POR = ∠POQ + ∠QOR = 60° + 60° = 120°

In ΔOPR, ∠OPR + ∠ORP + ∠POR = 180°

2∠OPR + 120° = 180°

2∠OPR = 60°

`∠OPR = 60^circ/2 = 30^circ`

Step 4:

∠QPR = ∠OPR – ∠OPQ

This is incorrect as ∠QPR is part of ΔPQR

In ΔPQR, PQ = QR, so it’s an isosceles triangle.

∠QPR = ∠QRP

∠PQR = ∠PQO + ∠OQR = 60° + 60° = 120°

In ΔPQR, ∠QPR + ∠QRP + ∠PQR = 180°

2∠QPR + 120° = 180°

2∠QPR = 60°

`∠QPR = 60^circ/2 = 30^circ`

The angles are ∠QOR = 60°, ∠OPR = 30°, ∠OPQ = 60° and ∠QPR = 30°.