Question

In the circle with centre O, PQ = QR = RS and ∠QOS = 116°. Calculate ∠POQ, ∠OQR, ∠OQS and ∠SQR.

Answer 1

Given:

• Center: O
• Chords: PQ = QR = RS ⇒ All three sides of triangle PQRS are equal
• Angle ∠QOS = 116°
• Need to find:
  • ∠POQ
  • ∠OQR
  • ∠OQS
  • ∠SQR

Step 1: Use Isosceles Triangles

Since PQ = QR = RS, triangles:

• ΔPOQ, ΔQOR and ΔROS are all isosceles.
• So, central angles ∠POQ = ∠QOR = ∠ROS

Step 2: Use central angles around a point

Total angle around point O = 360°

∠POQ + ∠QOR + ∠ROS = 360°

Given: ∠QOS = 116° = ∠QOR + ∠ROS

So, ∠POQ = 360° – 116° = 244°

Now since ∠POQ = 244° and ∠QOR = ∠ROS, then:

∠QOR = ∠ROS

= `116^circ/2`

= 58°

Step 3: Use triangle ΔOQR to find ∠OQR

In ΔQOR, use angle sum = 180°

• ∠OQR = 58°   ...(Central angle)
• OQ = OR   ...(Radii ⇒ Isosceles triangle)

So base angles are equal:

∠OQR = ∠ORQ

= `(180^circ - 58^circ)/2`

= `122^circ/2`

= 61°

Step 4: Find ∠OQS

This is part of triangle ΔQOS

In ΔQOS, with ∠QOS = 116° and it’s isosceles (OQ = OS):

So base angles:

∠OQS = ∠OSQ

= `(180^circ - 116^circ)/2`

= `64^circ/2`

= 32°

Step 5: Find ∠SQR

In triangle ΔQRS, all sides are equal ⇒ equilateral triangle? No, we already found angles.

From above:

• ∠OQR = 61°
• ∠OQS = 32°

So, ∠SQR = ∠OQR – ∠OQS

= 61° – 32°

= 29°

• ∠POQ = 58°
• ∠OQR = 61°
• ∠OQS = 32°
• ∠SQR = 29°