Question

Chord PS = chord SR = chord PT in the circle with centre O and ∠PQR = 50°. Calculate:

1. ∠PQS
2. ∠TQR
3. ∠QOR
4. ∠POR
5. ∠POS

Answer 1

Step 1: ∠PQS

Since SR = PS, triangle ΔPSR is isosceles

Arc PR is made up of 3 equal arcs (since PS = SR = PT)

So arc PS, SR and PT are each `1/3` of arc PR

Since ∠PQR = 50° and PQ subtends arc PR, then by angle at the circle:

`∠PQR = 1/2 xx arc  PR`

⇒ arc PR = 100°

Then arc PS = SR

= `100^circ/2`

= 50°

Arc RS = 50°

⇒ Angle ∠PQS subtends arc RS:

`∠PQR = 1/2 xx arc  RS`

= `1/2 xx 50^circ`

= 25°

Step 2: ∠TQR

PT = PS = SR

⇒ So arc TR = 150°   ...(3 segment of 50° each)

Angle ∠TQR subtends arc TR

`∠TQR = 1/2 xx arc  TR`

= `1/2 xx 150^circ`

= 75°

Step 3: ∠QOR

Central angle subtending arc QR

From earlier: arc QR = 50°

So, ∠QOR = Angle at center = 80°.

Step 4: ∠POR

Arc PR = 100°

So, ∠POR = Central angle subtending arc PR = 100°

Step 5: ∠POS

Arc PS = 50°

So, ∠POS = Angle at center = 50°.