Question

PA and PB are tangents drawn to a circle with centre O. If ∠AOB = 120° and OA = 10 cm, then 

1. Find ∠OPA.  (1)
2. Find the perimeter of ΔOAP.  (3)
3. Find the length of chord AB.  (1)

Answer 1

i. Given AOB = 120°, OA = 10 cm

Since PA and PB are tangents to the circle at A.

∴ ∠OAP = 90°

∠OBP = 90°

In quadrilateral APBО

∠OAP + ∠APB + ∠PBO + ∠AOB = 360°

90° + ∠AРВ + 90° + 120° = 360°

∠APB = 60°

∠OPA = `1/2` ∠APB

= `1/2 xx 60^circ`

= 30°

ii. In ΔΟΡΑ (right-angled)

tan 30° = `(OA)/(AP)`

`1/sqrt(3) = 10/(AP)`

∴ AP = `10sqrt(3)` cm

By Pythagoras theorem,

(OP)² = (OA)² + (AP)²

= `(10)^2 + (10sqrt(3))^2`

= 100 + 300

= 400

OP = 20 cm

Perimeter of ΔOAP,

⇒ OA + AP + OP

⇒ `10 + 10sqrt(3) + 20`

= `30 + 10sqrt(3)`

= `10sqrt(3) (sqrt(3) + 1)` cm

iii. In ΔAPB,

∠APB = 60°

∠PAB = ∠PBA   ...(∵ AP = BP)

By angle sum property,

∠APB + ∠PAB + ∠PBA = 180°

60° + ∠PАВ + ∠PАВ = 180°

2∠PAB = 180° – 60°

∠PAB = `120/2`

∠PAB = 60°

Since, ∠APB = ∠PAB = ∠PBA = 60°

So ΔAPB is a equilateral triangle.

Then, AP = BP = AB = `10sqrt(3)` cm

Hence, the length of chord AB is `10sqrt(3)` cm.