Question

In the given figure, O is the centre of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P.

AB = 9 cm, BP = 16 cm, ∠PTB = 50° ∠OBA = 45°

Find:

1. Length of PT 
2. ∠BAT
3. ∠BOT
4. ∠ABT

Answer 1

a. 

PT² = AP × BP

= (AB + BP) × BP

= (9 + 16) × 16

= 25 × 16

= 400

∴ PT = `sqrt(400)`

= 20 cm

b. Angles in alternate segments are equal

∴ ∠BAT = ∠PTB

= 50°

c. OT ⊥ PQ

∴ ∠OTP = 90°

`\implies` ∠OTP + ∠PTB = 90°

`\implies` ∠OTB + ∠50° = 90°

`\implies` ∠OTB = 90° – 50° = 40°

Now, In ΔBOT,

OB = OT(Radii of same circle)

∠OTB = ∠OBT = 40°

and ∠BOT + ∠OTB + ∠OBT = 180°

∠BOT + 40° + 40° = 180°

∠BOT = 180° – 80°

= 100°

d. In ΔABT,

∠ABT = ∠OBA + ∠OBT

= 45° + 40°

= 85°