Question

In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ADB = 30° and ∠CBD = 60°, calculate:

1. ∠QAB, 
2. ∠PAD, 
3. ∠CDB.

Answer 1

i. PAQ is a tangent and AB is the chord.

∠QAB = ∠ADB = 30°  ...(Angles in the alternate segment)

ii. OA = OD   ...(Radii of the same circle)

∴ ∠OAD = ∠ODA = 30°

But, OA ⊥ PQ

∴ ∠PAD = ∠OAP – ∠OAD

= 90° – 30°

= 60°

iii. BD is the diameter.

∴ ∠BCD = 90°  ...(Angle in a semi-circle)

Now in ΔBCD,

∠CDB + ∠CBD + ∠BCD = 180°

`=>` ∠CDB + 60° + 90° = 180° 

`=>` ∠CDB = 180° – (60° + 90°) 

`=>` ∠CDB = 180° – 150° = 30°