Question

In the following figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Answer 1

Given, in the figure BD = OD, CD ⊥ AB.

In ΔOBD, BD = OD   ...[Given]

OD = OB  ...[Both are the radius of circle]

∴ OB = OD = BD

Thus, ΔODB is an equilateral triangle.

∴ ∠BOD = ∠OBD = ∠ODB = 60°

In ΔMBC and ΔMBD,

MB = MB   ...[Common side]

 ∠CMB = ∠BMD = 90°

And CM = MD  ...[In a circle, any perpendicular drawn on a chord also bisects the chord]

∴ ΔMBC ≅ ΔMBD   ...[By SAS congruence rule]

∴ ∠MBC = ∠MBD   ...[By CPCT]

⇒ ∠MBC = ∠OBD = 60°  ...[∵ ∠OBD = 60°]

Since, AB is a diameter of the circle.

∴ ∠ACB = 90°

In ΔACB, ∠CAB + ∠CBA + ∠ACB = 180°  ...[By angle sum property of a triangle]

⇒ ∠CAB + 60° + 90° = 180°

⇒ ∠CAB = 180° – (60° + 90°) = 30°