Question

In figure, O is center of circle, BD = OD and CD ⊥ AB, then find ∠CAB

Answer 1

We know that BD = GD

∵ GD = OB (radius of circle)

∴ BD = OD = OB

Thus, OBD will be equilateral triangle

Each angle of equilateral triangle is 60°

∴ ∠BOD = ∠ODB = ∠OBD = 60°

By arc BD, angle subtended at center ∠BOD will be double the angle subtended at remaining part of circle 

i.e., ∠DAB

∠DAB = \(\frac { 1 }{ 2 }\) × 60°

= 30°

We know that AO, ∠CAD is of

∴ ∠CAB = ∠DAB = 30°

Thus, ∠CAB = 30°