Question

In the given figure, O is the centre of the circle, BD = OD and CD$\perp$AB. Find $\angle$CAB.

Answer 1

ANSWER:
In the given figure, BD = OD and CD ⊥ AB.Join AC and OC.
In ∆ ODE and ∆ DBE,
∠ DOE = ∠ DBE (given)
∠ DEO = ∠ DEB = 90 ∘
OD = DB (given)
∴ By AAS conguence rule, ∆ ODE ≌ ∆ BDE,
Thus, OE = EB
...(1)
Now, in ∆ COE and ∆ CBE ,
CE = CE (common)
∠ CEO = ∠ CEB = 90 ∘
OE = EB (from (1))
∴ By SAS conguence rule, ∆ COE ≌ ∆ CBE ,
Thus, CO = CB
...(2)
Also, CO = OB = OA (radius of the circle)
...(3)
From (2) and (3),
CO = CB = OB
∴ ∆ COB is equilateral triangle.
∴ ∠ COB = 60 ∘
...(4)
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the
remaining part of the circle.
Here, arc CB subtends ∠ COB at the centre and ∠ CAB at A on the circle.
∴ ∠ COB = 2∠ CAB
⇒ ∠ CAB=60°2=30°
(from (4))
Hence, ∠ CAB = 30 ∘ .