Question

ED is the tangent to the circle with centre O. ∠BCD = 52°. Then, ∠CAB equals

(a) 38°

(b) 76°

(c) 52°

(d) 46°

Answer 1

Correct option is (c) 52°

Given, ∠BCD = 52^∘

Now, ∠OCD = 90^∘ ....(Right angle is formed between the radius an the tangent a the point on circle)

⇒ ∠BCD + ∠OCB = 90^∘

⇒ ∠OCB = 90^∘ − 52^∘ = 38^∘

Now, in △OCB, we have

OB = OC

⇒ ∠OCB = ∠OBC = 38^∘ ....(Isosceles triangle property)

Sum of angles of triangle = 180^∘

⇒ ∠OCB + ∠OBC + ∠COB = 180^∘

⇒ 38^∘ + 38^∘ + ∠COB = 180^∘

⇒ ∠COB = 180^∘ − 76^∘

⇒ ∠COB = 104^∘

Now, ∠CAB = 21​∠COB 

= \(\frac 12\)​(104^∘)

= 52^∘

Angle formed at the circle is half the angle subtended at the center by the segment.