Question

AB is a diameter of a circle with the center O. C is a point on the circumference of the circle, such that ∠CBA = 32°. Find ∠CAB.

A) 28°
B) 32°
C) 58°
D) 96°

Answer 1

You have to know that triangles in a circle that go through the diameter is a right angle triangle: angle ACB = 90 degrees. So CAB = 180 - 90 - 32= 58

Answer 2

∠CAB = 58°. Therefore, The correct option is C) 58°

Explanation:

For better understanding of the solution, see the attached figure of the problem :

In ΔCOB, OC = OB ( Radius of the same circle)

⇒ ∠OCB = ∠OBC ( Angles opposite to equal sides are equal)

Since, ∠OBC = 32°

⇒ ∠OBC = 32°

Now, Using angle sum property,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ 32° + 32° + ∠BOC = 180°

⇒ ∠BOC = 116°

Now, ∠AOC + ∠BOC = 180° ( Linear pair angles sum is supplementary)

⇒ ∠AOC = 180 - 116

⇒ ∠AOC = 64°

Now, In ΔAOC, OA = OC ( Radius of same circle)

⇒ ∠OCA = ∠OAC ( Angles opposite to equal sides are equal)

Using angles sum property,

∠OCA + ∠OAC + ∠AOC = 180°

⇒ 2∠OAC = 180 - 64

⇒ ∠OAC = 58°

Hence, ∠CAB = 58°