Question

In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130o. Find:

1) ∠DAB

2) ∠DBA

Answer 1

Given: AB is the diameter of the circle with centre O, ∠BCD = 130°

To find: ∠DAB, ∠DBA

1) Clearly, ABCD is a cyclic quadrilateral.

We know, the sum of a pair of opposite angles of a cyclic quadrilateral is 180°

∴ ∠DAB + ∠DCB = 180°

⇒ ∠DAB + 130° = 180°

⇒ ∠DAB = 180° - 130° = 50°

2) Consider ΔDAB,

Here, ∠ADB = 90° …..[Since angle in a semi-circle is a right angle]

So, by angle sum property of a triangle,

∠DAB + ∠DBA + ∠ADB = 180°

⇒ 50° + ∠DBA + 90° = 180°

⇒ ∠DBA = 180° - 140°  = 40°