Question

AD is a diameter of the circle. If ∠BCD = 150°, Calculate ()∠BAD ()∠ADB​

Answer 1

Given :

AD is a diameter of the circle and ∠BCD = 150°

To Find :

(i) ∠BAD

(ii) ∠ADB​

Solution :

(i) Join BD  

Now, ABCD is a cyclic quadrilateral

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

∴ ∠BAD + ∠BCD = 180°

=> ∠BAD + 150° = 180°

=>     ∠BAD        = 180° - 150°

∴       ∠BAD        = 30°

Hence, ∠BAD = 30°

(ii) ∠ABD = 90° (Angle in a semi-circle)

We know, sum of the three angles of a triangle is 180°.

Therefore, In △ABD,  

∠ABD + ∠BAD + ∠ADB = 180°

  90° + 30° + ∠ADB = 180°  

=>      ∠ADB             = 180° – 120°

∴       ∠ADB             = 60°

Hence, ∠ADB = 60°

Note - Refer to the diagram attached.