Question

3) In fig AD is a diameter
the circle. It LB CD =150°. Calculate
of
1. LBAD
ii LADB​

Answer 1

Answer:

(i) Join BD

Now, ABCD is a cyclic quadrilateral

∴ ∠BAD + ∠BCD = 180° (Opposite angles of a cyclic quadrilateral)

=> ∠BAD + 150° = 180°

=> ∠BAD = 180° - 150° = 30°

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

Now, In △ABD, we have

∠ABD + ∠BAD + ∠ADB = 180°

90° + 30° + ∠ADB = 180°

=> ∠ADB = 180° – 120° = 60°