Question

In fig , AD is a diameter of the circle . If ∠ = 150° , calculate (i) ∠ (ii) ∠​

Answer 1

Answer:

Solution :- (i) Join BD Now, ABCD is a cyclic quadrilateral ∴ ∠BAD + ∠BCD = 180° ( Opposite angles of a cyclic quadrilateral) => ∠BAD ...

Step-by-step explanation:

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Answer 2

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°

Step-by-step explanation:

1 angle is 30°

2 angle is 60°

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