In fig , AD is a diameter of the circle . If ∠ = 150° , calculate (i) ∠ (ii) ∠
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(i) 30° (ii) 60°
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Step-by-step explanation:
(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°
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