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Question
A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130º. Find ∠BAC.
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Solution
Draw a quadrilateral ABCD inscribed in a circle having centre O.
Given, ∠ADC = 130°
Since, ABCD is a quadrilateral inscribed in a circle, therefore ABCD becomes a cyclic quadrilateral.
∵ Since, the sum of opposite angles of a cyclic quadrilateral is 180°.
∴ ∠ADC + ∠ABC = 180°
⇒ 130° + ∠ABC = 180°
⇒ ∠ABC = 50°
Since, AB is a diameter of a circle, then AB subtends an angle to the circle is right angle.
∴ ∠ACB = 90°
In ΔABC, ∠BAC + ∠ACB + ∠ABC = 180° ...[By angle sum property of a triangle]
⇒ ∠BAC + 90° + 50° = 180°
⇒ ∠BAC = 180° – (90° + 50°)
= 180° – 140°
= 40°
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