Q18 of 63 Page 156

One side AB of a quadrilateral is a diameter of its circumscribed circle and ∠ADC = 140°. Then, ∠BAC is equal to—

Given AB of a quadrilateral is a diameter of its circumscribed circle.


And ∠ADC = 140°



Here, O is the centre of the circle and ADCBOA is the semi circle.


∴ ∠ADB = 90°


Given that ∠ADC = 140°


⇒∠ADB + ∠BDC = 140°


⇒ 90° + ∠BDC = 140°


⇒∠BDC = 140° - 90°


∴ ∠BDC = 50°


We know that angles in the same segment are equal.


⇒∠BDC = ∠BAC


∴ ∠BAC = 50°

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