ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ADC = 140 �, then BAC is equal to:
Given: ABCD is a cyclic quadrilateral.
∠ADC = 140°
Since sum of opposite angles of a cyclic quadrilateral is 180°,
∴ ∠ADC + ∠ABC = 180°
⇒ 140° + ∠ABC = 180°
⇒ ∠ABC = 180° - 140°
⇒ ∠ABC = 40°
Since, diameter subtends a right angle to the circle,
∴ ∠ACB = 90°
Now, in triangle ACB; by angle sum property of a triangle, sum of all angles of a triangle is 180°.
∴ ∠CAB + ∠ABC + ∠ACB = 180°
⇒ ∠CAB + 40° + 90° = 180°
⇒ ∠CAB = 180° - 90° - 40°
⇒ ∠CAB = 50°
∴ ∠CAB = ∠50°