A quadrilateral ABCD is inscribed in a circle such that AB is diameter and ADC = 130º. Find BAC.
40°
Solution:
Given: Quadrilateral ABCD inscribed in a circle
ADC = 130º
Quadrilateral ABCD is a cyclic Quadrilateral. Sum of opposite angles in a cyclic quadrilateral is 180°
ADC + ABC = 180°
130° + ABC = 180°
ABC = 180° – 130°
ABC = 50° …(i)
AB is a diameter of a circle (given)
ACB = 90° (angle in a semicircle is 90º)
In ABC,
ABC + ACB + BAC = 180°(angle sum property of a triangle)
Putting all the values,
50° + 90° + BAC = 180°
BAC = 180° –50° – 90°
BAC = 40°