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  • A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and angle ADC equal to 130 degree. Find angle BAC.

A quadrilateral ABCD is inscribed in a circle such that AB is the diameter and \angleADC = 130º. Find \angleBAC.

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Given: Quadrilateral ABCD inscribed in a circle

\angleADC = 130º

\therefore Quadrilateral ABCD is a cyclic Quadrilateral. The sum of opposite angles in a cyclic quadrilateral is 180°   

\Rightarrow \angleADC + \angleABC = 180°

\Rightarrow 130° + \angleABC = 180°

\Rightarrow \angleABC = 180° – 130°

\Rightarrow \angleABC = 50°  …(i)

AB is the diameter of a circle (given)

\therefore ACB = 90°  (angle in a semicircle is 90º)

In \triangleABC,

\Rightarrow \angleABC + \angleACB + \angleBAC = 180°(angle sum property of a triangle)

Putting all the values,

\Rightarrow 50° + 90° + \angleBAC = 180°

\Rightarrow \angleBAC = 180° –50° – 90°

\therefore \angleBAC = 40°

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