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  • ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to: (A) 80º (B) 50º (C) 40º (D) 30º

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and \angleADC = 140º, then \angleBAC is equal to:

(A) 80º

(B) 50º

(C) 40º

(D) 30º

Answers (1)

(B) 50°

Solution:

Given, ABCD is cyclic Quadrilateral and \angleADC = 140°

We know that the sum the opposite angles in a cyclic quadrilateral is 180°.

\angleADC + \angleABC = 180°

140° + \angleABC = 180°

\angleABC = 180° – 140°

\angleABC = 40°

Since, \angleACB is an angle in semi circle

\angleACB = 90°

In \triangleABC

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

\angleBAC + 90° + 40° = 180°

\angleBAC = 180° – 130° = 50°

Therefore option (B) is correct.

Posted by

infoexpert24

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