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  • In triangle ABC, BC = AB and angle B = 80^{\circ}. Then angle A is equal to (A) 80^{\circ}\\ (B) 40^{\circ}\\ (C) 50^{\circ}\\ (D) 100^{\circ}\\

In \triangleABC, BC = AB and \angleB = 80^{\circ}. Then \angleA is equal to
\\(A) 80^{\circ}\\ (B) 40^{\circ}\\ (C) 50^{\circ}\\ (D) 100^{\circ}\\

Answers (1)

            [C]

Solution.         We know, angles opposite to equal sides of a triangle are equal

Here, AB = BC and \angleC is opposite to side AB and \angleA is opposite to side BC

\therefore \angleC = \angleA = x (Assume)

\angleB = 80° (Given)

\because \angleA + \angleB + \angleC = 180°

\Rightarrow x + 80° + x = 180°

\Rightarrow 2x = 100°

\Rightarrow x = 50°

Hence option (C) is correct.

Posted by

infoexpert24

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