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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 $\triangle A B C, B C=A B$ and $\angle B=80^{\circ}$. Then $\angle A$ is equal to
(A) $40^{\circ}$
(B) $80^{\circ}$
(C) $50^{\circ}$
(D) $60^{\circ}$

Answers (1)

Solution.

We know that angles opposite to equal sides of a triangle are equal

Here, $A B=B C$ and $\angle C$ is opposite to side $A B$ and $\angle A$ is opposite to side $B C$

$
\begin{aligned}
& \therefore \angle \mathrm{C}=\angle \mathrm{A}=\mathrm{x}(\text { Assume }) \\
& \angle \mathrm{B}=80^{\circ} \text { (Given) } \\
& \because \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
& \Rightarrow \mathrm{x}+80^{\circ}+\mathrm{x}=180^{\circ} \\
& \Rightarrow 2 \mathrm{x}=100^{\circ} \\
& \Rightarrow \mathrm{x}=50^{\circ}
\end{aligned}
$

Hence option (C) is correct.

Posted by

infoexpert24

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