Question

In the given figure, $\Delta ABC$ is an isosceles triangle with $\angle A={40}^0$ and AB = AC. The measure of the remaining angles is

• A. ${70}^0$ Each
• B. ${70}^0,{60}^0$
• C. ${60}^0$ Each
• D. ${50}^0,{40}^0$

Answer 1

The correct option is A

${70}^0$ Each

Given that, $\Delta ABC$ is an isosceles triangle with $\angle A={40}^0$ and AB = AC.

Since AB = AC

$\Rightarrow \angle B=\angle C$ [Base angles of an isosceles triangle]

$\angle A+\angle B+\angle C={180}^0$ [Sum of all the angles of a triangle is ${180}^0$]

${40}^0+2\angle B={180}^0$ [$\because \angle B=\angle C$]

$\Rightarrow 2\angle B={180}^0-{40}^0$

$\Rightarrow \angle B=\frac{{140}^0}{2}={70}^0$

$\Rightarrow$ The measure of the remaining each angle is ${70}^0$