Question

The vertical angle of an isosceles triangle is ${100}^{\circ }.$ Find its base angles.

Answer 1

In $\Delta ABC,AB=ACand\angle A={100}^{\circ }$

But AB = AC (In isosceles triangle)

$\angle C=\angle B$

(Angles opposite to equal sides)

$\angle A+\angle B+\angle C={180}^{\circ }$

(Sum of angles of a triangle)

$\Rightarrow {100}^{\circ }+\angle B+\angle B={180}^{\circ }$ ($\because \angle C=\angle B$)

$\Rightarrow 2\angle B={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

$\angle B=\frac{{80}^{\circ }}{2}={40}^{\circ }$

$\angle C=\angle B={40}^{\circ }$

Hence $\angle B={40}^{\circ },\angle C={40}^{\circ }$