Question

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is

• A. ${100}^{\circ }$
• B. ${120}^{\circ }$
• C. ${110}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is B

${120}^{\circ }$

(b),

$In\Delta ABC,$

$\angle A=2(\angle B+\angle C)$

$=2\angle B+2\angle C$

$Adding2\angle A\text{to both sides,}$

$\angle A+2\angle A=2\angle A+2\angle B+2\angle C$

$\Rightarrow 3\angle A=2(\angle A+\angle B+\angle C)$

$\Rightarrow 3\angle A=2\times {180}^{\circ }$

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

$\Rightarrow 3\angle A={360}^{\circ }\Rightarrow \angle A=\frac{{360}^{\circ }}{3}={120}^{\circ }$

$\therefore \angle A={120}^{\circ }$