For perimeter $15$, $\angle B = 60^{\circ}$, $\angle C = 50^{\circ}$, third angle $\angle A$ is

For perimeter $15$, $\angle B = 60^{\circ}$, $\angle C = 50^{\circ}$, third angle $\angle A$ is
  1. $70^{\circ}$
  2. $60^{\circ}$
  3. $110^{\circ}$
  4. $80^{\circ}$

Solution

$180 - 60 - 50 = 70^{\circ}$.

Asked in: MH-SSC-9

Practice more Constructions of Triangles questions on Aicharya