Question

In a $\Delta ABC$, if $\angle A={55}^{\circ },\angle B={40}^{\circ }$, find $\angle C$

Answer 1

$\because$ sum of three angles of a triangle is ${180}^{\circ }$

$\therefore$ In $\Delta ABC$, $\angle A={55}^{\circ },\angle B={40}^{\circ }$

But $\angle A+\angle B+\angle C={180}^{\circ }$ (sum of angles of a triangle)

$\Rightarrow {55}^{\circ }+{40}^{\circ }+\angle C={180}^{\circ }\Rightarrow {95}^{\circ }+\angle C={180}^{\circ }\therefore \angle C={180}^{\circ }-{95}^{\circ }={85}^{\circ }$