Question

Let's write the measurement of the angles of triangle $ABC$ if $AB=BC$ and $\angle BAC+\angle ACB={50}^o$

Answer 1

$AB=AC$

$\angle BAC+\angle ACB={50}^{\circ }$

$\because AB=AC$

$\therefore$ It is an isosceles triangle.

$\angle BAC=\angle ACB$

$\Rightarrow \angle BAC+\angle BAC={50}^{\circ }$

$\Rightarrow 2\angle BAC={50}^{\circ }$

$\Rightarrow \angle BAC={25}^{\circ }=\angle ACB$

Sum of all angles of triangle$={180}^{\circ }$

$\therefore \angle ABC={180}^{\circ }-{50}^{\circ }={130}^{\circ }$