Question

Given an equilateral triangle ABC and a point O inside it, with $\angle BOC={99}^{\circ }$ and $\angle AOC={129}^{\circ }$, Let the largest angle of the triangle with side lengths equal to AO,BO and CO be $x^{\circ }$. Find x?

Answer 1

Since, the sum of all the angles around a point = ${360}^{\circ }$

$\therefore x^o+{99}^o+{129}^o={360}^o$

$x^o+22s^o={360}^o$

$\therefore x={360}^o-{228}^o$

$x={132}^o$