Question

Consider the figure:

If BC = AC and angle ACB = 60^o, then find the value of $\angle$A + $\angle$B + $\angle$x + $\angle$y .

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Answer 1

Given $\angle$C = 60^o and BC = AC,
Therefore $\angle A=\angle B$ (base angles are equal in an isosceles triangle)
$\therefore$ by angle sum property of triangle, $\angle A=\angle B={60}^o$

We know that, $\angle$x +$\angle$y = $\angle$A +$\angle$B (Exterior angle property)

$\therefore \angle A+\angle B+\angle x+\angle y={60}^{\circ }+{60}^{\circ }+{120}^{\circ }={240}^{\circ }$