Question

If there is an isosceles triangle with at least one angle equal to ${60}^{\circ }$, then it must be a/ an ___triangle.

Answer 1

Case 1:
If $\angle A$ is ${60}^{\circ }$ and other two angles are equal.
$\angle A+\angle B+\angle C={180}^{\circ }$
${60}^{\circ }+x+x={180}^{\circ }$
$x=\frac{{180}^{\circ }-{60}^{\circ }}{2}={60}^{\circ }$
So, $\angle A=\angle B=\angle C={60}^{\circ }$
Hence it is an equilateral triangle.

Case 2:
If either one of the equal angles is ${60}^{\circ }$
$\angle B=\angle C={60}^{\circ }$ (Isosceles triangle)
$\angle A+\angle B+\angle C={180}^{\circ }$
$x+{60}^{\circ }+{60}^{\circ }={180}^{\circ }$
$x={180}^{\circ }-{60}^{\circ }-{60}^{\circ }={60}^{\circ }$
$\angle A=\angle B=\angle C={60}^{\circ }$
Hence it is an equilateral triangle.