# How many obtuse angles can an equilateral triangle have?

Nov 24, 2015

None. An equilateral triangle has three equal sides, and thus three equal angles.

We can prove this using the law of cosines with the SSS case.

$a = b = c$

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \angle C$

becomes

${a}^{2} = {a}^{2} + {a}^{2} - 2 a \cdot a \cdot \cos \angle A$

$- {a}^{2} = - 2 {a}^{2} \cos \angle A$

$1 = 2 \cos \angle A$

$\frac{1}{2} = \cos \angle A$

$\textcolor{b l u e}{\angle A = {60}^{\circ}}$

Since only one side $a$ corresponds to only one $\angle A$, and since sides $a = b = c$, we have $\angle A = \angle B = \angle C$.

Acute angles cannot be greater than ${90}^{\circ}$, and obtuse angles cannot be less than ${90}^{\circ}$. All triangles can only have ${\theta}_{\text{tot}} = {180}^{\circ}$.

Therefore, with three acute angles, there are no obtuse angles.