Why is it impossible to have an obtuse equilateral triangle?

Answer 1

This is not a formal proof, but should sufficiently show how to go about proving it.

In a triangle, there can be no more than one obtuse angle, because the sum of the angles equal 180°. If angle A is obtuse (greater than 90°), then the sum of the other two angles (B + C) must be less than 90°, so each angle B & C must be less than 90°.

An equilateral triangle has 3 equal sides. From the Law of Sines, the sines of the angles (opposite a side) are in the same proportion as the proportions of the lengths of sides.

Example: if angle A is opposite side a, and angle B is opposite side b. then:

• a : b = sinA : sinB, or rearranging: a/sinA = b/sinB.

Since the triangle is equilateral, then a = b, and sinA = sinB. Let's assume we can get an equilateral triangle which has one obtuse angle. So assume angle A is obtuse. Angle B does not have to equal angle A, because there are two angles which have the same sine: sin(A) = sin(180°-A). Since A is obtuse, neither B or C can be obtuse (from the first paragraph), so angle B cannot equal A, so it must equal (180°-A).

Let's find angle C: Sum up the angles: A + B + C = 180°. Substitute (180°-A) for B: A + (180°-A) + C = 180°, so we find that angle C = 0°, and it is not a triangle. But let's go on. We know that, since it's equilateral, then side c equals a & b, so sin(C) must equal sin(A) and sin(B). Since A is obtuse, then B & C cannot be obtuse, so angles B & C are equal, which means that angle B is also 0°, and angle A must be 180° to satisfy the sum of the angles equal 180°. So we now have a straight line, and the Law of Sines doesn't hold anymore since we have 0 : 0 : 0 {Sin(180°) : Sin(0°) : Sin(0°)}, and you cannot have a ratio with all zeros and be meaningful.