What is the measure of the vertex angle of an isosceles triangle if one of its base angles measures 70 degrees and its congruent sides each measure 15 cm?

Answer 1

When finding the angles, the length of the sides is irrelevant in this case.

Let the triangle be ABC with ∠A the vertex and BC the base; the real question is whether you have the isosceles triangle "drawn" and labelled with the equal sides:

• either side of the "vertex" making the equal angles ∠B and ∠C

The equal sides are AB and AC; the base being the odd length means the angles at each end of it are the same, thus:

vertex_angle = 180o - 2 x 70o

= 40o

• the base and one side to the vertex equal (say sides AB and BC) and the other side different (AC) making the equal angles ∠A and ∠C

• 70o angle is between the sides of equal length (∠B):

The vertex is one of the two equal angles:

vertex_angle = (180o - 70o) ÷ 2

= 55o

• 70o angle is between the odd side and the base (∠C):

The vertex angle (∠A) is the same as the given angle (∠C), that is 70o

Isosceles triangles are often drawn in the first case, but it is not necessarily so!