How do you find the perimeter of an isosceles triangle with base of 8 and vertex angle of 20°?

Feb 10, 2016

The perimeter is 54.07

Explanation:

I am not entirely sure of what you are allowed to use to answer this problem (presumably you can use a calculator because I did not find an easier way to answer this).

Basically, trigonometry.

First step is to find the base angles. Since a triangle has 180 degrees and the vertex is 20, the 2 equal angles together are 160 degrees, thus each base angle is 80 degrees.

Since the triangle is isosceles, we can draw an altitude/angle bisector/median from the vertex angle down to the base. Since it acts as an angle bisector, the two vertex becomes two 10 degree angles.

Combine that with the fact that it is an altitude and we have formed two 10-80-90 triangles.

Combine that fact with the fact that it is a median and the base is bisected to a length of 4 for each of the smaller triangles.

We can use trigonometry to now find the hypotenuse of the small triangle.
Using the 80 degree angle, we can say that cos 80 = $\frac{a \mathrm{dj}}{h y p}$
Therefore cos 80 = $\frac{4}{h y p}$

Rearrange to get hypotenuse alone:
$h y p$ = 4/(cos (80)

Plug that into your calculator to get that the hypotenuse is 23.035

This is the same as the other hypotenuse because they are congruent sides

Add up all the sides to get 23.035 + 23.035 + 8 = 54.07