Find the side of an isosceles triangle if its base is 16 and the base angle is 30 degrees.

Let ABC be an isosceles triangle given by the condition: AB = BC, AC = 16 cm, angle A = angle C = 30 degrees.
From the top B we will draw the height ВН to the base of the AC. Since ABC is isosceles, then BH is both height and median, then:
HA = HC = AC / 2 = 16/2 = 8 (cm).
The ВН height divides the ABC triangle into two equal triangles ВНA and ВНС.
Consider the BHA triangle: BHA angle = 90 degrees (since BH is the height), HAB angle = 30 degrees, AB — hypotenuse, BH and HA = 8 cm — legs.
Since the angle HAB = 30 degrees, the leg ВН is equal to half of the hypotenuse AB:
ВН = AB / 2.
By the Pythagorean theorem:
AB ^ 2 = BH ^ 2 + HA ^ 2.
We replace ВН with AB / 2, and instead of HA we substitute the value HA = 8 cm:
AB ^ 2 = (AB / 2) ^ 2 + 8 ^ 2;
AB ^ 2 = AB ^ 2/4 + 64;
AB ^ 2 = (AB ^ 2 + 256) / 4;
4AB ^ 2 = AB ^ 2 + 256;
4AB ^ 2 – AB ^ 2 = 256;
3AB ^ 2 = 256;
AB ^ 2 = 256/3;
AB = √ (256/3);
AB = 16 / √3 = 16√3 / 3 (cm).
Answer: AB = BC = 16√3 / 3 cm.