The angle opposite the base of an isosceles triangle is 120 °. The height drawn to the side is 8 cm.

The angle opposite the base of an isosceles triangle is 120 °. The height drawn to the side is 8 cm. Find the base of this triangle.

Let ABC be an isosceles triangle given by condition, AC — base, AB and BC — lateral sides. Since the angle B is equal to 120 degrees, ABC is an obtuse triangle and the height AD, drawn from the vertex A to the side of BC, will lie on the continuation of the straight line CB.
Angles ABD and ABC are adjacent, therefore:
angle ABD + angle ABC = 180 degrees;
angle ABD + 120 = 180;
angle ABD = 180 – 120 = 60 (degrees).
In triangle ABD, angle ABD = degrees, angle ADB = 90 degrees. DAB angle = 180 – 90 – 60 = 30 (degrees).
The leg BD of triangle ABD lies against an angle of 30 degrees, so BD = AB / 2. By the Pythagorean theorem:
AB ^ 2 = BD ^ 2 + AD ^ 2;
AB ^ 2 = (AB / 2) ^ 2 + 8 ^ 2;
AB ^ 2 = AB ^ 2/4 + 64;
AB ^ 2 = (AB ^ 2 + 256) / 4;
4AB ^ 2 = AB ^ 2 + 256;
3AB ^ 2 = 256;
AB ^ 2 = 256/3;
AB = √256 / 3 = 16 / √3 = 16√3 / 3 (cm).
Then the leg BD is equal to:
BD = AB / 2 = (16√3 / 3) / 2 = 16√3 / 6 = 8√3 / 3 (cm).
Since the triangle ABC is isosceles, then BC = AB = 16√3 / 3 cm.Let’s find the length DC:
DC = BD + BC;
DC = 8√3 / 3 + 16√3 / 3 = (8√3 + 16√3) / 3 = 24√3 / 3 = 8√3 (cm).
In triangle ADC, we find the hypotenuse AC, which is the base of the isosceles triangle ABC, given by the condition:
AC = √ (AD ^ 2 + DC ^ 2);
AC = √ (8 ^ 2 + (8√3) ^ 2) = √ (64 + 64 * 3) = √ (64 + 192) = √256 = 16 (cm).
Answer: AC = 16 cm.