The height CM of triangle ABC divides its side AB into segments AM and BM. Find side BC

The height CM of triangle ABC divides its side AB into segments AM and BM. Find side BC, if AM is 15 cm, BM is 5 cm, angle A is 30 degrees.

Since CM is the height, we have two right-angled triangles: CMA and CMB.

In the SMA triangle: CM – leg opposite to angle A, AM – adjacent. The ratio of the opposite leg to the adjacent leg is the tangent of the angle. Consequently:

tg A = CM / AM;

CM = AM * tg A = 15 * tan 30 ° = 15 * √3 / 3 = 5√3 cm.

In the triangle CMB CM and BM – legs, BC – hypotenuse. The sum of the squares of the legs is equal to the square of the hypotenuse, therefore:

BC ^ 2 = CM ^ 2 + BM ^ 2 = (5√3) ^ 2 + 5 ^ 2 = 25 * 3 + 25 = 100;

BC = √100 = 10 cm.