In triangle ABC, angle C = 90, height CD = 12, AC = 20. find leg BC.

In the triangle ABC we find the leg ВС, if it is known:

Angle C = 90 °,
Height CD = 12;
AC = 20.
Solution:

1) Consider triangle ADC. The triangle is a rectangle because the height is perpendicular to the hypotenuse AB. Hence, the angle D = 90 °.

2) If the hypotenuse AC and leg CD are known in the triangle ADC, then you can find sin A.

sin A = CD / AC = 12/20 = (3 * 4) / (4 * 5) = (3 * 1) / (1 * 5) = 3/5 = 0.6;

3) Consider the triangle ABC.

cos A = √ (1 – sin ^ 2 a) = √ (1 – 0.6 ^ 2) = √ (1 – 0.36) = √0.64 = 0.8;

tg a = sin a / cos a = 0.6 / 0.8 = 6/8 = 3/4;

Then:

tg a = BC / AC;

Hence:

BC = tg a * AC = 3/4 * 20 = 20/4 * 3 = 5 * 3 = 15.

Answer: BC = 15.