In a right-angled triangle ABC, the angle is A = 90 degrees, AB = 20 cm, and the height AD is 12 cm. Find AC and CosC.

In a right-angled triangle ABC it is known:

Angle A = 90 °;
AB = 20 cm;
Height AD = 12 cm.
Find AC and Cos C.

Decision:

1) Consider triangle BAD with right angle D.

BD = √ (AB ^ 2 – AD ^ 2) = √ (20 ^ 2 – 12 ^ 2) = √ ((20 – 12) * (20 + 12)) = √ (8 * 32) = √8 * 8 * 4) = √ ((4 * 4) = √ (8 * 2) ^ 2 = √16 ^ 2 = 16;

2) sin b = AD / AB;

Substitute the known values and calculate the sine of angle B.

sin B = 12/20 = (4 * 3) / (4 * 5) = (1 * 3) / (1 * 5) = 3/5 = 0.6;

3) Since sin B = cos C, then cos C = 0.6.

4) sin C = √ (1 – 0.6 ^ 2) = √ (1 – 0.36) = √0.64 = 0.8;

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

5) tg C = AB / AC;

AC = AB / tg A = 20 / (4/3) = 20 * 3/4 = 20/4 * 3 = 5 * 3 = 15;

Answer: AC = 15 and tg C = 4/3.