In triangle ABC, the angle ∠C is 90 °, AC = 4, cos ∠B = 0.6. Find BC.

Given:

ABC – right triangle;

Angle C = 90 °;

AC = 4;

cos b = 0.6;

Find BC.

Solution:

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

2) tg b = sin b / cos b = 0.8 / 0.6 = 8/6 = (4 * 2) / (2 * 3) = 4/3;

3) tg b = BC / AC;

From here we express BC.

BC = AC * tg b;

Substitute the known values into the formula and calculate the BC leg.

BC = 4 * 4/3 = 16/3.

Answer: BC = 16/3.