In triangle ABC, angle C is straight. Find BC if cos of angle B = √2 / 2 and AB = 14.

It is known:

ABC – right triangle;

∠ C = 90 °;

cos ∠B = √2 / 2;

Hypotenuse AB = 14.

Find the BC leg of the ABC triangle.

1) Find sin B by the formula sin ^ 2 B + cos ^ 2 B = 1;

sin ^ 2 B = 1 – cos ^ 2 B;

sin B = √ (1 – cos ^ 2 B);

Substitute the known values and calculate sin B.

sin B = √ (1 – (√2 / 2) ^ 2) = √ (1 – 2/4) = √ (1 – 1/2) = √ (1/2) = 1 / √2 = 1 * √ 2 / (√2 * √2) = √2 / √4 = √2 / 2;

2) sin B = BC / AB (the ratio of the opposite leg to the hypotenuse).

Let us express from this the formula of the leg.

BC = AB * sin B = 14 * √2 / 2 = 14/2 * √2 = 7√2;

Answer: BC = 7√2.