In triangle ABC, angle C is 90 degrees AB = 16 AC = 4√7 cos B -?

In the triangle ABC we find cos B, if it is known:

Angle C = 90 °;
AB = 16;
AC = 4√7.
Solution:

cos B = BC / AB;

BC – unknown, but other leg and hypotenuse are known.

Let’s apply the Pythagorean theorem.

AB ^ 2 = AC ^ 2 + BC ^ 2;

BC ^ 2 = AB ^ 2 – AC ^ 2;

BC = √ (AB ^ 2 – AC ^ 2) = √ (16 ^ 2 – (4√7) ^ 2) = √ (16 ^ 2 – 4 ^ 2 * 7) = √ (16 ^ 2 – 16 * 7 ) = √ (16 * (16 – 7)) = √16 * √9 = 4 * 3 = 12;

2) Now find cos B.

cos B = BC / AB = 12/16 = (3 * 4) / (4 * 4) = (3 * 1) / (4 * 1) = 3/4 = 0.75;

As a result, we got cos B = 0.75.

Answer: cos B = 0.75.