In the triangle ABC AC = BC, AB = 5, BH – height, AH = 3. Find cos B.

Since ВН is the height of the ABC triangle, the ABН triangle is rectangular.

In a right-angled triangle, the cosine of an acute angle is the ratio of the adjacent leg to the hypotenuse.

CosBAH = AH / AB = 3/5.

Since, by condition, AC = BC, then the triangle ABC is isosceles, and then the angle CAB = CBA, and hence CosCBA = CosCAB = 3/5.

Answer: The cosine of angle B is 3/5.

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