In triangle ABC angle C = 90, AB-12cm, AC = 5cm. Find the cosine and tangent of angle B.

Since the angle BCA = 90 in the ABC triangle, the ABC triangle is rectangular.

In a right-angled triangle, the sine of an acute angle is the ratio of the length of the opposite leg to the length of the hypotenuse.

SinABC = AC / AB = 5/12.

Determine the cosine of the angle ABC.

Cos2BAC = 1 – Sin2BAC = 1 – 25/144 = 119/144.

CosBAC = √119 / 12.

Determine the tangent of the angle BAC.

tgBAC = SinBAC / CosBAC = (5/12) / (√119 / 12) = 5 / √119.

Answer: The cosine of angle B is √119 / 12, the tangent of angle B is 5 / √119.