In triangle ABC, angle C is 90 CH height AC 10 AH8 find cosB.

Consider the triangle ABC, <C = 90º, AB – hypotenuse, AC and CB – legs, CH – height lowered to the hypotenuse. AC = 10 cm, AH = 8 cm.
cos A = AC / AB
sin A = CB / AB
Consider the triangle ACN, <H = 90º, AC – hypotenuse, AH and CH – legs, AC = 10 cm, AH = 8 cm.
By the Pythagorean theorem:
CH = √ (10²-8²) = 6 cm.
cos A = AH / AC
sin A = CH / AC
Let’s return to the triangle ABC, equate the cosines and sines of angle A for triangles ABC and ASN:
AH / AC = AC / AB = cos A
Hence: AB = AC * AC / AH = 10 * 10/8 = 12.5 cm.
CH / AC = CB / AB = sin A
Hence: SV = CH * AB / AC = 6 * 12.5 / 10 = 7.5 cm.
cos B = CB / AB = 7.5 / 12.5 = 0.6
Answer: cos B = 0.6.