In a right-angled triangle ABC, the angle C = 90, the height CD is 12 cm and cuts off the segment AD equal to 16 cm

In a right-angled triangle ABC, the angle C = 90, the height CD is 12 cm and cuts off the segment AD equal to 16 cm from the hypotenuse. Find BC, sine and cosine of angle B.

In the right-angled triangle ACD, we find the hypotenuse of the AC:
AC = √ (AD² + CD²) = √ (256 + 144) = √400 = 20 (cm).
Consider two right-angled triangles ABC and ACD, they are similar (acute angle A is common).
From the similarity with a triangle, we write down the aspect ratio:
AB / AC = AC / AD → AB = AC * AC / AD = 20 * 20/16 = 25 (cm).
In the right-angled triangle ABC we find the leg BC:
ВС = √ (AB² – AC²) = √ (625 – 400) = √225 = 15 (cm).
Let’s use the definitions and find the sine and cosine of the angle B.
Sin B = AC / AB = 20/25 = 0.8.
Cos B = BC / AB = 15/25 = 0.6.
Answer: BC – 15 cm, sin B = 0.8, cos B = 0.6.