In triangle ABC, AB = BC, and the height AH divides the side BC into segments BH = 64 and CH = 16. Find cos∠B.

In triangle ABC it is known:

AB = BC;
The height AH divides the BC side into segments;
BH = 64;
CH = 16.
Find cos ∠B.

Solution:

1) BC = BH + CH = 64 + 16 = 70 + 10 = 80;

2) Since the ABC triangle is isosceles, then:

AB = BC = 80;

3) Calculate the height of AN.

AH ^ 2 = BH * CH;

AH = √ (BH * CH) = √ (64 * 16) = √64 * √16 = 8 * 4 = 32;

3) Consider a triangle ACN with a right angle H.

AC = √ (32 ^ 2 + 16 ^ 2) = √ (1024 + 256) = √1280 = √ (256 * 5) = 16√5;

4) Theorem of cosines.

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * cos b;

16 * 16 * 5 = 80 * 80 + 80 * 80 – 2 * 80 * 80 * cos b;

16 * 80 = 80 * 80 + 80 * 80 – 2 * 80 * 80 * cos b;

16 = 80 + 80 – 2 * 80 * cos b;

1 = 5 + 5 – 2 * 5 * cos b;

1 – 10 = -10 * cos b;

cos b = -9 / (- 10);

cos b = 0.9.