In an acute-angled triangle ABC, the height AH is 19√21 and the side of AB is 95, find cos B.

It is known:

ABC – acute-angled triangle;

AH is the height of the triangle;

AH = 19√21;

AB = 95.

Find cos B.

1) The height AH is perpendicular to the BC side.

2) Consider the triangle AНВ. The triangle is rectangular. Angle H = 90 °.

3) Find the ВН leg of the AВН triangle.

BH = √ (AB ^ 2 – AH ^ 2) = √ (95 ^ 2 – (19√21) ^ 2) = √ (95 * 95 – 19 * 19 * 21) = √ (19 ^ 2 * (5 * 5 – 21)) = √ (19 ^ 2 * 4) = 19 * 2 = 38;

3) Find cos B.

cos B = BH / AB (the ratio of the adjacent leg to the hypotenuse of the ABH triangle);

cos B = 38/95 = 1 / (2.5) = 1 / (5/2) = 2/5 = 0.4;

Answer: cos B = 0.4.