In triangle ABC, angle C = 90. AB = 17, tgA = 5/3. Find the height CH

Knowing that tg A = 5/3, we can find cos A and sin A:

1 + tg2A = 1 / cos2A;

1 / cos2A = 1 + 25/9 = 34/9;

cos2A = 1 / (34/9) = 9/34;

cos A = 3 / √34;

cos2A + sin2A = 1;

sin2A = 1 – cos2A = 1 – 9/34 = 25/34;

sin A = 5 / √34.

For triangle ABC: AB is the hypotenuse, AC is the leg adjacent to the angle A. The ratio of the adjacent leg to the hypotenuse is equal to the cosine of the angle, which means:

cos A = AC / AB;

AC = AB * cos A = 17 * 3 / √34 = 51 / √34.

For triangle ANS: CH and AN – legs, AC – hypotenuse. CH – leg opposite to angle A. The ratio of the opposite leg to the hypotenuse is equal to the sine of the angle:

sin A = CH / AC;

CH = AC * sin A = (51 / √34) * (5 / √34) = 51 * 5/34 = 255/34 = 7.5 is the desired height.