In triangle ABC C = 90 CH-height angle A = 30 AB = 22 find AH

In a right-angled triangle ABC, the CB leg lies opposite an angle of 30, then BC = AB / 2 = 22/2 = 11 cm.

Then AC ^ 2 = AB ^ 2 – CB ^ 2 = 22 ^ 2 – 11 ^ 2 = 484 – 121 = 363.

Let the length of the segment BH = X cm, then the length of the segment AH = (22 – X) cm.

Let us express the height of CH from two right-angled triangles ACH and BCH.

CH ^ 2 = AC ^ 2 – AH ^ 2 = 363 – (22 – X) ^ 2 = 363 – 484 + 44 * X – X ^ 2 = 44 * X – X2 – 121.

CH ^ 2 = BC ^ 2 – BH ^ 2 = 121 – X ^ 2.

Let’s equate both equalities.

44 * X – X ^ 2 – 121 = 121 – X ^ 2.

44 * X = 242.

X = 242/44 = 5.5 cm.

AH = 22 – 5.5 = 16.5 cm.

Answer: The length of the segment AH is 16.5 cm.