In triangle ABC, BM is the median and BH is the height. It is known that AC = 76, HC = 19

In triangle ABC, BM is the median and BH is the height. It is known that AC = 76, HC = 19 and ∠ACB = 80. Find the corner AMB.

Since BM is the median of the triangle ABC, it divides the AC side into equal segments. AM = CM = AC / 2 = 76/2 = 38 cm.

The segment CH, according to the condition, is 19 cm, then MH = CM – CH = 38 – 19 = 19 cm.

Since the ВН height divides the base of the MС of the MВС triangle in half, the MВС triangle is isosceles, ВM = BC, and therefore the angle BMC = BCM = 80.

The angle AMB and CMB are adjacent angles, the sum of which is 180, then the angle AMB = 180 – 80 = 100.

Answer: Angle AMB is 100.