The height AM of the triangle ABC divides the side BC into segments BM

| September 25, 2021 | education

The height AM of the triangle ABC divides the side BC into segments BM and MС. Find the sides of the triangle if AM = 3cm, angle B = 60, angle C = 45

Triangles AMC and AMB are rectangular, since they are formed by the height AM of triangle ABC.

In a right-angled triangle AFM, one of the acute angles is 45, then the triangle AFM is rectangular and isosceles, CM = AM = 3 cm.Then, by the Pythagorean theorem, AC ^ 2 = AM ^ 2 + CM ^ 2 = 9 + 9 = 18.

AC = 2 * √2 cm.

In a right-angled triangle ABM Sin60 = AM / AB, then AB = AM / Sin60 = 3 / (√3 / 2) = 6 / √3 = 2 * √3 cm.

Angle BAM = (90 – 60) = 30, then the leg BM located opposite it is equal to half AB.

BM = AB / 2 = 2 * √3 / 2 = √3 cm.

BC = CM + BM = 3 + √3 cm.

The lengths of the sides of the triangle are equal: 2 * √2 cm, 2 * √3 cm, 3 + √3 cm.



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