In the triangle ABC, BC = 6cm, the angle ACB = 120 degrees, the line BM

In the triangle ABC, BC = 6cm, the angle ACB = 120 degrees, the line BM is perpendicular to the plane (ABC). Find the distance from point M to straight line AC, if BM = 3cm.

Let’s build the height BH to the AC side of the ABC triangle. Since the angle of ACB is obtuse, the height of the AC will drop to the continuation of the AC side.

The angle BCH is adjacent to the angle ACB, the sum of which is 180, then the angle BCH = 180 – 120 = 60.

In a right-angled triangle BCH Sin60 = BH / BC.

BH = BC * Sin60 = 6 * √3 / 2 = 3 * √3 cm.

The height BH is the projection of the inclined MH onto the plane of the triangle, and since BH is perpendicular to the AC, then the MH is perpendicular to the AC.

In a right-angled triangle MBN, according to the Pythagorean theorem, ML ^ 2 = BH ^ 2 + BM ^ 2 = 27 + 9 = 36.

MH = 6 cm.

Answer: From point M to straight line AC 6 cm.

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