In the triangle ABC, the median AM is four times smaller than the side AB and makes an angle of 60

In the triangle ABC, the median AM is four times smaller than the side AB and makes an angle of 60 degrees with it. Find the angle MAC.

The median splits a triangle into two triangles with equal areas.

Therefore, Sabm = Samc (1), Sabc = 2 * Sabm (2).

Let AB = x, then AM = x / 4.

From (1) it follows:

½ * AB * AM * sin60 = ½ * AC * AM * sinα, where α = MAC angle.

½ * x * x / 4 * √3 / 2 = ½ * AC * x / 4 * sinα.

AC = (√3 * x) / (2 * sinα).

From (2) it follows:

½ * AB * AC * sin (60 + α) = 2 * ½ * AB * AM * sin60.

AC * sin (60 + α) = 2 * AM * sin60.

(√3 * x) / (2 * sinα) * sin (60 + α) = x / 2 * √3 / 2.

1 / sinα * sin (60 + α) = 1/2.

2 * sin (60 + α) – sinα = 0.

2 * sin60 * cosα + 2 * sinα * cos60 – sinα = 0.

2 * √3 / 2 * cosα + sinα – sinα = 0.

√3 * cosα = 0.

α = angle MAC = 90º.