In a triangle ABC AC = AB = 10cm, angle A = 30 degrees, BK is perpendicular to the plane

In a triangle ABC AC = AB = 10cm, angle A = 30 degrees, ВK is perpendicular to the plane of the triangle and is equal to 5√6cm. find the distance from point K to AC

From vertex B of triangle ABC, let us lower the height to the side of AC.

In the formed right-angled triangle АНВ, the angle H is straight, and the angle A = 30 according to the condition.

The ВН leg lies opposite an angle of 30 degrees, and, accordingly, is equal to half the length of the AB hypotenuse.

ВН = AB / 2 = 10/2 = 5 cm.

Consider a right-angled triangle ВНК, whose angle B is a straight line by condition.

Then, by the Pythagorean theorem, KH ^ 2 = KB ^ 2 + BH ^ 2 = (5 * √6) ^ 2 + 5 ^ 2 = 150 + 25 = 175.

KН = 5 * √7 cm.

Answer: The distance from point K to AC is 5 * √7 cm.