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

In a triangle ABC AC = CB = 10cm, angle A = 30 degrees, BK is the perpendicular at the plane of the triangle and is 5 cm. Find the distance from K to AC.

ΔМКВ – rectangular with a right angle B. КМ will be sought by the Pythagorean theorem.

KM ^ 2 = KB ^ 2 + MB ^ 2

ВM is perpendicular to AS – according to the three perpendicular theorem.

We find the MВ from ΔAMВ.

MВ = AB * sin 30.

AB = 2 * AO (CO – median, bisector, isosceles height ΔABС)

AO = AC * cos 30 = 10 * √3 / 2 = 5√3 (cm).

AB = 2 * 5√3 = 10√3 (cm).

MВ = 10√3 * sin 30 = 10√3 * ½ = 5√3 (cm)

MK ^ 2 = 5 ^ 2 + (5√3) ^ 2 = 25 + 25 * 3 = 25 + 75 = 100.

MK = √100 = 10 (cm).

Answer. 10 cm.