From a point outside the plane, two inclined ones are drawn to the plane at an angle of 60 ° and 45 ° to the plane.

From a point outside the plane, two inclined ones are drawn to the plane at an angle of 60 ° and 45 ° to the plane. The length of the first inclined 12√2 cm. Find the length of the second inclined.

Given:
angle KAO = 60
angle RCM = 45
AK = 12√2cm
Find: AM-?
Solution:
Triangle AOK is rectangular.
angle O = 90.
angle K = 180-angle O-angle A = 180-90-60 = 30.
Let’s apply the property that the leg, which lies opposite the angle of 30 degrees, is equal to half of the hypotenuse.
AO = 1/2 * AK = (1/2) * 12√2 = 6√2cm.
The AOM triangle is rectangular.
angle O = 90.
angle M = 180-angle O-angle A = 180-90-45 = 45.
angle M = angle A = 45, which means the triangle AOM is rectangular and isosceles.
Then, AO = ОМ = 6√2cm.
We apply the Pythagorean theorem and find the hypotenuse AM:
AM ^ 2 = AO ^ 2 + OM ^ 2 = (6√2) ^ 2 + (6√2) ^ 2 = 36 * 2 + 36 * 2 = 72 + 72 = 144.
AM = √144 = 12cm.
Answer: AM = 12cm.