From a point to this plane, two inclined 8 cm long were drawn; these oblique form angles of 30 degrees with the given plane. Calculate the distance between the ends of the inclined ones if the angle between the projections of the inclined ones onto this plane is 120 degrees.
If those inclined to the plane are equal, then their projections onto the plane are also equal. Then OС = OB, since AB = AC = 8 cm by condition.
Consider a right-angled triangle AOB and determine the length of the leg OB.
CosB = OB / AB.
OB = CosB * OB = (√3 / 2) * 8 = 4 * √3.
ОВ = ОВ = 4 * √3.
In an isosceles triangle BOС, draw the height OH to the base of the BC, which in an isosceles triangle is the bisector and median, then BH = CH.
In a right-angled triangle OCH, the angle СOН = 120/2 = 60.
Then SinCOH = CH / OC.
CH = SinCOH * OC = (√3 / 2) * 4 * √3 = 6 cm.
Then BC = 2 * CH = 2 * 6 = 12 cm.
Answer: The distance between the ends of the slopes is 12 m.
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