In a regular triangular pyramid DABC, the height DH is equal to the side of the base.

| April 25, 2021 | education

In a regular triangular pyramid DABC, the height DH is equal to the side of the base. Point K is the middle of the side edge DA. Find the angle of inclination of the straight line KH to the plane of the base of the pyramid.

Let the height DH of the pyramid be equal to X cm, then, by condition, AB = BC = AC = X cm.

At the base of the pyramid is an equilateral triangle, then its height is AM = X * √3 / 2 cm.

Point H divides the segment AM in the ratio of 2/1, then AH = 2 * HM.

AH + HM = X * √3 / 2 = 3 * HM.

HM = X * √3 / 6 cm.

AH = X * √3 / 2 – X * √3 / 6 = X * √3 / 3 cm.

From point K we draw a perpendicular KP. Since point K is the middle of AD, then KP is the middle line of the triangle ADH, then KP = DH / 2 = X / 2, PH = AH / 2 = X * √3 / 6.

In a right-angled triangle KPH tgPHK = KP / PH = (X / 2) / (X * √3 / 6) = 6/2 * √3 = √3.

Angle AHK = arctg√3 = 60.

Answer: The tilt angle is 60.



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