The base of the pyramid is a right-angled triangle with an acute angle a (alpha). Two side faces of the pyramid containing the sides of this angle are perpendicular to the base plane, and the third is inclined to it at an angle of b (veta). The height of the pyramid is N. Find the volume of the pyramid.
Since the faces ACD and ABD are perpendicular to the plane of the base, the edge AD is perpendicular to the sides of AC and AB.
The angle ACB = 90, since the triangle is rectangular, then the angle ACD is the angle of inclination of the BCD face to the base of the pyramid. Angle АСD = β0.
Then in a right-angled triangle АСD tgβ = АД / АС.
AC = AD / tanβ = H / tanβ = H * ctgβ.
In a right-angled triangle ABC tgα = BC / AC.
ВС = АС * tgα = Н * ctgβ * tgα.
Determine the area of the triangle ABC.
Sас = АС * ВС / 2 = Н * ctgβ * Н * ctgβ * tanα / 2 = Н2 * ctg2β * tanα / 2 cm2.
Then V = Saс * АD / 3 = (Н2 * ctg2β * tgα / 2) * Н / 3 = Н3 * ctg2β * tanα / 6 cm3.
Answer: The volume of the pyramid is equal to H3 * ctg2β * tgα / 6 cm3.
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