Applications of trigonometry in solid geometry, Oblique prisms, Oblique pyramids

Applications of Trigonometry

Applications of Trigonometry in Solid Geometry

Oblique prism

As an example of an oblique prism consider a parallelepiped whose lengths of the base edges and the lateral edges sharing a common vertex are, a, b and c. Edges, a and b of bases, are perpendicular to each while the lateral edge c, forms the angle a with each of them, see the figure below. other,

Determine the volume, the surface area of the parallelepiped and the angle lateral edge to base.

The volume V = B · h = a · b · h.

The altitude of the parallelepiped is the intersection

line of the planes, A₁EN and A₁FN which are

perpendicular to the base edges, a and b and thus to

the plane of base.

Created triangles A₁AE and A₁AF are congruent as

they share the hypotenuse and both have the same

angle a.

Thus, A₁E = A₁F and therefore triangles, A₁EN and

A₁FN are congruent, so that EN = FN. AN is the bisector of the right angle DAB.

As by hypothesis, the base edges at the vertex A are perpendicular and the lateral edge c forms equal angles a with the base edges, then must be a > 45°, or 2a > 90°, and therefore, - cos2a > 0.

Note that, if we follow changes (decrease) of the angle a while the height h tends to zero, finally, the end points A₁ and N will coincide, and the lateral edge c equals its orthogonal projection AN (bisector of the right angle, 2a = 90°).

So the volume of the parallelepiped

Surface area S = 2B + S_lat. = 2ab + 2ac · sina + 2bc · sina = 2 · [ab + c · (a + b) · sina].

Inclination of the lateral edge to base,

Oblique Pyramid

As an example we consider the pyramid with square base. Two opposite faces are isosceles triangles one of which forms with the base interior angle b, and the other forms with the plane of the base an exterior angle a. The height of the pyramid is h.

Determine the volume of the pyramid and angles that form two other lateral faces with the plane of the base.

The plane VFG (F and G are midpoints of sides of

base) is perpendicular to the plane of the base and

passes through the height h of the pyramid.

Since a is an exterior angle of the triangle VFG, the

foot N of the altitude h falls at the extension of the

side FG.

The volume of the pyramid

The angle of inclination of the lateral sides VAB and VCD to the base we determine from the triangle VEN which is perpendicular to the plane of base, and whose leg is parallel with the edge AD of the base, thus

Example: Bases of a triangular prism are isosceles triangles, ABC and A₁B₁C₁, with equal sides AB = AC and two equal angles a. The normal projection of the vertex of the upper base falls at the center of incircle of the radius r of the lower base. Through the base edge AC and the vertex B₁ laid is a plane inclined to the lower base at angle a. See the figure below.

Determine the surface area of the pyramid ABCB₁ cut off from the prism and the volume of the prism.

Solution: The section plane ACB₁ cut off from the prism the

pyramid ABCB₁ the foot of the height of which is at the center

O of incircle, so that all lateral faces are inclined to the base

at equal angles a.

Therefore, we use the following formula for the surface area

Example: Base of a pyramid is a right triangle. The pyramid’s altitude foot coincides with the intersection point of the hypotenuse and the bisector of the right angle. The lateral edge at the right angle is inclined to the base at an angle a, see the figure below. Determine volume of the pyramid and the lateral faces to base dihedral angles if given is the right angle’s bisector t_c which with hypotenuse forms an angle of 45° + a.

Solution: The volume of the pyramid

In DNBC the angle BCN = 45°

and angle NBC = 180° - [ 45° + a + 45°] = 90° - a,

according to the sine law

Triangles NEC and NCD are congruent. The lateral faces to base angles, NEV = NDV = j .

They are the corresponding angles of two congruent triangles, VNE and VND.

Geometry and use of trigonometry contents - B