Applications of Trigonometry
Oblique or Scalene Triangle
The cosine law (rule) or the law of cosines
Calculating angles of an oblique triangle
Solving oblique triangles examples
Oblique or Scalene Triangle
From the congruence of triangles follows that an oblique triangle is determined by three of its parts, as are
- two sides and the included angle (SAS),
- two angles and the included side (ASA),
- three sides (SSS) and
- two sides and the angle opposite one of them (SSA), which does not always determine a unique triangle.
By using definitions of  trigonometric functions of an acute angle and Pythagoras’ theorem, we can examine mutual relationships of sides and corresponding angles of an oblique triangle.
The cosine law
 From the right triangles, BCD and ABD in the figure

Expressing the same way the squares of the altitudes as common legs of another two pairs of the right triangles, obtained are
As in every above formula included are all three sides and the angle that correspond to the side to be calculated, of the given oblique triangle, it follows that the cosine law can be used in the cases if known are,
- two sides and the included angle (SAS) and
- three sides (SSS).
Example:   Given is a triangle with sides, a = 64 cm, c = 29 cm and angle b = 147°, find the side b, and angles a and g.
Solution:  Given a = 64 cm, c = 29 cm and b = 147°.    b, a and g = ?   Applying the cosine law

 Then, using the sine law As  a + b + g = 180°,   g = 180° - (a + b) = 10°8′20″
Calculating angles of an oblique triangle
The formulas to calculate angles of an oblique triangle are derived from the cosine law, thus
Example:   In a triangle side, a = 17 cm, b = 7 cm and c = 13 cm, find angles a , b and g.
Solution:  Given a = 17 cm, b = 7 cm and c = 13 cm.  a , b and g = ?  Plug the given values into formulas
As  a + b + g = 180°    then   g = 180° - (a + b) = 180° - 135°14′30″ = 44°45′30″g = 44°45′30″.
Solving oblique triangles examples
Example:   From a point A on a river bank observer sees the top of an electric transmission tower at the angle of elevation j = 18°. Bottom C, of the tower, and points A and B, whose distance is 120 m, forms on the ground triangle with angles a = 75° and b = 62°, as is shown in the figure below.
Find the height h of the tower.
 Solution: From the triangle ABC by use of the sine law,

Example:   Given are sides, a, b and c of a triangle. Find the length of the median ta.
Solution:  Median ta is a half of the diagonal AD of the parallelogram ABCD
since a + b + g = 180°   =>   b + g = 180° - a according to
the cosine law,   (2ta)2 = b2 + c2 - 2bc · cos(180° - a)
then    4ta2 = b2 + c2 + 2bc · cos a,
from given triangle   a2 = b2 + c2 - 2bc · cos a,
and by adding these two equations,     4ta2 + a2 = 2(b2 + c2)
 obtained is,
Example:   An oblique triangle with side c = 5 cm and the angles on its ends, a = 22° and b = 125°, rotates around the given side. Determine the volume and the surface of the solid generated by the rotation.
Solution:  The volume of the solid of revolution equals the difference between the bigger cone of altitude c + and the smaller cone of the altitude h (a cone dent).
 Thus, the volume of the solid of revolution equals the  volume of the cone with base in the vertex B of the triangle and the height c. While, the surface of the solid of revolution equals the sum of the lateral surfaces of both cones. S = Slat.b. + Slat.a.   = rpb + rpa = rp(b + a)
Therefore, we need to calculate the slant height a and b (i.e., sides of the triangle) and radius r
Since  a + b + g = 180°   =>   g = 180° - (a + b),  and by using the sine law
 In the right DACD,
by plugging the value for r into the formula for the volume and the formula for the surface,
Example:   Base of a pyramid is an isosceles triangle whose equal sides are b and their corresponding (opposite) angles b (b > 4). All lateral edges of the pyramid are equal and make angle a with the base.
The pyramid is cut by the plane which passes through the altitude of the pyramid and the vertex of the angle b of the base, see the below figure. Determine the area of the section.
Solution:  The area of the section equals the area of triangle CDE, that is,  A = (1/2) · CD · h
Since the lateral edges are equal, the height h of the pyramid intersects the base at its circumcentre.
Thus, in right triangle AEF, the height h and radius r of the circumcircle are its legs, that is
 From the oblique triangle whose angles are CAD = 180° - 2b   and  DCA = 90° - b, ADC = 180° - [(180° - 2b) + (90° - b)] = 3b - 90°, using the sine law Let plug the calculated values into the formula for the section

Example:   The crane in the below figure is loaded by the force FL = 25000 N.  Crane arms are inclined at angles a = 38° and b = 57° to the horizontal. Find forces Fk1 and Fk2 in arms of the crane and force components, FH and FN in the point A.
Solution:  Angles in the triangle BCD,  angle BDC = 90° - b, angle DCB = b - a and angle CBD = 90° + a, thus, using the sine law
Example:   A ray of light in air is approaching the glass layer surface at the angle of an incidence a = 62°.
Find the parallel shift s of the ray when exiting the layer thick d = 15 mm, if the index of refraction for glass n = 1.75.
Solution:   It is experimentally proved that the ratio between the projection of the part of an incident ray and the part of a refracted ray, approaching the boundary of two different media, called the index of refraction is constant. n,
Geometry and use of trigonometry contents - B