Applications of Trigonometry
Oblique or Scalene Triangle The sine law (rule) or the law of sines Solving oblique triangles - use of the sine law and the cosine law
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 sine law
From the right triangles, ACD and BCD in the figure,
hc = a sinb   and   hc = b sina,
so that,               a sinb = b sina,
 therefore,  Expressing the same way the altitudes, hb and ha as common legs of another pairs of two right triangles in the given triangle, we get
hb = a sing   and   hb = c sina,
so that,            a sing = c sina,
 or and,     ha = b sing   and   ha = c sinb,
so that,            b sing = c sinb,
 or These relations are called the sine law and in words:
Sides of a triangle are to one another in the same ratio as sine of the corresponding (opposite) angles.
As these ratios express relations of any of two sides and their opposite angles, it follows that the sine law can be applied to solve an oblique triangle in the cases when given are,
- two angles and the included side (ASA) and
- two sides and the angle opposite one of them (SSA), that does not specify a triangle uniquely.
Example:   In the oblique triangle ABC side a = 6 cm, angles, a = 38° and g = 120°, find the remaining sides b and c and angle b.
Solution:  Given a = 6 cm, a = 38° and g = 120°.    b, c and b = ?
As  a + b + g = 180°    then     b = 180° - (a + g) = 180° - 158° = 22°,  and since  Example:   In the oblique triangle with sides, b = 7 cm and c = 4 cm and angle b = 115°, find the side a and angles, a and g.
Solution:  Given b = 7 cm, c = 4 cm and b = 115°.    a, a and g = ?
Sequence of calculating the particular part of the triangle depends of the given parts, that is, use the ratio that includes all three known parts and the one needed to be calculated, so   Example:   In the triangle with sides, b = 7 cm and c = 4 cm and angle g = 31°11′27″, find the side a and angles, a and b.
Solution:  Given b = 7 cm, c = 4 cm and g = 31°11′27″.    a, a and b = ?
As shows the below figure it is possible construct two triangles with given parts, one acute, the other obtuse.
That is, the problem has two solutions, one triangle with angle b and the other with angle b = 180° - b.  a and a′ from, a + b + g = 180° and a′ + b′ + g = 180°, so,  a = 180° - (b + g) = 180° - 96°11′27″ = 83°48′33″
and   a = 180° - (b′ + g) = 180° - (180° - b + g) = b - g = 65° - 31°11′27″ = 33°48′33″. Solving the oblique triangle - use of the sine law and the cosine law Example:   From the distance d = 180 m observer sees the top of a castle at the angle of elevation a = 39° and its bottom at angle b = 36°. Find the height h of the castle, see the figure below.
 Solution:   a) From the right triangle ABC, and from the oblique triangle ACD using the sine law  b) The same problem can be solved using only a right triangles. Comparing solutions for h obtained in a) and b) it follows,  which is known difference formula for sine.
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:   Two parallel sides of a trapezoid a = 12 cm and c = 5 cm, and side b = 9 cm, which makes angle b = 63° with the side a, see the figure below. Find the length of the fourth side d, remaining angles, and the area of the trapezoid.
Solution:  In triangle ADE,      d2 = (a - c)2 + b2 - 2(a - c) · b · cosb
 d2 = 72 + 92 - 2 · 7 · 9 · cos63°,    d = Ö72.8  = 8.53 cm. since     b + g = 180°  =>   g = 180° - b = 117° and       a + d = 180°  =>   d = 180° - a = 109°58′15″.  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 for the surface,    Geometry and use of trigonometry contents - B 