Applications of Trigonometry
Oblique or Scalene Triangle
Area of a triangle, the radius of the circumscribed circle and the radius of the inscribed circle
The radius of the circumscribed circle or circumcircle
Area of a triangle in terms of the inscribed circle or incircle
The radius of the inscribed circle
Oblique or scalene triangle examples
Oblique or Scalene Triangle
Area of a triangle, the radius of the circumscribed circle and the radius of the inscribed circle
Rectangular in the figure below is composed of two pairs of congruent right triangles formed by the given oblique triangle. Therefore, the area of a triangle equals the half of the rectangular area,
In the right triangles in the right diagram,
ha = b · sing,   hb = c · sina,   hc = a · sinb,
and by plugging into above formulas for the area
the area of a triangle in terms of an angle and the sides adjacent to it.
If, in the above formulas for the area, we substitute each side applying the sine law, that is
obtained is the area of a triangle in terms of a side and all its angles,
The radius of the circumscribed circle or circumcircle
Using known relation, which states that the angle subtended by a chord at the circumference is half the angle subtended at the center, from the right triangle in the below diagram follows,
the radius of the circumscribed circle, or
a = 2R · sina,   b = 2R · sinb,   c = 2R · sing.
Plugging the sides into  A = (1/2) ab sing  obtained is
 A = 2R2 · sina · sinb · sing
the area of a triangle in terms of the radius of circumcircle and angles.
If, in the above formula for the area which includes two adjacent sides and the angle between them, a given angle is substituted as follows
 obtained is the area of a triangle in terms of sides and the radius of the circumcircle.
Area of a triangle in terms of the inscribed circle (or incircle) radius
The oblique triangle ABC in the figure below consists of three triangles, ABO, BCO and ACO with the same altitude r therefore, its area can be written as
 where is the semi-perimeter
 then A = r · s the area of a triangle in terms of the
The parts of a triangle denoted as in the diagram relates as follows,
xA + xB = c,    xA + xC = b,    xB + xC = a  then,   2xA = - (xB + xC) + b + c = - a + b + c,
2xB = a - (xA + xC) + c a - b + c  and   2xC = a + b - (xA + xB) + b + c = a + b - c,
again using    a + b + c = 2s   it follows that   xA = s - a,    xB = s - b  and   xC = s - c.
Then, from the right triangles in the diagram,
Equating obtained formulas with the half-angle formulas, as for example
 or the radius of the inscribed circle.
Plugging given r into the formula for the area of a triangle  A = r · s  yields
 Heron's formula.
Oblique or scalene triangle examples
Example:   Determine the area of an isosceles triangle of which, the line segment that joints the midpoint of one of its equal sides by the midpoint of the base equals the half of the radius R of the circumcircle.
Solution:  In the similar triangles  ABC  and  CDE,    b/2 = R/2    =>   b = R.
 Area of the triangle
Equating obtained formula with the known formula for the area of a triangle in terms of the radius of the
 circumcircle
 so, the area of the triangle
Example:   Given is a triangle with sides, a, b and c, and angles, a, b and g. In the triangle inscribed is a triangle whose vertices lie in foots of the altitudes of the given triangle, as is shown in the figure down.
Determine sides, angles and the area of the inscribed triangle.
Solution:  Quadrilateral ODBF is cyclic (since the sum of the opposite angles is 180°) that is around which a circle can be circumscribed.
Thus, angle ODF = angle OBF = 90° - a since they are inscribed angles subtended by the same arc OF.
As well,    angle ODE = angle OCE = 90° - a     so    angle ODF + angle ODE = d = 180° - 2a
 Note that two angles with mutually perpendicular sides are equal. Also,   angle OFD = angle OBD = 90° - g since they are inscribed angles subtended by the same arc OD. As well,  angle OFE = angle EAO = 90° - g so  angle OFD + angle OFE = j = 180° - 2g. On a similar way can be proved that  e = 180° - 2b.
Triangles, BDF and ABC are similar since, angle DFB = 90° - angle OFD = 90° - (90° - g) = g therefore
Geometry and use of trigonometry contents - B