Triangles are classified on the basis of angles as acute, right and obtuse triangles. Right triangles are used to define trigonometric ratios and apply them in solving a right triangle. As the name "Trigonometry" is a Greek word meaning measurement of triangles, the study of trigonometry cannot be restricted to right triangles alone. The study of Oblique triangles done on the relationships of angles and sides of a non right triangle takes care of this and a range of formulas exist for measuring solving non right triangles.

An Oblique triangle is a non right triangle, meaning it is either an acute or obtuse triangle.

Oblique Triangle Definition

Formulas are derived establishing the relationships between the lengths of sides and trigonometric functions of angles.

These formulas are used to solve Oblique triangles using the measures given. Formula are derived to find the area of a triangle in terms of the sides and trigonometric functions of angles are useful in finding the area of oblique triangle.
ΔABC is any triangle with sides a, b and c representing the measures of the sides opposite to the angles A, B and C respectively.

Oblique Triangle Formula
Formulas for find the area of ΔABC.

1. Heron's formula:
Area of ΔABC Δ = $\sqrt{s(s-a)(s-b)(s-c)}$ where s = $\frac{a+b+c}{2}$ is the semi perimeter of ΔABC.

2. Formulas using two sides and the included angle
Area of ΔABC Δ = $\frac{1}{2}$ ab Sin C

= $\frac{1}{2}$ bc Sin A

= $\frac{1}{2}$ ca Sin B

Formulas used to solve oblique triangles

1. Law of sines
$\frac{a}{SinA}$ = $\frac{b}{SinB}$ = $\frac{c}{SinC}$

2. Law of Cosines
a2 = b2 + c2 - 2bc Cos A
b2 = c2 + a2 - 2ca Cos B
c2 = a2 + b2 - 2ab Cos C

The versions of Law of Cosines, solved for angles in terms of sides are as follows:

Cos A = $\frac{b^{2}+c^{2}-a^{2}}{2bc}$

Cos B = $\frac{c^{2}+a^{2}-b^{2}}{2ca}$

Cos C = $\frac{a^{2}+b^{2}-c^{2}}{2ab}$
It is sufficient to remember one formula for each of Area and Law of Cosines, as the other two can be got by applying symmetry, i.e replacing a with b, b with c and c with a.
Let us solve few examples using the formulas given above to compute the area of an Oblique Triangle.

Solved Examples

Question 1: Given the sides of ΔABC, a = 15 b = 9 and c = 7. Check whether ΔABC is an Oblique Triangle and find its area.
a2 = 2252, b2 = 81 and c2 = 49

     a2 ≠ b2 + c2. Hence ΔABC is an oblique triangle.

     As the lengths of the three sides of the triangle are given we can use Heron's formula to compute the area of ΔABC.

     s = $\frac{a+b+c}{2}$ = $\frac{15+9+7}{2}$ = $\frac{31}{2}$ = 15.5

     Area of ΔABC =  $\sqrt{s(s-a)(s-b)(s-c)}$

                            = $\sqrt{15.5\times 0.5\times 6.5\times 8.5}$ = 20.7 sq.units (rounded to the tenth of a sq.unit).

Question 2: Find the area of ΔABC if a = 12 cm,  b = 16 cm and C = 55º  rounded to the tenth of a
We can use the formula, Δ = $\frac{1}{2}$ ab Sin C to calculate the area of ΔABC.

                                            Δ = $\frac{1}{2}$ x 12 x 16 Sin 55 = 96 Sin 55

     Using Calculator Sin 55 ≈ 0.8192

     Hence  Area of ΔABC Δ ≈ 96 x 0.8192 = 78.6432 = 78.6 sq cm.

Question 3: The forest department proposes to fence a triangular area to form a Deer Park to conserve Deers. The two sides of the triangular region measured 10 miles and 12 miles each making an angle of 118º. What would be the area of the proposed Deer Park nearest to a square mile?

Area of Oblique Triangle
Area of the triangle = $\frac{1}{2}$ ab Sin C     where a and b are the measures of two adjacent sides and C is the included angle.

    For the given triangular region a = 10 m, b = 12 m and m ∠C = 118º.

    Area of the proposed Deer Park = $\frac{1}{2}$ x 10 x 12 x Sin 118

                                                        = 60 Sin 118 ≈ 53 sq.miles

Oblique triangles are solved using either Law of sines or Law of Cosines. We need three measures to solve an Oblique triangle and the formula to be used depends upon what three measures are given. The following table maps the formula to be used to what is known about the triangle.
Solving an Oblique Triangle
Given Formula to be used first
Two angles and any one side Law of Sines
Two sides and an angle opposite
to one of them.
Law of Sines
Two sides and the included angle Law of Cosines
Three sides Law of Cosines

Solved Examples

Question 1: Solving an oblique triangle given two angles and a side.

Solve for the unknown measures in ΔABC.
m ∠C = 180 - (m ∠A + m ∠B)
          = 180 - (70 + 80) = 30º.
Solve Oblique Triangle

Using sine rule,

$\frac{a}{SinA}$ = $\frac{b}{SinB}$ = $\frac{c}{Sinc}$

$\frac{6}{Sin70}$ = $\frac{b}{Sin80}$ = $\frac{c}{sin30}$

b = Sin 80 x $\frac{6}{Sin70}$ ≈ 6.28

c = Sin 30 x $\frac{6}{Sin70}$ ≈ 3.19

Question 2: Solving an oblique triangle given two sides and an angle opposite to one of them

Use sine rule to find the angle measure of ∠C

$\frac{a}{SinA}$ = $\frac{c}{SinC}$

$\frac{8}{sin72}$ = $\frac{6}{sinC}$

Solving Oblique Triangles

Sin C = $\frac{6}{8}$ x sin 72 ≈ 0.7133

m ∠C = Sin-1 (0.7133)

    = 45.5º     (Using inverse function in calculator)

m ∠B = 180 - (m ∠A + m ∠C)

         = 180 - (72 + 45.5) = 62.5º

Again applying sine rule

$\frac{a}{sinA}$ = $\frac{b}{sinB}$

$\frac{8}{Sin72}$ = $\frac{b}{sin62.5}$

b = sin 62.5 x $\frac{8}{Sin72}$ ≈ 7.46.

For this type of situations in some cases there might be two solutions. Consider
also the supplement of the first angle solved for the second solution as
Sin (180 - θ) = Sin θ.

In the above example, 180 - m ∠C = 180 - 45.5 = 134.5. This measure when added to
m ∠A, 72 + 134.5 > 180, thus making the third angle not possible. Hence there is
only one solution for the above situation.


Question 3: Solving an oblique triangle given two sides the included angle

Solve for ΔPQR
Oblique Triangle Solved Problems
Using Cosine rule,
p2 = q2 + r2 - 2qr Cos P
     = 162 + 122 - 2 x 16 x 12 Cos 102º
     = 256 + 144 - 384 Cos 102º ≈ 479.84
p = √479.84 = 21.9 in
Now the measure of ∠Q can be found using Sine Rule.

$\frac{p}{SinP}$ = $\frac{q}{SinQ}$

$\frac{21.9}{Sin102}$ = $\frac{16}{SinQ}$

Sin Q = $\frac{16}{21.9}$ x Sin 102 ≈ 0.7146

m ∠Q = Sin-1 0.7146 = 45.6º

m ∠R = 180 - (m ∠P + m ∠Q) = 180 - (102 + 45.6) = 32.4º.

Question 4: Solving an oblique triangle given three sides

Solve for the angles of the ΔABC.
Examples of Oblique Triangles

The Cosine rule can be solved for Cos A as,

Cos A = $\frac{b^{2}+c^{2}-a^{2}}{2bc}$

          = $\frac{33^{2}+37^{2}-28^{2}}{2\times 33\times 37}$

          ≈ 0.6855

m ∠A = Cos-1 0.2727  ≈ 46.7º

Now sine rule can be used to find m ∠B.

$\frac{a}{sinA}$ = $\frac{b}{sinB}$

$\frac{28}{Sin46.7}$ = $\frac{33}{SinB}$

Sin B = $\frac{33}{28}$ Sin 46.7 = 0.8581

m ∠B = Sin-1 0.8581 = 59.1º

m ∠C = 180 - (m ∠A + m ∠B)

          = 180 - (46.7 + 59.1) = 74.2º