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.

### 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.

** Solution: **

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}$ =

$\frac{6}{sinC}$

** Solution: **

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

** Solution: **

Using Cosine rule,

p^{2} = q^{2} + r^{2} - 2qr Cos P

= 16^{2} + 12^{2} - 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.

** Solution: **

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