The sine-rule and the cosine-rule for general triangles

Can the sine, cosine and tangent also be applied to general triangles? Yes, but the formulas are more complicated. Let us briefly discuss those formulas.

Consider a general triangle, e.g. the one below. Note that we labelled sides and angles such that side aa is opposite of angle α\alpha, side bb is opposite of β\beta, and side cc is opposite of angle γ\gamma. This is important for the formulas below to work.

Theorem 1: The sine-rule
sin⁡(α)a=sin⁡(β)b=sin⁡(γ)c\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}
Theorem 2: The cosine-rules
a2=b2+c2−2bc⋅cos⁡(α)a^2=b^2+c^2-2bc \cdot\cos(\alpha)b2=a2+c2−2ac⋅cos⁡(β)b^2=a^2+c^2-2ac \cdot\cos(\beta)c2=a2+b2−2ab⋅cos⁡(γ)c^2=a^2+b^2-2ab \cdot\cos(\gamma)
Example 1

Determine side length cc and the angles α\alpha and β\beta in the triangle below.

Solution

First cc: Using the cosine rule, we get

c2=52+72−2⋅5⋅7⋅cos⁡(130∘)=74−70⋅cos⁡(130∘)=118.99\begin{array}{lll} c^2&=&5^2+7^2-2\cdot 5\cdot 7\cdot \cos(130^\circ)\\ & = & 74-70\cdot \cos(130^\circ)\\ & = & 118.99\\ \end{array}

Thus, we have c=10.9‾c=\underline{10.9}. To find the angles, we can apply the sine-rule:

sin⁡(130∘)10.9=sin⁡(α)7\frac{\sin(130^\circ)}{10.9}=\frac{\sin(\alpha)}{7}

It follows

sin⁡(α)=7⋅sin⁡(130∘)10.9=0.492\sin(\alpha)=7\cdot \frac{\sin(130^\circ)}{10.9} = 0.492

and thus

α=arcsin⁡(0.492)=29.47∘‾\alpha = \arcsin(0.492)=\underline{29.47^\circ}

Finally, we have

β=180∘−130∘−29.47∘=20.53∘‾\beta = 180^\circ-130^\circ-29.47^\circ=\underline{20.53^\circ}

You can think of the cosine-rules as a kind of generalised Pythagoras, where a "correction term" has to be subtracted because we are not dealing with a right-angled triangle. In fact, if for example α=90cˆirc\alpha=90\^circ, then we get indeed the theorem of Pythagoras

a2=b2+c2−2bc⋅cos⁡(90∘)⏟=0=b2+c2\begin{array}{lll} a^2&=&b^2+c^2-2bc\cdot \underbrace{\cos(90^\circ)}_{=0}\\ &=&b^2+c^2 \end{array}

It is important to realise that we do not really use those laws, as we could always divide the general triangle in two right-angled triangles, as we already done many time in the previous sections. However, these laws offer some shortcuts, and calculations are quicker.

Exercise 1
Q1

Consider the general triangles shown below. Find the missing angles and side lengths using the sine-rule or the cosine-rule. You might have to try out which of those two rules work better for each problem.

Q2*

Proof the sine-law by dividing the general triangle into two right-angled triangles.

Q3*

Proof the cosine-law by dividing the general triangle into two right-angled triangles.

Solution