Solving any triangle: sine and cosine rules

Standard notation for a triangle

In a triangle $ABC$, $a$ is the side opposite vertex $A$, $b$ is opposite $B$ and $c$ is opposite $C$. Capital letters name the vertices (and their angles), and lower-case letters name the lengths. "Opposite" means that the side does not touch that vertex.

The two rules relate these six measurements. Solving a triangle means finding all the ones that are missing. Remember also that the three angles add up to $180^{\circ }$ and that the longest side is always opposite the largest angle.

The sine rule

The idea is that, in the same triangle, a larger angle corresponds to a larger side, and the ratio between each side and the sine of its opposite angle is constant. It is proved by drawing the height $h$ from $C$: it equals $h=b\sin A$ and also $h=a\sin B$, from which $\frac{a}{\sin A}=\frac{b}{\sin B}$.

To use it you need two "side and opposite angle" pairs, one complete and one with only one unknown. That is why it works in two situations:

• Two angles and a side (any side). The third angle comes from the sum of $180^{\circ }$, and you then have the necessary pairs.
• Two sides and an angle opposite one of them. Here you must be careful, because there may be two triangles.

The ambiguous case

Suppose you know angle $A$ and sides $b$ (the one touching $A$) and $a$ (the one opposite $A$). Imagine side $b$ fixed along a ray from $A$, with its end at $C$. Side $a$ is a segment of fixed length that starts at $C$ and must reach the other ray starting at $A$. Depending on its length, it can touch that ray at none, one or two points.

The height from $C$ is $h=b\sin A$ and marks the boundary:

In the case of two triangles, the sine rule gives $\sin B=\frac{b\sin A}{a}$, and this equation has two solutions between $0^{\circ }$ and $180^{\circ }$: $$B_1={\sin }^{-1}\left(\frac{b\sin A}{a}\right)\text{and}{B}_2=180^{\circ }-B_1$$ Both are valid because $\sin B_1=\sin (180^{\circ }-B_1)$. The calculator only gives you $B_1$ (acute): you must work out the obtuse $B_2$ yourself. Then discard any solution for which $A+B$ is $180^{\circ }$ or more. When $a\ge b$, this always happens with $B_2$, which is why only one solution remains.

The cosine rule

The sine rule needs a complete "side and opposite angle" pair. When you do not have one, you need another tool.

Notice that the angle on the right-hand side is always the opposite of the side on the left-hand side, and that the other two sides are the ones that form it. If $A=90^{\circ }$, the cosine term is 0 and $a^2=b^2+c^2$ remains: Pythagoras' theorem is a special case. The cosine term acts as a "correction" to Pythagoras: if $A$ is acute, something is subtracted and $a^2$ is less than $b^2+c^2$; if $A$ is obtuse, the cosine is negative and $a^2$ is greater.

It is used in these two situations:

• Two sides and the angle between them (the one they form). This gives the third side.
• All three sides. Any angle is obtained by rearranging for the cosine.

Choosing between the two rules

Look at the data you have and what you are looking for:

In many problems the steps are chained: the cosine rule gives the first missing quantity and then the sine rule completes the rest. When you need to find an angle and are unsure, use the cosine: its inverse function returns angles between $0^{\circ }$ and $180^{\circ }$, so it identifies obtuse angles without ambiguity. With the sine, an acute result can hide an obtuse one.