The law of sines

An oblique triangle is a triangle with no right angle. SOH-CAH-TOA does not apply to it directly. This page gives the law of sines, which solves an oblique triangle when you know two angles and a side. It also handles the tricky case of two sides and an angle opposite one of them.

Names and the law

As in Solving right triangles, call the angles ${A}$, ${B}$, ${C}$ and the sides opposite them ${a}$, ${b}$, ${c}$.

Side ${a}$ is opposite angle ${A}$, and so on. The dashed line ${h}$ is the height from ${C}$.

In any triangle, each side divided by the sine of its opposite angle gives the same number:

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

In words: the longer the side, the larger the angle across from it. You can also turn the fractions over. This form is handy for finding an angle:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

Why it is true

Drop the height ${h}$ from ${C}$ to side ${AB}$, as in the figure. It makes two right triangles.

In the left one, ${h}$ is opposite ${A}$ and ${b}$ is the hypotenuse, so ${\sin A=\dfrac{h}{b}}$. That gives ${h=b\sin A}$.
In the right one, ${h}$ is opposite ${B}$ and ${a}$ is the hypotenuse, so ${h=a\sin B}$.

Both equal ${h}$, so ${b\sin A=a\sin B}$. Dividing both sides by ${\sin A\sin B}$ gives ${\dfrac{b}{\sin B}=\dfrac{a}{\sin A}}$. The same idea with a different height brings in ${c}$. (If an angle is obtuse, the height falls outside the triangle, but the same steps work.)

Two angles and a side

If you know two angles, you know the third, because the angles of a triangle add up to ${180^\circ}$. Then the law of sines gives the other sides. This works whether the known side is between the angles (called ASA) or not (AAS).

Example 1: Solve the triangle with ${A=40^\circ}$, ${B=60^\circ}$, and ${a=10}$. Round to two decimal places.

${A=40^\circ}$, ${B=60^\circ}$, and ${a=10}$.

Solution:

Angle ${C}$: ${C=180^\circ-40^\circ-60^\circ=80^\circ}$.

Side ${b}$: Use ${\dfrac{b}{\sin B}=\dfrac{a}{\sin A}}$, and multiply by ${\sin B}$:

$b=\dfrac{10\sin 60^\circ}{\sin 40^\circ}\approx 13.47$

Side ${c}$: In the same way,

$c=\dfrac{10\sin 80^\circ}{\sin 40^\circ}\approx 15.32$

Check: The largest angle, ${C=80^\circ}$, is across from the longest side, ${c}$. Good.

Example 2: Two observers stand ${500}$ m apart on a straight shore, at ${A}$ and ${B}$. They sight a boat ${C}$. The angle between the shore and the line to the boat is ${52^\circ}$ at ${A}$ and ${61^\circ}$ at ${B}$. How far is the boat from ${A}$?

Two observers at ${A}$ and ${B}$ sight the boat ${C}$.

Solution:

The angle at the boat is ${C=180^\circ-52^\circ-61^\circ=67^\circ}$. The known side ${AB=500}$ is opposite ${C}$. The distance from ${A}$ to the boat is side ${b}$, opposite ${B}$:

$b=\dfrac{500\sin 61^\circ}{\sin 67^\circ}\approx 475.1$ m

The ambiguous case: two sides and an angle

Now suppose you know two sides and an angle that is not between them, such as ${A}$, ${a}$, and ${b}$. This is called SSA. It is tricky: there may be no triangle, one triangle, or two.

The figures show why, for an acute angle ${A}$. Side ${b}$ is fixed. Side ${a}$ hangs from ${C}$ and swings like a pendulum, along the dashed arc. Let ${h=b\sin A}$ be the height of ${C}$ above the base line.

(a) ${a<h}$: no triangle

(b) ${a=h}$: one right triangle

(d) ${a\ge b}$: one triangle

In practice, you do not need the figure. Use the law of sines to find ${\sin B}$, and let the numbers tell you:

If ${\sin B>1}$, there is no triangle.
If ${\sin B\le 1}$, there are two angles with that sine: ${B}$ and ${180^\circ-B}$. Each one gives a triangle if the angles still add up to less than ${180^\circ}$.

Watch out: The calculator gives only the acute angle for ${\sin^{-1}}$. In the SSA case, always test the obtuse angle ${180^\circ-B}$ too.

Example 3: Solve the triangle with ${A=40^\circ}$, ${a=5}$, and ${b=12}$.

Solution:

Use ${\dfrac{\sin B}{b}=\dfrac{\sin A}{a}}$:

$\sin B=\dfrac{12\sin 40^\circ}{5}\approx 1.543$

A sine can never be more than ${1}$. So there is no triangle with these measurements. (Side ${a}$ is too short to reach.)

Example 4: Solve the triangle with ${A=30^\circ}$, ${a=6}$, and ${b=10}$.

Solution:

$\begin{align*}\sin B&=\dfrac{10\sin 30^\circ}{6}\\&=\dfrac{5}{6}\approx 0.8333\end{align*}$

So ${B\approx 56.44^\circ}$, or ${B\approx 123.56^\circ}$ (that is, ${180^\circ-56.44^\circ}$). Test each:

Triangle 1: ${B\approx 56.44^\circ}$. The angles add up to ${180^\circ}$, so ${C\approx 93.56^\circ}$. Then

$c=\dfrac{6\sin 93.56^\circ}{\sin 30^\circ}\approx 11.98$

Triangle 2: ${B\approx 123.56^\circ}$. The angles add up to ${180^\circ}$, so ${C\approx 26.44^\circ}$. Then

$c=\dfrac{6\sin 26.44^\circ}{\sin 30^\circ}\approx 5.34$

Both are real triangles:

${B\approx 56.4^\circ}$

${B\approx 123.6^\circ}$

Example 5: Solve the triangle with ${A=50^\circ}$, ${a=12}$, and ${b=10}$.

Solution:

$\sin B=\dfrac{10\sin 50^\circ}{12}\approx 0.6384$

So ${B\approx 39.67^\circ}$ or ${B\approx 140.33^\circ}$. But ${50^\circ+140.33^\circ}$ is more than ${180^\circ}$, so the second is impossible. There is one triangle:

${C\approx 180^\circ-50^\circ-39.67^\circ}$

${C\approx 90.33^\circ}$

$c=\dfrac{12\sin 90.33^\circ}{\sin 50^\circ}\approx 15.66$

Summary

Law of sines: ${\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}}$.
Use it when you know two angles and a side (AAS or ASA). Find the third angle first.
With two sides and an angle opposite one of them (SSA), there may be no triangle, one, or two. Find ${\sin B}$; if it is at most ${1}$, test both ${B}$ and ${180^\circ-B}$.

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