Math Core

Right-triangle trigonometry only works when one angle is $90^\circ$ . Real triangles, like the one formed by two lighthouses and a ship or by three stakes on a hillside, usually have no right angle at all. The law of sines and the law of cosines let you solve any triangle from three pieces of information, as long as at least one of them is a side.

Naming a triangle

Label the vertices $A$ , $B$ and $C$ , and use the matching lowercase letter for the side opposite each vertex. So side $a$ sits across from angle $A$ , side $b$ across from $B$ , and side $c$ across from $C$ . "Solving a triangle" means finding all three sides and all three angles.

Side a is opposite angle A, and so on. The dashed segment h is the height from C.

The law of sines

Drop the height $h$ from $C$ to side $c$ , as in the picture. The height splits the triangle into two right triangles. In the left one, $\sin A = \dfrac{h}{b}$ , so $h = b \sin A$ . In the right one, $\sin B = \dfrac{h}{a}$ , so $h = a \sin B$ . Both expressions equal $h$ , so

b \sin A = a \sin B \quad\Longrightarrow\quad \frac{a}{\sin A} = \frac{b}{\sin B}.

Dropping the height from a different vertex brings $c$ and $\sin C$ into the same chain.

Law of sines

In any triangle,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.

Use it when you know an angle and its opposite side, plus one more side or angle: the cases AAS, ASA and SSA.

Worked example: Two angles and a side (AAS)

In triangle $ABC$ , $A = 40^\circ$ , $B = 65^\circ$ and $a = 12$ . Solve the triangle. Round sides to the nearest tenth.

The angles add to $180^\circ$ , so $C = 180^\circ - 40^\circ - 65^\circ = 75^\circ$ .

You know the pair $a$ and $A$ , so every other side comes from $\dfrac{12}{\sin 40^\circ}$ :

\begin{aligned} b &= \frac{12 \sin 65^\circ}{\sin 40^\circ} \approx 16.9, \\ c &= \frac{12 \sin 75^\circ}{\sin 40^\circ} \approx 18.0. \end{aligned}

Check: the largest angle, $C$ , faces the largest side, $c$ . Good.

The law of cosines

The law of sines is useless when you don't know any angle-and-opposite-side pair, for example when you know two sides and the angle between them (SAS) or all three sides (SSS). For those cases you need the law of cosines, which is the Pythagorean theorem with a correction term.

Law of cosines

In any triangle,

\begin{aligned} c^2 &= a^2 + b^2 - 2ab\cos C, \\ a^2 &= b^2 + c^2 - 2bc\cos A, \\ b^2 &= a^2 + c^2 - 2ac\cos B. \end{aligned}

Each version pairs one angle with its opposite side on the left. Use it for SAS (find the third side) and SSS (find an angle).

If $C = 90^\circ$ , then $\cos C = 0$ and the formula becomes $c^2 = a^2 + b^2$ . When $C$ is obtuse, $\cos C$ is negative, so the correction term makes $c$ longer than the Pythagorean value, exactly as you would expect for a side across from a wide angle.

Worked example: Two sides and the included angle (SAS)

A triangle has $b = 9$ , $c = 14$ and $A = 52^\circ$ . Find $a$ to the nearest tenth.

\begin{aligned} a^2 &= 9^2 + 14^2 - 2(9)(14)\cos 52^\circ \\ &= 277 - 252\cos 52^\circ \\ &\approx 121.85, \end{aligned}

so $a \approx 11.0$ . To finish solving, you now have a known pair ( $a$ and $A$ ), so you could switch to the law of sines for the remaining angles.

Worked example: Three sides (SSS)

A triangle has sides $a = 7$ , $b = 9$ and $c = 12$ . Find its largest angle.

The largest angle is across from the longest side, so find $C$ . Solve the law of cosines for $\cos C$ :

\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49 + 81 - 144}{126} = \frac{-14}{126} \approx -0.1111.

So $C = \cos^{-1}(-0.1111) \approx 96.4^\circ$ . The negative cosine tells you right away that the angle is obtuse.

Tip

When you need an angle, prefer the law of cosines if you can. The inverse cosine returns angles from $0^\circ$ to $180^\circ$ , so it handles obtuse angles correctly. The inverse sine only returns angles up to $90^\circ$ .

The ambiguous case (SSA)

Knowing two sides and an angle that is not between them can produce two triangles, one triangle, or none. Suppose $A = 35^\circ$ , $b = 12$ and $a = 8$ . Picture side $b$ fixed, and side $a$ swinging from $C$ like a compass until it hits the base.

A side of length 8 swung from C meets the base twice, giving two different triangles.

The height from $C$ is $h = b \sin A = 12 \sin 35^\circ \approx 6.88$ . Compare $a$ with $h$ and $b$ :

Situation (with $A$ acute)	Number of triangles
$a < h$	0 (side $a$ is too short to reach)
$a = h$	1 (a right triangle)
$h < a < b$	2
$a \ge b$	1

If $A$ is obtuse, there is exactly one triangle when $a > b$ and none otherwise.

Worked example: Finding both triangles

Solve for $B$ when $A = 35^\circ$ , $a = 8$ and $b = 12$ .

Here $6.88 < 8 < 12$ , so expect two triangles. The law of sines gives

\sin B = \frac{12 \sin 35^\circ}{8} \approx 0.8604.

Your calculator reports $B_1 = \sin^{-1}(0.8604) \approx 59.4^\circ$ . The supplement has the same sine, so $B_2 = 180^\circ - 59.4^\circ = 120.6^\circ$ is also possible. Check that each fits with $A$ : $35^\circ + 120.6^\circ < 180^\circ$ , so both work. The third angles are $C_1 \approx 85.6^\circ$ and $C_2 \approx 24.4^\circ$ .

Common mistake

Whenever you use the inverse sine to find an angle, ask whether the supplement $180^\circ - \theta$ also works. Forgetting the second triangle is the most common mistake with the law of sines. The supplement is valid exactly when it plus the known angle is less than $180^\circ$ .

Area of any triangle

The height in the first picture was $h = b \sin A$ , and the base was $c$ . So the area is $\dfrac{1}{2} \cdot c \cdot b \sin A$ . In general:

Area formulas

Two sides and the included angle: $\text{Area} = \dfrac{1}{2}ab\sin C$ (or any two sides with the angle between them).

Three sides (Heron's formula): with $s = \dfrac{a + b + c}{2}$ ,

\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}.

For the SAS triangle above, $\text{Area} = \dfrac{1}{2}(9)(14)\sin 52^\circ \approx 49.6$ square units.

Practice

Practice 1

In triangle $ABC$ , $A = 30^\circ$ , $B = 45^\circ$ and $a = 10$ . Find $b$ . Round to the nearest hundredth.