The Law of Cosines and Triangle Area, precalculus lesson

Two straight trails leave a junction $60^\circ$ apart. You walk $5$ kilometers along one, a friend walks $8$ kilometers along the other, and you want to know how far apart you are now. The law of sines cannot start here, because no side is known together with the angle across from it. The law of cosines can, and the same few ideas give the area of the triangle you have marked out.

Key terms

Included angle

Definition The included angle of two sides of a triangle is the angle at the vertex where those two sides meet. In triangle $ABC$ , with sides $a$ , $b$ and $c$ opposite the angles $A$ , $B$ and $C$ , the angle $C$ is included between sides $a$ and $b$ , the angle $A$ between $b$ and $c$ , and the angle $B$ between $a$ and $c$ .

In plain words Think of two sides as the arms of a hinge. The included angle tells you how far the hinge is open, and that decides how far apart the free ends are.

Example Join sides of lengths $5$ and $8$ at $C$ and open the hinge. With these two lengths fixed, the third side $c$ and the area depend only on $C$ . The table uses the law of cosines and the area formula from the sections below.

$C$	Third side $c$	Area
$30^\circ$	$4.44$	$10.00$
$60^\circ$	$7.00$	$17.32$
$90^\circ$	$9.43$	$20.00$
$120^\circ$	$11.36$	$17.32$
$150^\circ$	$12.58$	$10.00$

The third side keeps growing as the hinge opens, while the area is largest at $90^\circ$ .

The law of cosines

Definition In any triangle $ABC$ with sides $a$ , $b$ and $c$ opposite the angles $A$ , $B$ and $C$ ,

c^2 = a^2 + b^2 - 2ab \cos C,

and in the same way $a^2 = b^2 + c^2 - 2bc \cos A$ and $b^2 = a^2 + c^2 - 2ac \cos B$ . Solved for the angle, it reads

\cos C = \frac{a^2 + b^2 - c^2}{2ab}.

To see why it holds, place $C$ at the origin, $B$ at $(a, 0)$ and $A$ at $(b \cos C, b \sin C)$ . The distance formula and the identity $\cos^2 C + \sin^2 C = 1$ give

\begin{aligned} c^2 &= (b \cos C - a)^2 + (b \sin C)^2 \\[0.4em] &= a^2 + b^2 - 2ab \cos C. \end{aligned}

In plain words It is the Pythagorean theorem with a correction term. When $C = 90^\circ$ the term $2ab \cos C$ is zero. When $C$ is obtuse, $\cos C$ is negative and the opposite side comes out longer than in a right triangle, and when $C$ is acute it comes out shorter.

Example The trails from the start give $a = 5$ , $b = 8$ and $C = 60^\circ$ , so

c^2 = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \tfrac{1}{2} = 49

and you are $c = 7$ kilometers apart.

SAS and SSS

Definition In the SAS case you know two sides and their included angle, with the angle strictly between $0^\circ$ and $180^\circ$ . In the SSS case you know all three sides, and they satisfy the triangle inequality below. Each case determines exactly one triangle. For SAS, find the third side with the law of cosines. For SSS, find two of the angles from

\begin{gathered} \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \\[0.8em] \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \end{gathered}

and then $C = 180^\circ - A - B$ . Arccosine is one-to-one with outputs in $[0^\circ, 180^\circ]$ , so an angle found from its cosine is never ambiguous, unlike an angle found from its sine. If you prefer the law of sines for a later angle in SAS, use it only for the angle opposite the shorter given side, which is always acute.

In plain words Two sides with the angle between them, or all three sides, pin down one triangle with no second answer. Start with the law of cosines, and whenever you solve for an angle, take the arccosine.

Example Solve the triangle with $a = 7$ , $b = 5$ and $c = 3$ . Start with the largest angle, which is opposite the longest side.

\begin{aligned} \cos A &= \frac{5^2 + 3^2 - 7^2}{2 \cdot 5 \cdot 3} \\[0.8em] &= \frac{-15}{30} = -\frac{1}{2}, \\[0.8em] A &= 120^\circ. \end{aligned}

Next,

\begin{aligned} \cos B &= \frac{7^2 + 3^2 - 5^2}{2 \cdot 7 \cdot 3} \\[0.8em] &= \frac{33}{42} = \frac{11}{14}, \\[0.8em] B &\approx 38.21^\circ, \end{aligned}

and the last angle is $C \approx 21.79^\circ$ , since the three angles add to $180^\circ$ .

Triangle inequality

Definition Three positive numbers $a$ , $b$ and $c$ are the side lengths of a triangle exactly when each one is less than the sum of the other two,

\begin{gathered} a < b + c, \\[0.4em] b < a + c, \\[0.4em] c < a + b. \end{gathered}

It is enough to check that the largest number is less than the sum of the other two. If the largest equals that sum, the three corners lie on one line, a flat case that is not a triangle. For a genuine triangle the law of cosines always gives a cosine strictly between $-1$ and $1$ .

In plain words Going straight from one corner to another is shorter than any detour through the third corner. Three sticks only close up into a triangle when the longest is shorter than the other two put together.

Example Sticks of lengths $4$ , $6$ and $11$ cannot form a triangle, since $11 > 4 + 6$ . The law of cosines notices too. Solving for the angle opposite $11$ gives

\begin{aligned} \cos C &= \frac{4^2 + 6^2 - 11^2}{2 \cdot 4 \cdot 6} \\[0.8em] &= -\frac{69}{48} \approx -1.44, \end{aligned}

and no angle has a cosine below $-1$ . With $4$ , $6$ and $10$ the same computation gives exactly $-1$ , an angle of $180^\circ$ , so those three sticks lie flat along one line.

Area from two sides and the included angle

Definition The area $K$ of a triangle with sides $a$ and $b$ and included angle $C$ is

K = \frac{1}{2} ab \sin C.

The other pairs of sides give the same area in the same way,

K = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B.

The reason is that the height from $A$ to the line through $B$ and $C$ equals $b \sin C$ , also when $C$ is obtuse because $\sin(180^\circ - C) = \sin C$ , so the formula is one half of the base $a$ times the height.

In plain words It is one half of base times height, where the sine of the angle between the two known sides supplies the height. An angle and its supplement have the same sine, so they give the same area.

Example For the trail triangle, $a = 5$ , $b = 8$ and $C = 60^\circ$ , so

\begin{aligned} K &= \frac{1}{2} \cdot 5 \cdot 8 \cdot \sin 60^\circ \\[0.8em] &= 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \\[0.8em] &\approx 17.32 \text{ square kilometers.} \end{aligned}

Heron’s formula

Definition Let a triangle have sides $a$ , $b$ and $c$ and semiperimeter

s = \frac{a + b + c}{2}.

Then its area is

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

When $a$ , $b$ and $c$ satisfy the triangle inequality, each of the factors $s - a$ , $s - b$ and $s - c$ is positive, so the number under the root is positive.

In plain words Three sides fix the triangle, so they also fix its area. Heron’s formula gets the area straight from the sides, with no angle and no height needed.

Example The trail triangle has sides $5$ , $8$ and $7$ , so $s = 10$ and

\begin{aligned} K &= \sqrt{10 \cdot 5 \cdot 2 \cdot 3} \\[0.4em] &= \sqrt{300} = 10\sqrt{3}. \end{aligned}

This matches the area found from one half of $ab \sin C$ .

Bearings

Definition A bearing gives a direction in the plane as an angle measured from north or south. A three-figure bearing is measured clockwise from north, from $000^\circ$ up to but not including $360^\circ$ , so east is $090^\circ$ , south is $180^\circ$ and west is $270^\circ$ . A quadrant bearing such as N $40^\circ$ E starts from north or south and turns the stated acute angle toward east or west. So N $40^\circ$ E is the three-figure bearing $040^\circ$ , and S $30^\circ$ W is $210^\circ$ .

In plain words A bearing is a compass heading written as an angle. To use bearings in a triangle, sketch a north arrow at each corner and work out the angles inside the triangle from the headings.

Example A boat leaves port, sails $12$ kilometers on bearing $050^\circ$ , then turns onto bearing $110^\circ$ and sails $8$ kilometers. The turn is $60^\circ$ to the right, so the angle inside the triangle at the turning point is $180^\circ - 60^\circ = 120^\circ$ . The distance $d$ back to port satisfies

\begin{aligned} d^2 &= 12^2 + 8^2 - 2 \cdot 12 \cdot 8 \cos 120^\circ \\[0.4em] &= 144 + 64 + 96 = 304, \end{aligned}

so $d = \sqrt{304} \approx 17.44$ kilometers. The angle $P$ at port must be acute, since the triangle already has a $120^\circ$ angle, so the law of sines finds it safely,

\begin{gathered} \sin P = \frac{8 \sin 120^\circ}{\sqrt{304}} \approx 0.397, \\[0.8em] P \approx 23.41^\circ. \end{gathered}

The boat turned clockwise, so seen from port it now lies on bearing $050^\circ + 23.41^\circ$ , or about $073^\circ$ .

Check yourself

Two sides of a triangle are $10$ and $14$ , and the angle between them is $35^\circ$ . Find the third side and the area.

Show the answer

By the law of cosines

\begin{aligned} c^2 &= 10^2 + 14^2 - 2 \cdot 10 \cdot 14 \cos 35^\circ \\[0.4em] &\approx 66.64, \end{aligned}

so $c \approx 8.16$ . The area is

K = \frac{1}{2} \cdot 10 \cdot 14 \cdot \sin 35^\circ \approx 40.15.

Can $3$ , $4$ and $8$ be the sides of a triangle? Find the area of the triangle with sides $13$ , $14$ and $15$ .

Show the answer

No, because $8 > 3 + 4$ . For the sides $13$ , $14$ and $15$ the semiperimeter is $s = 21$ , and Heron’s formula gives

K = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84.

Two boats leave the same dock. One travels $10$ kilometers on bearing $030^\circ$ and the other $6$ kilometers on bearing $100^\circ$ . How far apart are they?

Show the answer

The angle between their paths at the dock is $100^\circ - 30^\circ = 70^\circ$ , so

\begin{aligned} d^2 &= 10^2 + 6^2 - 2 \cdot 10 \cdot 6 \cos 70^\circ \\[0.4em] &\approx 94.96 \end{aligned}

and $d \approx 9.74$ kilometers.

The Law of Cosines and Triangle Area

Key terms

Included angle

The law of cosines

SAS and SSS

Triangle inequality

Area from two sides and the included angle

Heron’s formula

Bearings

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus