The Law of Cosines Advanced. This lesson goes beyond core Geometry. You can skip it.

Watch the video explainer

Learning goals

Derive the Law of Cosines from a signed projection and the Pythagorean theorem
Explain how the included angle changes the square of the opposite side
Use two sides and their included angle to find the third side
Recover and check the angles of a triangle from its three sides
Choose a triangle law from the given sides and angles

Measure the separation with a projection

In a nondegenerate triangle $ABC$ , use $a$ for the length of $BC$ , $b$ for $CA$ , and $c$ for $AB$ . Each lowercase letter names the side opposite the corresponding angle. The included angle between two sides is the angle at their shared endpoint. Sides $a$ and $b$ meet at $C$ .

Put $C$ at the origin and $A$ at $(9,0)$ . Compare two placements of $B$ at the same distance from $C$ .

The perpendicular foot is to the right of the origin.

In the acute placement, $B$ is at $(4,3)$ . The horizontal distance from $B$ to $A$ is $9-4$ units, and the vertical distance is $3$ units. The Pythagorean theorem gives $c^2$ .

\begin{aligned} c^2&=(9-4)^2+3^2\\ &=34. \end{aligned}

The perpendicular foot is to the left of the origin. The dashed horizontal line extends the base.

In the obtuse placement, $B$ is at $(-4,3)$ . Its horizontal coordinate is negative. The same subtraction now gives a larger horizontal distance.

\begin{aligned} c^2&=(9-(-4))^2+3^2\\ &=178. \end{aligned}

Expanding the square shows what changed. In both placements, the squares of the horizontal and vertical coordinates of $B$ add to $25$ , the square of $BC$ . The changing part is the product of $-2$ , the base length $9$ , and the signed horizontal coordinate.

\begin{aligned} 34&=9^2+25-2(9)(4),\\ 178&=9^2+25-2(9)(-4). \end{aligned}

The Law of Cosines from a signed projection#

Now put $C$ at $(0,0)$ and $A$ at $(b,0)$ . Rotate and, if necessary, reflect the triangle so $B=(x,y)$ lies above the horizontal axis. As in Triangle Area with Sine Advanced. This lesson goes beyond core Geometry. You can skip it., scaling the unit semicircle by the length $a$ gives the signed projection, the horizontal coordinate of $B$ .

x=a\cos C.

The horizontal and vertical distances between $A$ and $B$ are $|b-x|$ and $y$ . Squaring the horizontal distance gives $(b-x)^2$ , even when the perpendicular foot is beyond $A$ . If the foot coincides with $A$ , the horizontal distance is zero and the same squared-distance equation gives $c^2=y^2$ .

\begin{aligned} c^2&=(b-x)^2+y^2\\ &=b^2+(x^2+y^2)-2bx. \end{aligned}

The Pythagorean theorem gives $x^2+y^2=a^2$ for $BC$ . This relation also holds when $x=0$ , since then $a=y$ . Substitute this sum and the expression for $x$ .

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

The numerical placements used the same expansion. The sign and size of $x$ can change without changing any of its steps. This equation is the Law of Cosines. Relabel the triangle to use it with any side and its opposite angle.

Check your understanding

Set $C=(0,0)$ and put $A$ on the positive horizontal axis. Side $BC$ has length $12$ and angle $C$ is $120^\circ$ . What is the horizontal coordinate of $B$ ?

6

-6

-12

0

Find a third side from SAS

Two sides and their included angle form SAS data. The cosine equation uses those three measurements to give the square of the remaining side. Take the positive square root because a side has positive length. Use degree mode on your calculator and round only the final measurements.

Acute included angle. Cosine is positive, so the correction reduces the sum of the two side squares.
Obtuse included angle. Cosine is negative, so the correction increases that sum.
Right included angle. Cosine is zero, so the equation becomes the Pythagorean theorem.

Worked example 1 An acute opening

Problem. Two rods of lengths $7$ cm and $10$ cm meet at a hinge opened to $60^\circ$ . Find the exact distance $d$ between the free ends.

The hinge angle is between the known lengths.

Solution. The free ends are opposite the included angle. Since $\cos60^\circ=\tfrac12$ , the correction subtracts $70$ from the sum of the squares.

\begin{aligned} d^2&=7^2+10^2-70\\ &=79,\\ d&=\sqrt{79}\text{ cm}. \end{aligned}

Check your understanding

Two fixed sides meet at an acute angle. The hinge opens to an obtuse angle while the side lengths stay fixed. How does the square of the distance between the free ends change?

A It stays equal to the sum of the side squares. B It goes from above the sum of the side squares to below. C It goes from below the sum of the side squares to above. D It equals the difference of the side squares at both openings.

Worked example 2 Two sightings from a dock

Problem. From one dock, the straight distances to two buoys are $230$ m and $310$ m. The angle between the sight lines is $124^\circ$ . Treat all three points as lying in one horizontal plane. Find the straight distance $d$ between the buoys to the nearest meter.

The sought distance joins the buoys directly.

Solution. The measured angle is between the two known sides, and $d$ lies across from it.

\begin{aligned} d^2&=230^2+310^2\\ &\quad-2(230)(310)\cos124^\circ,\\ d&\approx478\text{ m}. \end{aligned}

The result fits between the difference of the measured lengths, $80$ m, and their sum, $540$ m.

Worked example 3 The right-angle boundary

Problem. Two straight paths of lengths $11$ m and $19$ m start at the same point and meet there at $90^\circ$ . Use the cosine equation to find the exact distance $d$ between their far ends. Explain why the Pythagorean theorem gives the same result.

Solution. The correction vanishes because $\cos90^\circ=0$ . The paths are the two legs, so the unknown side is the hypotenuse.

\begin{aligned} d^2&=11^2+19^2\\ &=482,\\ d&=\sqrt{482}\text{ m}. \end{aligned}

Check your understanding

In triangle $ABC$ , the lengths of $a$ and $b$ are 16 and 23. The angle at $C$ is $108^\circ$ . What is $c$ to the nearest tenth?

28.0

23.6

43.8

31.8

Find angles from SSS

With SSS data, all three sides are given. First check that the two shorter positive lengths add to more than the longest. This one comparison guarantees the triangle inequalities for positive lengths. Then isolate the cosine of the angle across from the side chosen on the left of the original equation.

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

As the ray from $C$ turns through the upper unit semicircle from $0^\circ$ to $180^\circ$ , the horizontal coordinate decreases from $1$ to $-1$ . Every value strictly between $-1$ and $1$ occurs at exactly one interior angle. In degree mode the inverse cosine, written $\arccos$ or $\cos^{-1}$ , returns that angle. An input of $-1$ or $1$ would produce a straight or zero angle, so the triangle would be degenerate.

If you need every angle, use the cosine equation for the largest angle and then for one of the others. Subtract both from $180^\circ$ to find the third. Starting opposite the longest side lets you see whether any angle is obtuse.

Worked example 4 Recover all three angles

Problem. In triangle $ABC$ , the side lengths are $8$ , $13$ , and $18$ units for $a$ , $b$ , and $c$ , respectively. Find the measures of $A$ , $B$ , and $C$ to the nearest tenth of a degree, and check the results against the sides.

The longest given side, AB, faces angle C.

Solution. The two shorter sides add to $21$ units, more than $18$ . The largest angle is $C$ , across from the longest side.

\begin{aligned} \cos C&=\frac{8^2+13^2-18^2}{2(8)(13)}\\ &=-\frac{91}{208},\\ C&\approx115.9^\circ. \end{aligned}

Use the sides touching $A$ in the denominator and subtract the square of the side opposite $A$ in the numerator.

\begin{aligned} \cos A&=\frac{13^2+18^2-8^2}{2(13)(18)},\\ A&\approx23.6^\circ. \end{aligned}

Subtract the unrounded values to obtain $B$ .

\begin{aligned} B&=180^\circ-C-A\\ &\approx40.5^\circ. \end{aligned}

The displayed angles add to $180.0^\circ$ . Their order from smallest to largest is $A$ , $B$ , $C$ , which matches the order $8$ , $13$ , $18$ of the opposite sides. The negative cosine computed for $C$ also agrees with its obtuse measure.

Check your understanding

The lengths of $a$ , $b$ , and $c$ are $12$ , $16$ , and $25$ . Which expression gives $C$ in degrees?

\arccos(-225/384)

\arccos(513/600)

\arccos(225/384)

\arcsin(-225/384)

Choose a law from the measurements

Start by pairing each known side with the angle across from it. The Law of Sines Advanced. This lesson goes beyond core Geometry. You can skip it. can use a known opposite pair as a ratio. The cosine equation can start from measurements that give no such pair.

Given data	First move
Two sides and their included angle (SAS)	Use the cosine law to find the third side.
Three sides (SSS)	Use the cosine law to find an angle.
Two angles and a side (ASA or AAS)	Use the angle sum if needed to complete an opposite pair. Then use the sine law.
Two sides and an angle opposite a known side (SSA)	Use the sine law and check the possible supplementary angles and remaining angle.

Once you have a new measurement, look at the updated data. For example, after finding the third side from SAS, all three sides are available, so the SSS procedure finds the remaining angles.

Worked example 5 Choose a law for two angles and a side

Problem. A survey triangle has angle measures $37^\circ$ at $A$ and $74^\circ$ at $B$ . Side $AB$ measures $21$ m. Choose the first triangle law that can find $AC$ , explain the choice, and calculate $AC$ to the nearest tenth of a meter.

The angle opposite AB follows from the two given angles.

Solution. Only one side is given. The angle sum completes the opposite pair for side $AB$ .

\begin{aligned} C&=180^\circ-37^\circ-74^\circ\\ &=69^\circ. \end{aligned}

Use the Law of Sines with $AB$ opposite $C$ and $AC$ opposite $B$ .

\begin{aligned} \frac{AC}{\sin74^\circ}&=\frac{21}{\sin69^\circ},\\ AC&=\frac{21\sin74^\circ}{\sin69^\circ}\\ &\approx21.6\text{ m}. \end{aligned}

Check your understanding

A triangle has sides $a$ and $c$ known. Angle $A$ is also known. Which first move matches the measurements?

A Use

A

as the included angle between

a

and

c

. B Use the sine law with

a

and

A

, then check possible values of

C

. C Use inverse cosine with the two known sides alone. D Subtract

A

from

180^\circ

to obtain

C

The Law of Cosines Advanced. This lesson goes beyond core Geometry. You can skip it.

Learning goals

Measure the separation with a projection

Find a third side from SAS

Find angles from SSS

Choose a law from the measurements

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress