# 13.5: The Law of Cosines

**At Grade**Created by: CK-12

**Objective**

Use the Law of Cosines equation to solve non right triangles and find the area of triangles using Heron’s Formula.

**Review Queue**

Find the value of \begin{align*}x\end{align*}

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## Using the Law of Cosines with SAS (to find the third side)

**Objective**

Use the Law of Cosines to determine the length of the third side of a triangle when two sides and the included angle are known.

**Guidance**

The Law of Cosines can be used to solve for the third side of a triangle when two sides and the included angle are known in a triangle. consider the non right triangle below in which we know \begin{align*}a, b\end{align*}

Now we can use the Pythagorean Theorem to relate the lengths of the segments in each of the right triangles shown.

Triangle 1: \begin{align*}x^2+k^2=a^2\end{align*}

Triangle 2: \begin{align*}(b-x)^2+k^2=c^2\end{align*}

Since both equations are equal to \begin{align*}k^2\end{align*}

\begin{align*}a^2-x^2 &=c^2-(b-x)^2 \\ a^2-x^2 &=c^2-(b^2-2bx+x^2) \\ a^2-x^2 &=c^2-b^2+2bx-x^2 \\ a^2 &=c^2-b^2+2bx \\ a^2+b^2-2bx &=c^2\end{align*}

Recall that we know the values of \begin{align*}a\end{align*}

\begin{align*} \cos C=\frac{x}{a} \quad \text{so} \quad x=a \cos C\end{align*}

Finally, we can replace \begin{align*}x\end{align*} in the equation to get the Law of Cosines: \begin{align*}a^2+b^2-2ab \cos C=c^2\end{align*}

Keep in mind that \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the sides of angle \begin{align*}C\end{align*} in the formula.

**Example A**

Find \begin{align*}c\end{align*} when \begin{align*}m \angle C=80^\circ, a = 6\end{align*} and \begin{align*}b = 12\end{align*}.

**Solution:** Replacing the variables in the formula with the given information and solve for \begin{align*}c\end{align*}:

\begin{align*}c^2 &=6^2+12^2-2(6)(12) \cos 80^\circ \\ c^2 & \approx 154.995 \\ c & \approx 12.4\end{align*}

**Example B**

Find \begin{align*}a\end{align*}, when \begin{align*}m \angle A=43^\circ\end{align*}, \begin{align*}b=16\end{align*} and \begin{align*}c=22\end{align*}.

**Solution:** This time we are given the sides surrounding angle \begin{align*}A\end{align*} and the measure of angle \begin{align*}A\end{align*}. We can rewrite the formula as: \begin{align*}a^2=c^2+b^2-2cb \cos A\end{align*}. Just remember that the length by itself on one side should be the side opposite the angle in the cosine ratio. Now we can plug in our values and solve for \begin{align*}a\end{align*}.

\begin{align*}a^2 &=16^2+22^2-2(16)(22) \cos 43^\circ \\ a^2 & \approx 225.127 \\ a & \approx 15 \end{align*}

**Example C**

Rae is making a triangular flower garden. One side is bounded by her porch and a second side is bounded by her fence. She plans to put in a stone border on the third side. If the length of the porch is 10 ft and the length of the fence is 15 ft and they meet at a \begin{align*}100^\circ\end{align*} angle, how many feet of stone border does she need to create?

**Solution:** Let the two known side lengths be \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and the angle between is \begin{align*}C\end{align*}. Now we can use the formula to find \begin{align*}c\end{align*}, the length of the third side.

\begin{align*}c^2 &=10^2+15^2-2(10)(15) \cos 100^\circ \\ c^2 & \approx 377.094 \\ c & \approx 19.4\end{align*}

So Rae will need to create a 19.4 ft stone border.

**Guided Practice**

1. Find \begin{align*}c\end{align*} when \begin{align*}m \angle C=75^\circ, a = 32\end{align*} and \begin{align*}b = 40\end{align*}.

2. Find \begin{align*}b\end{align*} when \begin{align*}m \angle B=120^\circ, a = 11\end{align*} and \begin{align*}c =17\end{align*}.

3. Dan likes to swim laps across a small lake near his home. He swims from a pier on the north side to a pier on the south side multiple times for a workout. One day he decided to determine the length of his swim. He determines the distances from each of the piers to a point on land and the angles between the piers from that point to be \begin{align*}50^\circ\end{align*}. How many laps does Dan need to swim to cover 1000 meters?

**Answers**

1. \begin{align*}c^2 &=32^2+40^2-2(32)(40) \cos 75^\circ \\ c^2 & \approx 1961.42 \\ c & \approx 44.3\end{align*}

2. \begin{align*}b^2 &=11^2+17^2-2(11)(17) \cos 120^\circ \\ b^2 & \approx 597 \\ b & \approx 24.4\end{align*}

3. \begin{align*}c^2 &=30^2+35^2-2(30)(35) \cos 50^\circ \\ c^2 & \approx 775.146 \\ c & \approx 27.84\end{align*}

Since each lap is 27.84 meters, Dan must swim \begin{align*}\frac{1000}{27.84} \approx 36\end{align*} laps.

**Problem Set**

Use the Law of Cosines to find the value of \begin{align*}x\end{align*}, to the nearest tenth, in problems 1 through 6.

For problems 7 through 10, find the unknown side of the triangle. Round your answers to the nearest tenth.

- Find \begin{align*}c\end{align*}, given \begin{align*}m \angle C=105^\circ\end{align*}, \begin{align*}a = 55\end{align*} and \begin{align*}b = 61\end{align*}.
- Find \begin{align*}b\end{align*}, given \begin{align*}m \angle B=26^\circ\end{align*}, \begin{align*}a = 33\end{align*} and \begin{align*}c = 24\end{align*}.
- Find \begin{align*}a\end{align*}, given \begin{align*}m \angle A=77^\circ\end{align*}, \begin{align*}b = 12\end{align*} and \begin{align*}c = 19\end{align*}.
- Find \begin{align*}b\end{align*}, given \begin{align*}m \angle B=95^\circ\end{align*}, \begin{align*}a = 28\end{align*} and \begin{align*}c = 13\end{align*}.
- Explain why when \begin{align*}m \angle C=90^\circ\end{align*}, the Law of Cosines becomes the Pythagorean Theorem.
- Luis is designing a triangular patio in his backyard. One side, 20 ft long, will be up against the side of his house. A second side is bordered by his wooden fence. If the fence and the house meet at a \begin{align*}120^\circ\end{align*} angle and the fence is 15 ft long, how long is the third side of the patio?

## Using the Law of Cosines with SSS (to find an angle)

**Objective**

Use the Law of Cosines to find the measure of an angle in a triangle in which all three side lengths are known.

**Guidance**

The Law of Cosines, \begin{align*}a^2+b^2-2ab \cos C\end{align*}, can be rearranged to facilitate the calculation of the measure of angle \begin{align*}C\end{align*} when \begin{align*}a, b\end{align*} and \begin{align*}c\end{align*} are all known lengths.

\begin{align*}a^2+b^2-2ab \cos C &=c^2 \\ a^2+b^2-c^2 &=2ab \cos C \\ \frac{a^2+b^2-c^2}{2ab} &=\cos C\end{align*}

which can be further manipulated to \begin{align*}C=\cos^{-1} \left(\frac{a^2+b^2-c^2}{2ab} \right)\end{align*}.

**Example A**

Find the measure of the largest angle in the triangle with side lengths 12, 18 and 21.

**Solution:** First, we must determine which angle will be the largest. Recall from Geometry that the longest side is opposite the largest angle. The longest side is 21 so we will let \begin{align*}c = 21\end{align*} since \begin{align*}C\end{align*} is the angle we are trying to find. Let \begin{align*}a =12\end{align*} and \begin{align*}b = 18\end{align*} and use the formula to solve for \begin{align*}C\end{align*} as shown. It doesn’t matter which sides we assign to \begin{align*}a\end{align*} and \begin{align*}b\end{align*}. They are interchangeable in the formula.

\begin{align*}m \angle C=\cos^{-1} \left(\frac{12^2+18^2-21^2}{2(12)(18)} \right) \approx 86^\circ\end{align*}

**Note:** Be careful to put parenthesis around the entire numerator and entire denominator on the calculator to ensure the proper order of operations. Your calculator screen should look like this:

\begin{align*}\cos^{-1}((12^2+18^2-21^2)/(2(12)(18)))\end{align*}

**Example B**

Find the value of \begin{align*}x\end{align*}, to the nearest degree.

**Solution:** The angle with measure \begin{align*}x^\circ\end{align*} will be angle \begin{align*}C\end{align*} so \begin{align*}c = 16, a = 22\end{align*} and \begin{align*}b = 8\end{align*}. Remember, \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are interchangeable in the formula. Now we can replace the variables with the known measures and solve.

\begin{align*}\cos^{-1} \left(\frac{22^2+8^2-16^2}{2(22)(8)} \right) \approx 34^\circ\end{align*}

**Example C**

Find the \begin{align*}m \angle A\end{align*}, if \begin{align*}a = 10, b = 15\end{align*} and \begin{align*}c = 21\end{align*}.

**Solution:** First, let’s rearrange the formula to reflect the sides given and requested angle:

\begin{align*}\cos A=\left(\frac{b^2+c^2-a^2}{2(b)(c)} \right)\end{align*}, now plug in our values \begin{align*}m \angle A=\cos^{-1} \left(\frac{15^2+21^2-10^2}{2(15)(21)} \right) \approx 26^\circ\end{align*}

**Guided Practice**

1. Find the measure of \begin{align*}x\end{align*} in the diagram:

2. Find the measure of the smallest angle in the triangle with side lengths 47, 54 and 72.

3. Find \begin{align*}m \angle B\end{align*}, if \begin{align*}a = 68, b = 56\end{align*} and \begin{align*}c = 25\end{align*}.

**Answers**

1. \begin{align*}\cos^{-1} \left(\frac{14^2+8^2-19^2}{2(14)(8)} \right) \approx 117^\circ\end{align*}

2. The smallest angle will be opposite the side with length 47, so this will be our \begin{align*}c\end{align*} in the equation.

\begin{align*}\cos^{-1} \left(\frac{54^2+72^2-47^2}{2(54)(72)} \right) \approx 41^\circ\end{align*}

3. Rearrange the formula to solve for \begin{align*}m \angle B, \cos B=\left(\frac{a^2+c^2-b^2}{2(a)(c)} \right); \cos^{-1} \left(\frac{68^2+25^2-56^2}{2(68)(25)} \right) \approx 52^\circ\end{align*}

**Problem Set**

Use the Law of Cosines to find the value of \begin{align*}x\end{align*}, to the nearest degree, in problems 1 through 6.

- Find the measure of the smallest angle in the triangle with side lengths 150, 165 and 200 meters.
- Find the measure of the largest angle in the triangle with side length 59, 83 and 100 yards.
- Find the \begin{align*}m \angle C\end{align*} if \begin{align*}a = 6, b = 9\end{align*} and \begin{align*}c=13\end{align*}.
- Find the \begin{align*}m \angle B\end{align*} if \begin{align*}a = 15, b = 8\end{align*} and \begin{align*}c = 9\end{align*}.
- Find the \begin{align*}m \angle A\end{align*} if \begin{align*}a = 24, b = 20\end{align*} and \begin{align*}c = 14\end{align*}.
- A triangular plot of land is bordered by a road, a fence and a creek. If the stretch along the road is 100 meters, the length of the fence is 115 meters and the side along the creek is 90 meters, at what angle do the fence and road meet?

## Heron’s Formula for the Area of a Triangle and Problem Solving with Trigonometry

**Objective**

Use Heron’s formula for area of a triangle when the side lengths are known and solve real world application problems using Law of Sines, Law of Cosines or the area formulas.

**Guidance**

Heron’s Formula, named after Hero of Alexandria 2000 years ago, can be used to find the area of a triangle given the three side lengths. The formula requires the semi-perimeter, \begin{align*}s\end{align*}, or \begin{align*}\frac{1}{2}(a+b+c)\end{align*}, where \begin{align*}a, b\end{align*} and \begin{align*}c\end{align*} are the lengths of the sides of the triangle.

Heron’s Formula: \begin{align*}\text{Area} =\sqrt{s(s-a)(s-b)(s-c)}\end{align*}

**Example A**

Use Heron’s formula to find the area of a triangle with side lengths 13 cm, 16 cm and 23 cm.

**Solution:** First, find the semi-perimeter or \begin{align*}s\end{align*}: \begin{align*}s=\frac{1}{2}(13+16+23)=26\end{align*}. Next, substitute our values into the formula as shown and evaluate:

\begin{align*}A=\sqrt{26(26-13)(26-16)(26-23)}=\sqrt{26(13)(10)(3)}=\sqrt{10140} \approx 101 \ cm^2\end{align*}

**Example B**

Alena is planning a garden in her yard. She is using three pieces of wood as a border. If the pieces of wood have lengths 4 ft, 6ft and 3 ft, what is the area of her garden?

**Solution:** The garden will be triangular with side lengths 4 ft, 6 ft and 3 ft. Find the semi-perimeter and then use Heron’s formula to find the area.

\begin{align*}s &=\frac{1}{2}(4+6+3)=\frac{13}{2} \\ A &=\sqrt{\frac{13}{2} \left(\frac{13}{2} -4 \right) \left(\frac{13}{2} -6 \right) \left(\frac{13}{2} -3 \right)}=\sqrt{\frac{13}{2} \left(\frac{5}{2} \right) \left(\frac{1}{2} \right) \left(\frac{7}{2} \right)}=\sqrt{\frac{455}{16}} \approx 28 \ ft^2\end{align*}

**Example C**

Caroline wants to measure the height of a radio tower. From some distance away from the tower, the angle of elevation from her spot to the top of the tower is \begin{align*}65^\circ\end{align*}. Caroline walks 100 m further away from the tower and measures the angle of elevation to the top of tower to be \begin{align*}48^\circ\end{align*}. How tall is the tower?

**Solution:** First, make a diagram to illustrate the situation.

We can use angle properties (linear pair and triangle sum) to find the angles shown in green in the diagram.

\begin{align*}180^\circ-65^\circ=115^\circ\end{align*} and \begin{align*}180^\circ-48^\circ-115^\circ=17^\circ\end{align*}

Next, we can use the Law of Sines in the obtuse triangle to find the hypotenuse in the right triangle:

\begin{align*}\frac{\sin 17^\circ}{100} &=\frac{\sin 48^\circ}{x} \\ x &=\frac{100 \sin 48^\circ}{\sin 17^\circ} \approx 254.18\end{align*}

Finally we can use the sine ratio in the right triangle to find the height of the tower:

\begin{align*}\sin 65^\circ=\frac{h}{254.18}, h=254.18 \sin 65^\circ \approx 230.37 \ m\end{align*}

**Guided Practice**

Use the most appropriate rule or formula (Law of Sines, Law of Cosines, area formula with sine or Heron’s formula) to answer the following questions.

1. Find the area of a triangle with side lengths 50 m, 45 m and 25 m.

2. Matthew is planning to fertilize his grass. Each bag of fertilizer claims to cover 500 sq ft of grass. His property of land is approximately in the shape of a triangle. He measures two sides of his yard to be 75 ft and 100 ft and the angle between them is \begin{align*}72^\circ\end{align*}. How many bags of fertilizer must he buy?

3. A pair of adjacent sides in a parallelogram are 3 in and 7 in and the angle between them is \begin{align*}62^\circ\end{align*}, find the length of the diagonals.

**Answers**

1. Heron’s Formula: \begin{align*}s=\frac{1}{2}(50+45+25)=60, A=\sqrt{60(60-50)(60-45)(60-25)} \approx 561 \ m^2\end{align*}.

2. Area formula with sine: \begin{align*}\frac{1}{2}(75)(100) \sin 72^\circ \approx 3566 \ ft^2\end{align*}, Number of bags \begin{align*}\frac{3566}{500} \approx 7.132 \approx 8\end{align*} bags. We round up because 7 bags is not quite enough.

3.

Law of Cosines to find the blue diagonal:

\begin{align*}c^2 &=3^2+7^2-2(3)(7) \cos 62^\circ \\ c &=\sqrt{38.28} \approx 6.19\end{align*}

To find the green diagonal we can use the Law of Cosines with the adjacent angle: \begin{align*}180^\circ-62^\circ-118^\circ\end{align*}:

\begin{align*}c^2 &=7^2+3^2-2(7)(3) \cos 118^\circ \\ c &=\sqrt{77.72} \approx 8.82\end{align*}

**Problem Set**

Use the Law of Sines, Law of Cosine, area of triangle with sine or Heron’s Formula to solve the real world application problems.

- Two observers, Rachel and Luis, are standing on the shore, 0.5 miles apart. They each measure the angle between the shoreline and a sailboat out on the water at the same time. If Rachel’s angle is \begin{align*}63^\circ\end{align*} and Luis’ angle is \begin{align*}56^\circ\end{align*}, find the distance between Luis and the sailboat to the nearest hundredth of a mile.
- Two pedestrians walk from opposite ends of a city block to a point on the other side of the street. The angle formed by their paths is \begin{align*}125^\circ\end{align*}. One pedestrian walks 300 ft and the other walks 320 ft. How long is the city block to the nearest foot?
- Two sides and the included angle of a parallelogram have measures 3.2 cm, 4.8 cm and \begin{align*}54.3^\circ\end{align*} respectively. Find the lengths of the diagonals to the nearest tenth of a centimeter.
- A bridge is supported by triangular braces. If the sides of each brace have lengths 63 ft, 46 ft and 40 ft, find the measure of the largest angle to the nearest degree.
- Find the triangular area, to the nearest square meter, enclosed by three pieces of fencing 123 m, 150 m and 155 m long.
- Find the area, to the nearest square inch, of a parallelogram with sides of length 12 in and 15 in and included angle of \begin{align*}78^\circ\end{align*}.
- A person at point \begin{align*}A\end{align*} looks due east and spots a UFO with an angle of elevation of \begin{align*}40^\circ\end{align*}. At the same time, another person, 1 mi due west of A looks due east and sights the same UFO with an angle of elevation of \begin{align*}25^\circ\end{align*}. Find the distance between \begin{align*}A\end{align*} and the UFO. How far is the UFO above the ground? Give answers to the nearest hundredth of a mile.
- Find the area of a triangular playground, to the nearest square meter, with sides of length 10 m, 15 m and 16 m.
- A yard is bounded on two sides with fences of length 80 ft and 60 ft. If these fences meet at a \begin{align*}75^\circ\end{align*} angle, how many feet of fencing are required to completely enclosed a triangular region?
- When a boy stands on the bank of a river and looks across to the other bank, the angle of depression is \begin{align*}12^\circ\end{align*}. If he climbs to the top of a 10 ft tree and looks across to other bank, the angle of depression is \begin{align*}15^\circ\end{align*}. What is the distance from the first position of the boy to the other bank of the river? How wide is the river? Give your answers to the nearest foot.

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Apr 23, 2013## Last Modified:

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