Trig The Law of Cosines

Section 3.3 The Law of Cosines

If we know two angles and one side of a triangle, we can use the Law of Sines to solve the triangle. We can also use the Law of Sines when we know two sides and the angle opposite one of them. But the Law of Sines is not helpful for the problem that opened this chapter, finding the distance from Avery to Clio. In this case we know two sides of the triangle, $a$ and $c\text{,}$ and the included angle, $B\text{.}$

$triangle$

To solve a triangle when we know two sides and the included angle, we will need a generalization of the Pythagorean theorem known as the Law of Cosines.

In a right triangle, with $C = 90\degree\text{,}$ the Pythagorean theorem tells us that

\begin{equation*} c^{2} = a^{2} + b^{2} \end{equation*}

If we allow angle $C$ to vary, but keep $a$ and $b$ the same length, the side $c$ will grow or shrink, depending on whether we increase or decrease the angle $C\text{,}$ as shown below.

$triangle$

The Pythagorean theorem is actually a special case of a more general law that applies to all triangles, no matter what the size of angle $C\text{.}$ The equation relating the three sides of a triangle is

\begin{equation*} c^{2} = a^{2} + b^{2} - 2ab \cos (C) \end{equation*}

You can see that when $C$ is a right angle, $\cos 90\degree = 0\text{,}$ so the equation reduces to the Pythagorean theorem.

We can write similar equations involving the angles or $A$ or $B\text{.}$ The three equations are all versions of the Law of Cosines.

Law of Cosines.

If the angles of a triangle are $A, B\text{,}$ and $C\text{,}$ and the opposite sides are respectively $a, b,$ and $c\text{,}$ then

\begin{gather*} \blert{a^{2} = b^{2} + c^{2} - 2bc \cos (A)}\\ \blert{b^{2} = a^{2} + c^{2} - 2ac \cos (B)}\\ \blert{c^{2} = a^{2} + b^{2} - 2ab \cos (C)} \end{gather*}

Note 3.40.

For a proof of the Law of Cosines, see Homework Problems 57 and 58.

Subsection Finding a Side

Now we can solve the problem of the distance from Avery to Clio. Here is the figure from Section 3.1 showing the location of the three towns.

$triangle$

Example 3.41.

How far is it from Avery to Clio?

Solution.

The angle $\angle ABC = 35\degree + 90\degree = 125\degree\text{.}$ Thus, in $\triangle ABC$ we have $a = 34,~ c = 48$ and $B = 125\degree\text{.}$ The distance from Avery to Clio is represented by $b$ in the figure.

$triangle$

We know two sides and the included angle, and we choose the version of the Law of Cosines that uses our known angle, $B\text{,}$ and substitute the other known values.

\begin{align*} b^{2} \amp = a^{2} + c^{2} - 2ac \cos (B) \\ b^{2} \amp = 34^{2} + 48^{2} - 2(34)(48) \cos (125\degree) \amp\amp \blert{\text{Simplify the right side.}}\\ b^{2} \amp = 3460 - 3264 \cos (125\degree) = 5332.153 \amp\amp \blert{\text{Take square roots.}} \\ b \amp = 73.02 \end{align*}

Avery is about 73 miles from Clio.

Caution 3.42.

When simplifying the Law of Cosines, be careful to follow the order of operations. In the previous example, the right side of the equation

\begin{align*} b^{2} \amp = 3460 - 3264 \cos (125\degree)\\ b^{2} \amp = 3460 - 3264(-0.5736) \end{align*}

has two terms, because 3264 is the coefficient of $\cos (125\degree)\text{,}$ so it would be incorrect to subtract 3264 from 3460; we should multiply $3264(-0.5736)$ first. If you are using a graphing calculator, you can enter the right side of the equation exactly as it is written.

Checkpoint 3.43.

In $\triangle ABC\text{,}$ $a = 11,~c = 23\text{,}$ and $B = 87\degree\text{.}$ Find $b\text{,}$ and round your answer to two decimal places.

Answer.

We use the Law of Cosines.

\begin{equation*} b^2 = 11^2 + 23^2 - 11(23)\cos (87\degree) = 623.52 \end{equation*}

so $b = \sqrt{623.52} = 24.97\text{.}$

Subsection Finding an Angle

We can also use the Law of Cosines to find an angle when we know all three sides of a triangle. Pay close attention to the algebraic steps used to solve the equation in the next example.

Example 3.44.

In the triangle at right, $a = 6,~ b = 7\text{,}$ and $c = 11\text{.}$ Find angle $C\text{.}$

$triangle$

Solution.

We choose the version of the Law of Cosines that uses angle $C\text{.}$

\begin{align*} c^{2} \amp = a^{2} + b^{2} - 2ab \cos (C) \amp\amp \blert{\text{Substitute the known values.}}\\ 11^{2} \amp = 6^{2} + 7^{2} - 2(6)(7) \cos (C) \amp\amp \blert{\text{Simplify each side.}}\\ 121 \amp = 36 + 49 - 84 \cos (C) \amp\amp \blert{\text{Isolate the cosine term.}}\\ 36 \amp = -84 \cos (C) \amp\amp \blert{\text{Solve for cos (C).}}\\ \dfrac{-3}{7} \amp = \cos (C) \amp\amp \blert{\text{Solve for C.}}\\ C \amp = \cos^{-1} \left(\dfrac{-3}{7}\right) = 115.4\degree \end{align*}

Angle $C$ is about $115.4\degree\text{.}$

Checkpoint 3.45.

In $\triangle ABC\text{,}$ $a = 5.3,~b = 4.7\text{,}$ and $c = 6.1\text{.}$ Find angle $B\text{,}$ and round your answer to two decimal places.

Answer.

Substitute the known values into the Law of Cosines to get

\begin{equation*} 4.7^2 = 5.3^2 + 6.1^2 - 2(5.3)(6.1)\cos(B) \end{equation*}

from which we find $\cos(B) = 0.6683\text{,}$ so $B = \cos^{-1}(0.6683) = 48.07\degree\text{.}$

Once we have calculated one of the angles in a triangle, we can use either the Law of Sines or the Law of Cosines to find a second angle. Here is how we would use the Law of Sines to find angle $A$ in the previous Example.

\begin{align*} \dfrac{\sin (A)}{a} \amp = \dfrac{\sin (C)}{c} \amp\amp \blert{\text{Substitute the known values.}}\\ \dfrac{\sin (A)}{6} \amp = \dfrac{\sin (115.4\degree)}{11} \amp\amp \blert{\text{Solve for} \sin A.}\\ \sin (A) \amp = 6 \cdot \dfrac{\sin (115.4\degree)}{11} \approx 0.4928 \end{align*}

Thus, $A = \sin^{-1}(0.4928) = 29.5\degree\text{.}$ There are two angles with sine $29.5\degree\text{,}$ but we know that $A$ is an acute angle because it is opposite the shortest side of the triangle. Finally,

\begin{equation*} B = 180\degree - (A + C) \approx 35.1\degree \end{equation*}

Alternatively, we can use the Law of Cosines to find angle $A\text{.}$

\begin{align*} a^{2} \amp = b^{2} + c^{2} - 2bc \cos (A) \amp\amp \blert{\text{Substitute the known values.}}\\ 6^{2} \amp = 7^{2} + 11^{2} - 2(7)(11) \cos (A) \amp\amp \blert{\text{Simplify each side.}}\\ 36 \amp = 49 + 121 - 154 \cos (A) \amp\amp \blert{\text{Isolate the cosine term.}}\\ -134 \amp = -154 \cos (A) \amp\amp \blert{\text{Solve for} \cos (A).}\\ \dfrac{67}{77} \amp = \cos (A) \amp\amp \blert{\text{Solve for A.}}\\ A \amp = \cos^{-1} \left(\dfrac{67}{77}\right) = 29.5\degree \end{align*}

Note 3.46.

You may notice that the Law of Cosines requires more calculation than the Law of Sines, but there is only one possible angle with the given cosine. If we use the Law of Sines there are two possible angles with the given sine, and either or both may fit the triangle.

Subsection Navigation

Even with the aid of GPS (Global Positioning System) instruments, aircraft pilots and ship captains need to understand navigation based on trigonometry.

Example 3.47.

The sailing club leaves the marina on a heading $15\degree$ east of north and sails for 18 miles. They then change course, and after traveling for 12 miles on a heading $35\degree$ east of north, they experience engine trouble and radio for help. The marina sends a speed boat to rescue them. How far should the speed boat go, and on what heading?

Solution.

We’d like to find the distance $b$ and the angle $\theta$ shown in the figure. In $\triangle ABC\text{,}$ we can calculate the angle at point $B$ where the sailing club changed course:

\begin{equation*} B = 180\degree - 35\degree + 15\degree = 160\degree \end{equation*}

We know $a = 12$ and $c = 18\text{.}$ First we use the Law of Cosines to find the distance $b\text{.}$

$triangle$

\begin{align*} b^{2} \amp = a^{2} + c^{2} - 2ac \cos (B) \amp\amp \blert{\text{Substitute the known values.}}\\ \amp = 12^{2} + 18^{2} - 2(12)18 \cos (160\degree) \amp\amp \blert{\text{Evaluate.}}\\ \amp = 873.95\amp\amp \blert{\text{Take positive square root.}}\\ b \amp = 29.56 \end{align*}

Next, we apply the Law of Cosines again to find $\angle A\text{.}$

\begin{align*} a^{2} \amp = b^{2} + c^{2} - 2bc \cos (A) \\ 12^{2} \amp = 29.56^{2} + 18^{2} - 2(29.56)(18) \cos (A) \amp\amp \blert{\text{Solve for} \cos (A).}\\ \cos (A) \amp = \dfrac{12^{2} - 29.56^{2} - 18^{2}}{-2(29.56)(18)} = 0.9903\\ C \amp = \cos^{-1} (0.9903) = 8\degree \end{align*}

Thus, $b = 29.56$ and $\theta = 8\degree + 15\degree = 23\degree\text{.}$ The speed boat should travel 29.56 miles on a heading $23\degree$ east of north.

Checkpoint 3.48.

Howard wants to fly from Anchorage to Nome, Alaska, a distance of 540 miles on a heading $57\degree$ west of north. After flying for some time, he discovers that his heading is in error, and he is actually flying $47\degree$ west of north. Howard corrects his flight plan and changes course when he is exactly 200 miles from Anchorage. What is his new heading, and how far is he from Nome?

$triangle$

Answer.

We use the Law of Cosines to find his distance from Nome:

\begin{equation*} d^2 = 540^2+200^2-2(200)(540)\cos (10\degree) \end{equation*}

Solving for $d$ gives 344.8. Next, use either law to find that angle $B$ is $164.22\degree\text{.}$ The new heading is $47\degree$ plus the supplement of $B\text{,}$ or $47\degree + 15.78\degree = 62.78\degree$ west of north, and the distance is 344.8 miles.

Subsection Which Law to Use

How can we decide which law, the Law of Sines or the Law of Cosines, is appropriate for a given problem?

If we are solving a right triangle, we don’t need the Laws of Sines and Cosines; all we need are the definitions of the trigonometric ratios.
But for oblique triangles, we can identify the following cases:

How to Solve an Oblique Triangle.

If we know:	We can use:
1. One side and two angles (SAA)	1. Law of Sines, to find another side
2. Two sides and the included angle (SAS)	2. Law of Cosines, to find the third side
3. Three sides (SSS)	3. Law of Cosines, to find an angle
4. Two sides and the angle opposite one of them (SSA, the ambiguous case)	4. Law of Sines, to find another angle, or Law of Cosines, to find another side

Example 3.49.

In the triangle at right, which law should you use to find $\angle B\text{?}$

$triangle$

Solution.

We know two sides of the triangle and the angle opposite one of them. We can use the Law of Sines to find $\angle B\text{.}$

\begin{align*} \dfrac{\sin (B)}{6} \amp = \dfrac{\sin (40\degree)}{10}\\ \sin (B) \amp = 0.3857 \end{align*}

There are two angles between $0\degree$ and $180\degree$ with sine $0.3857\text{,}$ namely $22.7\degree$ and its supplement, $157.3\degree\text{.}$ Both of these angles might produce a solution. However, in this case we notice that because $a \gt b\text{,}$ angle $B$ must be acute.

Checkpoint 3.50.

In the triangle at right, which part of the triangle can you find, and which law should you use?

$triangle$

Answer.

We know two sides and the included angle, so we can find side $c$ using the Law of Cosines.

In the previous Example we used the Law of Sines to find an angle. Because there are always two angles with a given sine, we should check whether both possible angles result in a triangle. But here is another approach: We can apply the Law of Cosines to find the third side first. With that method, we’ll need the quadratic formula.

Quadratic Formula.

The solutions of the quadratic equation $ax^{2} + bx + c = 0,~~a \not= 0,$ are given by

\begin{equation*} \blert{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}} \end{equation*}

Example 3.51.

In the triangle at right, use the Law of Cosines to find $c\text{.}$

$triangle$

Solution.

We will use the Law of Cosines with angle $A\text{.}$

\begin{align*} a^2 \amp = b^2 + c^2 -2bc \cos (A)\\ 10^{2} \amp = 6^{2} + c^{2} - 2(6)c \cos (40\degree)\\ 100 \amp = 36 + c^2 - 12 (0.7660) c \end{align*}

This is a quadratic equation in $c\text{,}$ so we rearrange the terms into standard form.

\begin{align*} c^{2} - 12(0.7660)c - 64 \amp = 0\\ c^{2} - 9.1925c - 64 \amp = 0 \end{align*}

Now we can solve for $c$ using the quadratic formula.

\begin{equation*} c = \dfrac{9.1925 \pm \sqrt{9.1925^2 - 4(1)(-64)}}{2(1)} \end{equation*}

There are two solutions, $13.8$ and $-4.6\text{.}$ We can ignore the negative solution, so $c = 13.8\text{.}$

Subsection Using the Law of Cosines for the Ambiguous Case

In Section 3.2 we encountered the ambiguous case:

If we know two sides $a$ and $b$ of a triangle and the acute angle $\alpha$ opposite one of them, there may be one solution, two solutions, or no solution, depending on the size of $a$ in relation to $b$ and $\alpha\text{,}$ as shown below.

The Ambiguous Case.

No solution: $a \lt b \sin (\alpha)$

$triangle$

$\alpha$ is too short to make a triangle.
One solution: $a = b \sin (\alpha)$

$triangle$

$\alpha$ is exactly the right length to make a right triangle.
Two solutions: $b \sin (\alpha) \lt a \lt b$

$triangle$
One solution: $a \gt b$

$triangle$

Note 3.52.

If $\alpha$ is an obtuse angle, things are simpler: there is one solution if $a \gt b\text{,}$ and no solution if $a \le b\text{.}$

Tp discover which case applies (no triangle, one triangle, or two triangles), we can begin by using the Law of Cosines to find the third side. As we saw in the previous Example, for this method we need the quadratic formula.

A quadratic equation can have one solution, two solutions, or no solution, depending on the value of the discriminant, $b^{2} - 4ac\text{.}$

If the quadratic equation has one positive solution, there is one triangle.
If the quadratic equation has two positive solutions, there are two triangles.
If the quadratic equation has no positive solutions, there is no triangle with the given properties.

Example 3.53.

In $\triangle ABC,~ B = 14.4\degree,~ a = 8\text{,}$ and $b = 3\text{.}$ Solve the triangle.

Solution.

We begin by finding the third side of the triangle. Using the Law of Cosines, we have

\begin{align*} b^{2} \amp = a^{2} + c^{2} - 2ac \cos (B) \amp\amp \blert{\text{Substitute the known values.}}\\ 3^{2} \amp = 8^{2} + c^{2} - 2(8)c \cos (14.4\degree) \amp\amp \blert{\text{Simplify.}}\\ 9 \amp = 64 + c^{2} - 16c (0.9686) \amp\amp \blert{\text{Write the equation in standard form.}}\\ 0 \amp = c^{2}-15.497c + 55 \amp\amp \blert{\text{Apply the quadratic formula.}} \end{align*}

\begin{align*} c \amp = \dfrac{15.497 \pm \sqrt{(-15.497)^{2} - 4(1)(55)}}{2(1)} \amp\amp \blert{\text{Simplify.}}\\ \amp = \dfrac{15.497 \pm 4.490}{2} = 5.503~ \text{ or }~ 9.994 \end{align*}

Because there are two positive solutions for side $c\text{,}$ either $c = 5.503$ or $c = 9.994\text{,}$ there are two triangles with the given properties. We apply the Law of Cosines again to find angle $C$ in each triangle.

For the triangle with $c = 5.503\text{,}$ we have

\begin{align*} c^{2} \amp = a^{2} + b^{2} - 2ab \cos (C) \amp\amp \blert{\text{Substitute the known values.}}\\ 5.503^{2} \amp = 8^{2} + 3^{2} - 2(8)(3) \cos (C) \amp\amp \blert{\text{Solve for} \cos (C).}\\ \cos (C) \amp = \dfrac{5.503^{2} - 64 - 9}{-48} = 0.8899\\ C \amp = \cos^{-1} (0.8899) = 27.1\degree \end{align*}

and $A = 180\degree - (14.4\degree + 27.1\degree) = 138.5\degree\text{.}$

For the triangle with $c = 9.994\text{,}$ we have

\begin{align*} c^{2} \amp = a^{2} + b^{2} - 2ab \cos (C) \amp\amp \blert{\text{Substitute the known values.}}\\ 9.994^{2} \amp = 8^{2} + 3^{2} - 2(8)(3) \cos (C) \amp\amp \blert{\text{Solve for} \cos (C).}\\ \cos (C) \amp = \dfrac{9.994^{2} - 64 - 9}{-48} = -0.5600\\ C \amp = \cos^{-1} (-0.5600) = 124.1\degree \end{align*}

and $A = 180\degree - (14.4\degree + 124.1\degree) = 41.5\degree\text{.}$

Both triangles are shown below.

$triangle$

Checkpoint 3.54.

Use the Law of Cosines to find all triangles $ABC$ with $A = 48\degree,~ a = 10\text{,}$ and $b = 15\text{.}$

Answer.

Using the Law of Cosines to find side $c\text{,}$ we get the equation

\begin{align*} a^2 \amp = b^2 + c^2 - 2bc \cos (A)\\ 100 = 225 + c^2 - 30c \cos(48\degree) \end{align*}

Putting this quadratic equation into standard form we get

\begin{equation*} c^2 - 20.07c + 125 = 0 \end{equation*}

The discriminant is

\begin{equation*} b^2 - 4ac = 20.07^2 - 4(125) = -97.2 \end{equation*}

Because the discriminant is negative, the equation has no solution, and there is no triangle satisfying the stated conditions.

Review the following skills you will need for this section.

Algebra Refresher 3.55.

Solve each quadratic equation.

$\displaystyle 2.5x^{2} + 6.2 = 166.2$
$\displaystyle 0.8x^{2} - 124 = 376$
$\displaystyle 2x(2x - 3) = 208$
$\displaystyle 3x(x + 5) = 900$
$\displaystyle 2x^{2} - 6x = 233.12$
$\displaystyle 0.5x^{2} + 1.5x = 464$

$\underline{\qquad\qquad\qquad\qquad}$

Algebra Refresher Answers

$\displaystyle \pm 8$
$\displaystyle \pm 25$
$\displaystyle 8,~ {-6.5}$
$\displaystyle 15,~ {-20}$
$\displaystyle 12.4,~ {-9.4}$
$\displaystyle 29,~ {-32}$

Subsection Section 3.3 Summary

Subsubsection Vocabulary

Quadratic equation
Quadratic formula
Discriminant

Subsubsection Concepts

The Law of Sines is not helpful when we know two sides of the triangle and the included angle. In this case we need the Law of Cosines.
Law of Cosines.

If the angles of a triangle are $A, B\text{,}$ and $C\text{,}$ and the opposite sides are respectively $a, b,$ and $c\text{,}$ then

\begin{gather*} a^{2} = b^{2} + c^{2} - 2bc \cos (A)\\ b^{2} = a^{2} + c^{2} - 2ac \cos (B)\\ c^{2} = a^{2} + b^{2} - 2ab \cos (C) \end{gather*}
We can also use the Law of Cosines to find an angle when we know all three sides of a triangle.
We can use the Law of Cosines to solve the ambiguous case.

How to Solve an Oblique Triangle.

If we know:	We can use:
1. One side and two angles (SAA)	1. Law of Sines, to find another side
2. Two sides and the angle opposite one of them (SSA, the ambiguous case)	2. Law of Sines, to find another angle, or Law of Cosines, to find another side
3. Two sides and the included angle (SAS)	3. Law of Cosines, to find the third side
4. Three sides (SSS)	4. Law of Cosines, to find an angle

Subsubsection Study Questions

The Law of Cosines is really a generalization of what familiar theorem?
If you know all three sides of a triangle and one angle, what might be the advantage in using the Law of Cosines o find another angle, instead of the Law of Sines?
State the quadratic formula from memory. Try to sing the quadratic formula to the tune of "Pop Goes the Weasel."
Francine is solving a triangle in which $a = 20,~ b = 16\text{,}$ and $A = 26\degree\text{.}$ She finds that $\sin B = 0.3507\text{.}$ How does she know that $B = 20.5\degree\text{,}$ and not $159.5\degree\text{?}$

Subsubsection Skills

Use the Law of Cosines to find the side opposite an angle #7-12
Use the Law of Cosines to find an angle #13-20
Use the Law of Cosines to find a side adjacent to an angle #21-26
Decide which law to use #27-34
Solve a triangle #35-42
Solve problems using the Law of Cosines #43-56

Exercises Homework 3.3

1.

Simplify $~~5^2 + 7^2 - 2(5)(7)\cos (\theta)$
Evaluate the expression in part (a) for $\theta = 29\degree$
Evaluate the expression in part (a) for $\theta = 151\degree$

2.

Simplify $~~26.1^2 + 32.5^2 - 2(26.1)(32.5)\cos (\phi)$
Evaluate the expression in part (a) for $\phi = 64\degree$
Evaluate the expression in part (a) for $\phi = 116\degree$

3.

Solve $~~b^2 = a^2 + c^2 - 2ac \cos (\beta)~~$ for $~~ \cos (\beta)$
For the equation in part (a), find $\cos \beta$ if $a = 5,~ b = 11\text{,}$ and $c = 8\text{.}$

4.

Solve $~~a^2 = b^2 + c^2 - 2bc \cos (\alpha)~~$ for $~~ \cos (\alpha)$
For the equation in part (a), find $\cos (\alpha)$ if $a = 4.6,~ b = 7.2\text{,}$ and $c = 9.4\text{.}$

5.

The equation $~~9^2 = b^2 + 4^2 - 2b(4) \cos (\alpha)~~$ is quadratic in $b\text{.}$ Write the equation in standard form.
Solve the equation in part (a) for $b$ if $\alpha = 48\degree\text{.}$

6.

The equation $~~6^2 = 5^2 + c^2 - 2(5)c \cos (\beta)~~$ is quadratic in $c\text{.}$ Write the equation in standard form.
Solve the equation in part (a) for $c$ if $\beta = 126\degree\text{.}$

Exercise Group.

For Problems 7–12, use the Law of Cosines to find the indicated side. Round to two decimal places.

7.

$triangle$

8.

$triangle$

9.

$triangle$

10.

$triangle$

11.

$triangle$

12.

$triangle$

Exercise Group.

For Problems 13–16,use the Law of Cosines to find the indicated angle. Round to two decimal places.

13.

$triangle$

14.

$triangle$

15.

$triangle$

16.

$triangle$

Exercise Group.

For Problems 17–20, find the angles of the triangle. Round answers to two decimal places.

17.

$a = 23,~ b = 14,~ c = 18$

18.

$a = 18,~ b = 25,~ c = 19$

19.

$a = 16.3,~ b = 28.1,~ c = 19.4$

20.

$a = 82.3,~ b = 22.5,~ c = 66.8$

Exercise Group.

For Problems 21–26, use the Law of Cosines to find the unknown side. Round your answers to two decimal places.

21.

$triangle$

22.

$triangle$

23.

$triangle$

24.

$triangle$

25.

$triangle$

26.

$triangle$

Exercise Group.

For Problems 27–34, which law should you use to find the labeled unknown value, the Law of Sines or the Law of Cosines? Write an equation you can solve to find the unknown value. For Problems 31-34, you may need two steps to find the unknown value.

27.

$triangle$

28.

$triangle$

29.

$triangle$

30.

$triangle$

31.

$triangle$

32.

$triangle$

33.

$triangle$

34.

$triangle$

Exercise Group.

For Problems 35–42,

Sketch and label the triangle.
Solve the triangle. Round answers to two decimal places.

35.

$B = 47\degree,~a = 23,~ c = 17$

36.

$C = 32\degree, ~a = 14,~ b = 18$

37.

$a = 8,~ b =7,~ c = 9$

38.

$a = 23,~ b = 34,~ c = 45$

39.

$b = 72,~ c = 98,~ B = 38\degree$

40.

$a = 28,~ c = 41,~A = 27\degree$

41.

$c =5.7,~A = 59\degree,~B = 82\degree$

42.

$b = 82,~ A = 11\degree,~C = 42\degree$

Exercise Group.

For Problems 43–52,

Sketch and label a triangle to illustrate the problem.
Solve the problem. Round answers to one decimal place.

43.

A surveyor would like to know the distance $PQ$ across a small lake, as shown in the figure. She stands at point $O$ and measures the angle between the lines of sight to points $P$ and $Q$ at $76\degree\text{.}$ She also finds $OP = 1400$ meters and $OQ = 600$ meters. Calculate the distance $PQ\text{.}$

$lake$

44.

Highway engineers plan to drill a tunnel through Boney Mountain from $G$ to $H\text{,}$ as shown in the figure. The angle at point $F$ is $41\degree\text{,}$ and the distances to $G$ and $H$ are 900 yards and 2500 yards, respectively. How long will the tunnel be?

$mountains$

45.

Two pilots leave an airport at the same time. One pilot flies $3\degree$ east of north at a speed of 320 miles per hour, the other flies $157\degree$ east of north at a speed of 406 miles per hour. How far apart are the two pilots after 3 hours? What is the heading from the first plane to the second plane at that time?

46.

Two boats leave port at the same time. One boat sails due west at a speed of 17 miles per hour, the other powers $42\degree$ east of north at a speed of 23 miles per hour. How far apart are the two boats after 2 hours? What is the heading from the first boat to the second boat at that time?

47.

Caroline wants to fly directly south from Indianapolis to Cancun, Mexico, a distance of 1290 miles. However, to avoid bad weather, she flies for 400 miles on a heading $18\degree$ east of south. What is the heading to Cancun from that location, and how far is it?

48.

Alex sails 8 miles from Key West, Florida on a heading $40\degree$ east of south. He then changes course and sails for 10 miles due east. What is the heading back to Key West from that point, and how far is it?

49.

The phone company wants to erect a cell tower on a steep hill inclined $26\degree$ to the horizontal. The installation crew plans to run a guy wire from a point on the ground 20 feet uphill from the base of the tower and attach it to the tower at a height of 100 feet. How long should the guy wire be?

50.

Sandstone Peak rises 3500 feet above the desert. The Park Service plans to run an aerial tramway up the north face, which is inclined at an angle of $68\degree$ to the horizontal. The base station will be located 500 feet from the foot of Sandstone Peak. Ignoring any slack in the cable, how long should it be?

51.

The sides of a triangle are 27 cm, 15 cm, and 20 cm. Find the area of the triangle. (Hint: Find one of the angles first.)

52.

The sides of a parallelogram are 10 inches and 8 inches, and form an angle of $130\degree\text{.}$ Find the lengths of the diagonals of the parallelogram.

Exercise Group.

For Problems 53–56, find $x\text{,}$ the distance from one vertex to the foot of the altitude.

53.

$triangle$

54.

$triangle$

55.

$triangle$

56.

$triangle$

Exercise Group.

Problems 57 and 58 prove the Law of Cosines.

$triangles$

57.

Copy the three figures above showing the three possibilities for an angle $C$ in a triangle: $C$ is acute, obtuse, or a right angle. For each figure, explain why it is true that $c^2 = (b - x)^2 + y^2\text{,}$ then rewrite the right side to get $c^2 = (x^2 + y^2) + b^2 - 2bx\text{.}$
For each figure, explain why it is true that $x^2 + y^2 = a^2\text{.}$
For all three figures, $a$ is the distance from the origin to the point $(x,y)\text{.}$ Use the definition of cosine to write $\cos (C)$ in terms of $a$ and $x\text{,}$ then solve your equation for $x\text{.}$
Start with the last equation from (a), and substitute expressions from (b) and (c) to conclude one case of the Law of Cosines.

58.

Demonstrate the other two cases of the Law of Cosines:

$\displaystyle a^2 = b^2 + c^2 - 2bc \cos (A)$
$\displaystyle b^2 = a^2 + c^2 - 2ac \cos (B)$

(Hint: See Problem 57 and switch the roles of $a$ and $c\text{,}$ etc.)

59.

Use the Law of Cosines to prove the projection laws:

\begin{align*} a \amp = b \cos (C) + c \cos (B)\\ b \amp = c \cos (A) + a \cos (C)\\ c \amp = a \cos (B) + b \cos (A) \end{align*}

Illustrate with a sketch. (Hint: Add together two of the versions of the Law of Cosines.)

60.

If $\triangle ABC$ is isosceles with $a = b\text{,}$ show that $c^2 = 2a^2(1 - \cos (C))\text{.}$

61.

Use the Law of Cosines to prove:

\begin{align*} 1 + \cos (A) \amp = \dfrac{(a + b + c)(-a + b + C)}{2bc}\\ 1 - \cos (A) \amp = \dfrac{(a - b + c)(-a + b + C)}{2bc} \end{align*}

62.

Prove that

\begin{equation*} \dfrac{\cos (A)}{a} + \dfrac{\cos (B)}{b} + \dfrac{\cos (C)}{c} = \dfrac{a^2 + b^2 + c^2}{2abc} \end{equation*}

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