Non-right Triangles: Law of Cosines

Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1. How far from port is the boat?

A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees. — Figure 1

Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.

Using the Law of Cosines to Solve Oblique Triangles

The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.

Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle $A B C$ is placed in the coordinate plane with vertex $A$ at the origin, side $c$ drawn along the x-axis, and vertex $C$ located at some point $(x, y)$ in the plane, as illustrated in Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.

A triangle A B C plotted in quadrant 1 of the x,y plane. Angle A is theta degrees with opposite side a, angles B and C, with opposite sides b and c respectively, are unknown. Vertex A is located at the origin (0,0), vertex B is located at some point (x-c, 0) along the x-axis, and point C is located at some point in quadrant 1 at the point (b times the cos of theta, b times the sin of theta). — Figure 2

We can drop a perpendicular from $C$ to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that

\cos θ = \frac{x (adjacent)}{b (hypotenuse)} and \sin θ = \frac{y (opposite)}{b (hypotenuse)}

In terms of $θ, x = b \cos θ$ and $y = b \sin θ .$ The $(x, y)$ point located at $C$ has coordinates $(b \cos θ,$ $b \sin θ) .$ Using the side $(x - c)$ as one leg of a right triangle and $y$ as the second leg, we can find the length of hypotenuse $a$ using the Pythagorean Theorem. Thus,

\begin{array}{l} a^{2} = {(x - c)}^{2} + y^{2} \\ = {(b \cos θ - c)}^{2} + {(b \sin θ)}^{2} & Substitute (b \cos θ) for x and (b \sin θ) for y . \\ = (b^{2} \cos^{2} θ - 2 b c \cos θ + c^{2}) + b^{2} \sin^{2} θ & Expand the perfect square . \\ = b^{2} \cos^{2} θ + b^{2} \sin^{2} θ + c^{2} - 2 b c \cos θ & Group terms noting that \cos^{2} θ + \sin^{2} θ = 1. \\ = b^{2} (\cos^{2} θ + \sin^{2} θ) + c^{2} - 2 b c \cos θ & Factor out b^{2} . \\ a^{2} = b^{2} + c^{2} - 2 b c \cos θ \end{array}

The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.

Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.

Law of Cosines

The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 3, with angles $α, β,$ and $γ,$ and opposite corresponding sides $a, b,$ and $c,$ respectively, the Law of Cosines is given as three equations.

\begin{array}{l} a^{2} = b^{2} + c^{2} - 2 b c \cos α \\ b^{2} = a^{2} + c^{2} - 2 a c \cos β \\ c^{2} = a^{2} + b^{2} - 2 a b \cos γ \end{array}

Figure 3

To solve for a missing side measurement, the corresponding opposite angle measure is needed.

When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.

\begin{array}{l} \begin{array}{l} \begin{array}{l} \cos α = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \end{array} \\ \cos β = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \\ \cos γ = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \end{array} \end{array}

Example 1

Finding the Unknown Side and Angles of a SAS Triangle

Find the unknown side and angles of the triangle in Figure 4.

A triangle with standard labels. Side a = 10, side c = 12, and angle beta = 30 degrees. — Figure 4

Solution

First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.

Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side $b,$ as we know the measurement of the opposite angle $β .$

\begin{array}{l} b^{2} = a^{2} + c^{2} - 2 a c \cos β \\ b^{2} = 10^{2} + 12^{2} - 2 (10) (12) \cos (30^{\circ}) \begin{matrix}  \end{matrix} & Substitute the measurements for the known quantities . \\ b^{2} = 100 + 144 - 240 (\frac{\sqrt{3}}{2}) & Evaluate the cosine and begin to simplify . \\ b^{2} = 244 - 120 \sqrt{3} \\ b = \sqrt{244 - 120 \sqrt{3}} & Use the square root property . \\ b \approx 6.013 \end{array}

Because we are solving for a length, we use only the positive square root. Now that we know the length $b,$ we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle $α,$ we have

\begin{array}{l} \frac{\sin α}{a} = \frac{\sin β}{b} \\ \frac{\sin α}{10} = \frac{\sin (30°)}{6.013} \\ \sin α = \frac{10 \sin (30°)}{6.013} & Multiply both sides of the equation by 10 . \\ α = \sin^{- 1} (\frac{10 \sin (30°)}{6.013}) \begin{matrix}  \end{matrix} & Find the inverse sine of \frac{10 \sin (30°)}{6.013} . \\ α \approx 56.3° \end{array}

The other possibility for $α$ would be $α = 180° - 56.3° \approx 123.7°.$ In the original diagram, $α$ is adjacent to the longest side, so $α$ is an acute angle and, therefore, $123.7°$ does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between $0°$ and $180°.$ Proceeding with $α \approx 56.3°,$ we can then find the third angle of the triangle.

γ = 180° - 30° - 56.3° \approx 93.7°

The complete set of angles and sides is

\begin{array}{l} α \approx 56.3° \begin{matrix}  \end{matrix} & a = 10 \\ β = 30° & b \approx 6.013 \\ γ \approx 93.7° & c = 12 \end{array}

Example 2

Solving for an Angle of a SSS Triangle

Find the angle $α$ for the given triangle if side $a = 20,$ side $b = 25,$ and side $c = 18.$

Solution

For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle $α,$ we have

\begin{array}{l} a^{2} = b^{2} + c^{2} −2 b c \cos α \\ 20^{2} = 25^{2} + 18^{2} −2 (25) (18) \cos α & Substitute the appropriate measurements . \\ 400 = 625 + 324 - 900 \cos α & Simplify in each step . \\ 400 = 949 - 900 \cos α \\ −549 = −900 \cos α & Isolate cos α . \\ \frac{−549}{−900} = \cos α \\ 0.61 \approx \cos α \\ \cos^{−1} (0.61) \approx α & Find the inverse cosine . \\ α \approx 52.4° \end{array}

See Figure 5.

A triangle with standard labels. Side b =25, side a = 20, side c = 18, and angle alpha = 52.4 degrees. — Figure 5

Analysis

Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.

Solving Applied Problems Using the Law of Cosines

Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.

Example 3

Using the Law of Cosines to Solve a Communication Problem

On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5,050 feet from the first tower and 2,420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.

Solution

For simplicity, we start by drawing a diagram similar to Figure 6 and labeling our given information.

A triangle formed between the two cell phone towers located on am east to west highway and the cellphone between and north of them. The side between the two towers is 6000 feet, the side between the left tower and the phone is 5050 feet, and the side between the right tower and the phone is 2420 feet. The angle between the 5050 and 6000 feet sides is labeled theta. — Figure 6

Using the Law of Cosines, we can solve for the angle $θ .$ Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let $a = 2,420, b = 5,050,$ and $c = 6,000.$ Thus, $θ$ corresponds to the opposite side $a = 2,420.$

\begin{matrix} a^{2} & = & b^{2} + c^{2} - 2 b c \cos θ \\ {(2, 420)}^{2} & = & {(5, 050)}^{2} + {(6, 000)}^{2} - 2 (5, 050) (6, 000) \cos θ \\ \begin{matrix} {(2, 420)}^{2} & - & {(5, 050)}^{2} - {(6, 000)}^{2} \end{matrix} & = & - 2 (5, 050) (6, 000) \cos θ \\ \frac{\begin{matrix} {(2, 420)}^{2} & - & {(5, 050)}^{2} - {(6, 000)}^{2} \end{matrix}}{- 2 (5, 050) (6, 000)} & = & \cos θ \\ \cos θ & \approx & 0.9183 \\ θ & \approx & \cos^{- 1} (0.9183) \\ θ & \approx & 23.3 ° \end{matrix}

To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 7. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.

The triangle between the phone, the left tower, and a point between the phone and the highway between the towers. The side between the phone and the highway is perpendicular to the highway and is y feet. The highway side is x feet. The angle at the tower, previously labeled theta, is 23.3 degrees. — Figure 7

Using the angle $θ = 23.3°$ and the basic trigonometric identities, we can find the solutions. Thus

\begin{array}{l} \begin{array}{l} \cos (23.3°) = \frac{x}{5,050} \end{array} \\ x = 5,050 \cos (23.3°) \\ x \approx 4,638.15 feet \\ \sin (23.3°) = \frac{y}{5,050} \\ y = 5,050 \sin (23.3°) \\ y \approx 1,997.5 feet \end{array}

The cell phone is approximately 4,638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.

Example 4

Calculating Distance Traveled Using a SAS Triangle

Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 8.

Solution

The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, $180° - 20° = 160° .$ With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.

\begin{array}{l} x^{2} = 8^{2} + 10^{2} - 2 (8) (10) \cos (160°) \\ x^{2} = 314.35 \\ x = \sqrt{314.35} \\ x \approx 17.7 miles \end{array}

The boat is about 17.7 miles from port.

Using Heron’s Formula to Find the Area of a Triangle

We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.

Example 5

Using Heron’s Formula to Find the Area of a Given Triangle

Find the area of the triangle in Figure 9 using Heron’s formula.

A triangle with angles A, B, and C and opposite sides a, b, and c, respectively. Side a = 10, side b - 15, and side c = 7. — Figure 9

Solution

First, we calculate $s .$

\begin{array}{l} s = \frac{(a + b + c)}{2} \\ s = \frac{(10 + 15 + 7)}{2} = 16 \end{array}

Then we apply the formula.

\begin{array}{l} Area = \sqrt{s (s - a) (s - b) (s - c)} \\ Area = \sqrt{16 (16 - 10) (16 - 15) (16 - 7)} \\ Area \approx 29.4 \end{array}

The area is approximately 29.4 square units.

Example 6

Applying Heron’s Formula to a Real-World Problem

A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See Figure 10 for a view of the city property.

A triangle formed by sides Rush Street, N. Wabash Ave, and E. Pearson Street with lengths 62.4, 43.5, and 34.1, respectively. — Figure 10

Solution

Find the measurement for $s,$ which is one-half of the perimeter.

\begin{array}{l} s = \frac{(62.4 + 43.5 + 34.1)}{2} \\ s = 70 m \end{array}

Apply Heron’s formula.

\begin{array}{l} Area = \sqrt{70 (70 - 62.4) (70 - 43.5) (70 - 34.1)} \\ Area = \sqrt{506,118.2} \\ Area \approx 711.4 \end{array}

The developer has about 711.4 square meters.

Key Equations

..
Law of Cosines	$\begin{array}{l} a^{2} = b^{2} + c^{2} - 2 b c \cos α \\ b^{2} = a^{2} + c^{2} - 2 a c \cos β \\ c^{2} = a^{2} + b^{2} - 2 a b c o s γ \end{array}$
Heron’s formula	$\begin{array}{l} Area = \sqrt{s (s - a) (s - b) (s - c)} \\ where s = \frac{(a + b + c)}{2} \end{array}$

Key Concepts

The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See Example 1 and Example 2.
The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See Example 3 and Example 4.
Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example 5 and See Example 6.

Section Exercises

Verbal

Exercise 1

If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?

Solution

two sides and the angle opposite the missing side.

Exercise 2

If you are looking for a missing angle of a triangle, what do you need to know when using the Law of Cosines?

Exercise 3

Explain what $s$ represents in Heron’s formula.

Solution

$s$ is the semi-perimeter, which is half the perimeter of the triangle.

Exercise 4

Explain the relationship between the Pythagorean Theorem and the Law of Cosines.

Exercise 5

When must you use the Law of Cosines instead of the Pythagorean Theorem?

Solution

The Law of Cosines must be used for any oblique (non-right) triangle.

Algebraic

For the following exercises, assume $α$ is opposite side $a, β$ is opposite side $b,$ and $γ$ is opposite side $c .$ If possible, solve each triangle for the unknown side. Round to the nearest tenth.

Exercise 6

$γ = 41.2°, a = 2.49, b = 3.13$

Exercise 7

$α = 120°, b = 6, c = 7$

Solution

11.3

Exercise 8

$β = 58.7°, a = 10.6, c = 15.7$

Exercise 9

$γ = 115°, a = 18, b = 23$

Solution

34.7

Exercise 10

$α = 119°, a = 26, b = 14$

Exercise 11

$γ = 113°, b = 10, c = 32$

Solution

26.7

Exercise 12

$β = 67°, a = 49, b = 38$

Exercise 13

$α = 43.1°, a = 184.2, b = 242.8$

Solution

$c = 257.3, 96.7$

Exercise 14

$α = 36.6°, a = 186.2, b = 242.2$

Exercise 15

$β = 50°, a = 105, b = 45$

Solution

not possible

For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth.

Exercise 16

$a = 42, b = 19, c = 30;$ find angle $A .$

Exercise 17

$a = 14, b = 13, c = 20;$ find angle $C .$

Solution

95.5°

Exercise 18

$a = 16, b = 31, c = 20;$ find angle $B .$

Exercise 19

$a = 13, b = 22, c = 28;$ find angle $A .$

Solution

26.9°

Exercise 20

$a = 108, b = 132, c = 160;$ find angle $C .$

For the following exercises, solve the triangle. Round to the nearest tenth.

Exercise 21

$A = 35°, b = 8, c = 11$

Solution

$B \approx 45.9°, C \approx 99.1°, a \approx 6.4$

Exercise 22

$B = 88°, a = 4.4, c = 5.2$

Exercise 23

$C = 121°, a = 21, b = 37$

Solution

$A \approx 20.6°, B \approx 38.4°, c \approx 51.1$

Exercise 24

$a = 13, b = 11, c = 15$

Exercise 25

$a = 3.1, b = 3.5, c = 5$

Solution

$A \approx 37.8°, B \approx 43.8, C \approx 98.4°$

Exercise 26

$a = 51, b = 25, c = 29$

For the following exercises, use Heron’s formula to find the area of the triangle. Round to the nearest hundredth.

Exercise 27

Find the area of a triangle with sides of length 18 in, 21 in, and 32 in. Round to the nearest tenth.

Solution

177.56 in²

Exercise 28

Find the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.

Exercise 29

$a = \frac{1}{2} m, b = \frac{1}{3} m, c = \frac{1}{4} m$

Solution

0.04 m²

Exercise 30

$a = 12.4 ft, b = 13.7 ft, c = 20.2 ft$

Exercise 31

$a = 1.6 yd, b = 2.6 yd, c = 4.1 yd$

Solution

0.91 yd²

Graphical

For the following exercises, find the length of side $x .$ Round to the nearest tenth.

Exercise 32

A triangle. One angle is 72 degrees, with opposite side = x. The other two sides are 5 and 6.5.

Exercise 33

A triangle. One angle is 42 degrees with opposite side = x. The other two sides are 4.5 and 3.4.

Solution

3.0

Exercise 34

A triangle. One angle is 40 degrees with opposite side = 15. The other two sides are 12 and x.

Exercise 35

A triangle. One angle is 65 degrees with opposite side = x. The other two sides are 30 and 23.

Solution

29.1

Exercise 36

A triangle. One angle is 50 degrees with opposite side = x. The other two sides are 225 and 305.

Exercise 37

A triangle. One angle is 123 degrees with opposite side = x. The other two sides are 1/5 and 1/3.

Solution

0.5

For the following exercises, find the measurement of angle $A .$

Exercise 38

A triangle. Angle A is opposite a side of length 2.3. The other two sides are 1.5 and 2.5.

Exercise 39

A triangle. Angle A is opposite a side of length 125. The other two sides are 115 and 100.

Solution

70.7°

Exercise 40

A triangle. Angle A is opposite a side of length 6.8. The other two sides are 4.3 and 8.2.

Exercise 41

A triangle. Angle A is opposite a side of length 40.6. The other two sides are 38.7 and 23.3.

Solution

77.4°

Exercise 42

Find the measure of each angle in the triangle shown in Figure 11. Round to the nearest tenth.

A triangle A B C. Angle A is opposite a side of length 10, angle B is opposite a side of length 12, and angle C is opposite a side of length 7. — Figure 11

For the following exercises, solve for the unknown side. Round to the nearest tenth.

Exercise 43

A triangle. One angle is 60 degrees with opposite side unknown. The other two sides are 20 and 28.

Solution

25.0

Exercise 44

A triangle. One angle is 30 degrees with opposite side unknown. The other two sides are 16 and 10.

Exercise 45

A triangle. One angle is 22 degrees with opposite side unknown. The other two sides are 20 and 13.

Solution

9.3

Exercise 46

A triangle. One angle is 88 degrees with opposite side = 9. Another side is 5.

For the following exercises, find the area of the triangle. Round to the nearest hundredth.

Exercise 47

A triangle with sides 8, 12, and 17. Angles unknown.

Solution

43.52

Exercise 48

A triangle with sides 50, 22, and 36. Angles unknown.

Exercise 49

A triangle with sides 1.9, 2.6, and 4.3. Angles unknown.

Solution

1.41

Exercise 50

A triangle with sides 8.9, 12.5, and 16.2. Angles unknown.

Exercise 51

A triangle with sides 1/2, 2/3, and 3/5. Angles unknown.

Solution

0.14

Extensions

Exercise 52

A parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal.

Exercise 53

The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal.

Solution

18.3

Exercise 54

The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100°. Find the length of the shorter diagonal.

Exercise 55

A regular octagon is inscribed in a circle with a radius of 8 inches. (See Figure 12.) Find the perimeter of the octagon.

An octagon inscribed in a circle. — Figure 12

Solution

48.98

Exercise 56

A regular pentagon is inscribed in a circle of radius 12 cm. (See Figure 13.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter.

A pentagon inscribed in a circle. — Figure 13

For the following exercises, suppose that $x^{2} = 25 + 36 - 60 \cos (52)$ represents the relationship of three sides of a triangle and the cosine of an angle.

Exercise 57

Draw the triangle.

Solution

A triangle. One angle is 52 degrees with opposite side = x. The other two sides are 5 and 6.

Exercise 58

Find the length of the third side.

For the following exercises, find the area of the triangle.

Exercise 59

A triangle. One angle is 22 degrees with opposite side = 3.4. Another side is 5.3.

Solution

7.62

Exercise 60

A triangle. One angle is 80 degrees with opposite side unknown. The other two sides are 8 and 6.

Exercise 61

A triangle. One angle is 18 degrees with opposite side = 12.8. Another side is 18.8.

Solution

85.1

Real-World Applications

Exercise 62

A surveyor has taken the measurements shown in Figure 14. Find the distance across the lake. Round answers to the nearest tenth.

A triangle. One angle is 70 degrees with opposite side unknown, which is the length of the lake. The other two sides are 800 and 900 feet. — Figure 14

Exercise 63

A satellite calculates the distances and angle shown in Figure 15 (not to scale). Find the distance between the two cities. Round answers to the nearest tenth.

Insert figure(table) alt text: A triangle formed by two cities on the ground and a satellite above them. The angle by the satellite is 2.1 degrees with opposite side unknown, which is the distance between the two cities. The lengths of the other sides are 370 and 350 km. — Figure 15

Solution

24.0 km

Exercise 64

An airplane flies 220 miles with a heading of 40°, and then flies 180 miles with a heading of 170°. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth.

Exercise 65

A 113-foot tower is located on a hill that is inclined 34° to the horizontal, as shown in Figure 16. A guy-wire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed.

Insert figure(table) alt text: Two triangles, one on top of the other. The bottom triangle is the hill inclined 34 degrees to the horizontal. The second is formed by the base of the tower on the incline of the hill, the top of the tower, and the wire anchor point uphill from the tower on the incline. The sides are the tower, the incline of the hill, and the wire. The tower side is 113 feet and the incline side is 98 feet. — Figure 16

Solution

99.9 ft

Exercise 66

Two ships left a port at the same time. One ship traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°. Find the distance between the two ships after 10 hours of travel.

Exercise 67

The graph in Figure 17 represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327° and the second boat is traveling at 4 miles per hour at a heading of 60°. Find the distance between the two boats after 2 hours.

Insert figure(table) alt text: A graph of two rays, which represent the paths of the two boats. Both rays start at the origin. The first goes into the first quadrant at a 60 degree angle at 4 mph. The second goes into the fourth quadrant at a 327 degree angle from the origin. The second travels at 18 mph. — Figure 17

Solution

37.3 miles

Exercise 68

A triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)?

Exercise 69

A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10° to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?

Solution

2371 miles

Exercise 70

Los Angeles is 1,744 miles from Chicago, Chicago is 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle.

Exercise 71

Philadelphia is 140 miles from Washington, D.C., Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle.

Solution

Angle BO is 9.1 degrees, angle PH is 150.2 degrees, and angle DC is 20.7 degrees.

Exercise 72

Two planes leave the same airport at the same time. One flies at 20° east of north at 500 miles per hour. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after 2 hours?

Exercise 73

Two airplanes take off in different directions. One travels 300 mph due west and the other travels 25° north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude?

Solution

292.4 miles

Exercise 74

A parallelogram has sides of length 15.4 units and 9.8 units. Its area is 72.9 square units. Find the measure of the longer diagonal.

Exercise 75

The four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral?

Solution

65.4 cm²

Exercise 76

The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral?

Exercise 77

Find the area of a triangular piece of land that measures 30 feet on one side and 42 feet on another; the included angle measures 132°. Round to the nearest whole square foot.

Solution

468 ft²

Exercise 78

Find the area of a triangular piece of land that measures 110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot.

Law of Cosines

states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle

Generalized Pythagorean Theorem

an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles