Learning Objectives

In this section, you will:

- Use the Law of Cosines to solve oblique triangles.
- Solve applied problems using the Law of Cosines.
- Use Heron’s formula to ﬁnd the area of a triangle.

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

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 (ABC) 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 (PageIndex{2}). Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.

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 heta=dfrac{x(adjacent)}{b(hypotenuse)}) and (sin heta=dfrac{y(opposite)}{b(hypotenuse)})

In terms of ( heta), (x=b cos heta) and (y=b sin heta). The ((x,y)) point located at (C) has coordinates ((b cos heta, b sin heta)). 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,

(egin{array}{ll} a^2={(x−c)}^2+y^2 [4pt] ;;;;; ={(b cos heta−c)}^2+{(b sin heta)}^2 & ext{Substitute }(b cos heta) ext{ for }x ext{ and }(b sin heta) ext{ for }y [4pt] ;;;;;; =(b^2{cos}^2 heta−2bc cos heta+c^2)+b^2 {sin}^2 heta & ext{Expand the perfect square.} [4pt] ;;;;; =b^2{cos}^2 heta+b^2{sin}^2 heta+c^2−2bc cos heta & ext{Group terms noting that }{cos}^2 heta+{sin}^2 heta=1 [4pt] ;;;;; =b^2({cos}^2 heta+{sin}^2 heta)+c^2−2bc cos heta & ext{Factor out }b^2 [4pt] end{array})

(a^2=b^2+c^2−2bc cos heta )

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.

The 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 (PageIndex{3}), with angles (alpha), (eta) and (gamma), and opposite corresponding sides (a), (b), and (c), respectively, the Law of Cosines is given as three equations.

[a^2=b^2+c^2−2bc cos alpha]

[b^2=a^2+c^2−2ac cos eta]

[c^2=a^2+b^2−2ab cos gamma]

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.

[cos alpha=dfrac{b^2+c^2−a^2}{2bc}]

[cos eta=dfrac{a^2+c^2−b^2}{2ac}]

[cos gamma=dfrac{a^2+b^2−c^2}{2ab}]

How to: Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle

- Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
- Apply the Law of Cosines to find the length of the unknown side or angle.
- Apply the Law of Sines or Cosines to find the measure of a second angle.
- Compute the measure of the remaining angle.

Example (PageIndex{1}): Finding the Unknown Side and Angles of a SAS Triangle

Find the unknown side and angles of the triangle in Figure (PageIndex{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 (eta).

(egin{array}{ll} b^2=a^2+c^2−2ac cos eta [4pt] b^2={10}^2+{12}^2−2(10)(12)cos(30°) & ext{Substitute the measurements for the known quantities.} [4pt] b^2=100+144−240 left(dfrac{sqrt{3}}{2} ight) & ext{Evaluate the cosine and begin to simplify.} [4pt] b^2=244−120sqrt{3} [4pt] b=sqrt{244−120sqrt{3}} & ext{Use the square root property.} [4pt] b≈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 (alpha), we have

(egin{array}{cc} dfrac{sin alpha}{a}=dfrac{sin eta}{b} [4pt] dfrac{sin alpha}{10}=dfrac{sin(30°)}{6.013} [4pt] sin alpha=dfrac{10sin(30°)}{6.013} & ext{Multiply both sides of the equation by }10. [4pt] alpha={sin}^{−1}left(dfrac{10sin(30°)}{6.013} ight) & ext{Find the inverse sine of } dfrac{10sin(30°)}{6.013}. [4pt] alpha≈56.3° end{array})

The other possibility for (alpha) would be (alpha=180°-56.3°≈123.7°). In the original diagram,(alpha) is adjacent to the longest side, so (alpha) 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 (alpha≈56.3°), we can then find the third angle of the triangle.

[egin{align*} gamma&= 180^{circ}-30^{circ}-56.3^{circ} &approx 93.7^{circ} end{align*}]

The complete set of angles and sides is

(alpha≈56.3°) (a=10)

(eta=30°) (b≈6.013)

(gamma≈93.7°) (c=12)

Exercise (PageIndex{1})

Find the missing side and angles of the given triangle: (alpha=30°), (b=12), (c=24).

- Answer
(a≈14.9), (eta≈23.8°), (gamma≈126.2°).

Example (PageIndex{2}): Solving for an Angle of a SSS Triangle

Find the angle (alpha) 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 (alpha), we have

(egin{array}{ll} a^2=b^2+c^2−2bc cos alpha [4pt] {20}^2={25}^2+{18}^2−2(25)(18)cos alpha & ext{Substitute the appropriate measurements.} [4pt] 400=625+324−900 cos alpha & ext{ Simplify in each step.} [4pt] 400=949−900 cos alpha [4pt] −549=−900 cos alpha & ext{Isolate }cos alpha. [4pt] −549−900=cos alpha [4pt] 0.61≈cos alpha [4pt] 0.61≈cos alpha & ext{Find the inverse cosine.} [4pt] alpha≈52.4° end{array})

See Figure (PageIndex{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.

Exercise (PageIndex{2})

Given (a=5), (b=7), and (c=10), find the missing angles.

**Answer**(alpha≈27.7°), (eta≈40.5°), (gamma≈111.8°)

## 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 (PageIndex{3A}): 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 (5050) feet from the first tower and (2420) 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 (PageIndex{6}) and labeling our given information.

Using the Law of Cosines, we can solve for the angle ( heta). 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=2420), (b=5050), and (c=6000). Thus, ( heta) corresponds to the opposite side (a=2420).

[egin{align*} a^2 & =b^2+c^2−2bc cos heta [4pt] {(2420)}^2 &={(5050)}^2+{(6000)}^2−2(5050)(6000) cos heta [4pt] cos heta &≈ 0.9183 [4pt] cos heta &≈ 0.9183 [4pt] heta &≈ {cos}^{−1}(0.9183) [4pt] heta &≈ 23.3° end{align*}]

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 (PageIndex{7}). This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.

Using the angle ( heta=23.3)° and the basic trigonometric identities, we can find the solutions. Thus

[egin{align*} cos(23.3°) &= dfrac{x}{5050} [4pt] x &= 5050cos(23.3°) [4pt] x &≈ 4638.15, feet[4pt] sin(23.3°) &= dfrac{y}{5050} [4pt] y &= 5050sin(23.3°) [4pt] y &≈1997.5 , feet end{align*}]

The cell phone is approximately (4638) feet east and (1998) feet north of the first tower, and (1998) feet from the highway.

Example (PageIndex{3B}): 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 (PageIndex{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.

[egin{align*} x^2 &= 8^2+{10}^2−2(8)(10)cos(160°) [4pt] x^2 &= 314.35 [4pt] x &= sqrt{314.35} [4pt] x&≈17.7, miles end{align*}]

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.

HERON’S FORMULA

Heron’s formula finds the area of oblique triangles in which sides (a), (b),and (c) are known.

[Area=sqrt{s(s−a)(s−b)(s−c)}]

where (s=dfrac{(a+b+c)}{2}) is one half of the perimeter of the triangle, sometimes called the semi-perimeter.

Example (PageIndex{4}): Using Heron’s Formula to Find the Area of a Given Triangle

Find the area of the triangle in Figure (PageIndex{9}) using Heron’s formula.

**Solution**

First, we calculate (s).

[egin{align*} s&= dfrac{(a+b+c)}{2} s&= dfrac{(10+15+7)}{2} &= 16 end{align*}]

Then we apply the formula.

[egin{align*} Area&= sqrt{s(s-a)(s-b)(s-c)} Area&= sqrt{16(16-10)(16-15)(16-7)} Area&approx 29.4 end{align*}]

The area is approximately (29.4) square units.

Exercise (PageIndex{3})

Use Heron’s formula to find the area of a triangle with sides of lengths (a=29.7) ft, (b=42.3) ft, and (c=38.4) ft.

- Answer
Area = (552) square feet

Example (PageIndex{5}): 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 (PageIndex{10}) for a view of the city property.

**Solution**

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

[egin{align*} s&= dfrac{(62.4+43.5+34.1)}{2} s&= 70; m ext {Apply Heron's formula.} Area&= sqrt{70(70-62.4)(70-43.5)(70-34.1)} Area&= sqrt{506,118.2} Area&approx 711.4 end{align*}]

The developer has about (711.4) square meters.

Exercise (PageIndex{4})

Find the area of a triangle given (a=4.38) ft , (b=3.79) ft, and (c=5.22) ft.

**Answer**about (8.15) square feet

Media

Access these online resources for additional instruction and practice with the Law of Cosines.

- Law of Cosines
- Law of Cosines: Applications
- Law of Cosines: Applications 2

## Key Equations

Law of Cosines | (a^2=b^2+c^2−2bc cos alpha) (b^2=a^2+c^2−2ac cos eta) (c^2=a^2+b^2−2ab cos gamma) |

Heron’s formula | (Area=sqrt{s(s−a)(s−b)(s−c)}) where (s=dfrac{(a+b+c)}{2}) |

## 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 (PageIndex{1}) and Example (PageIndex{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 (PageIndex{3}) and Example (PageIndex{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 (PageIndex{5}) and See Example (PageIndex{6}).

## 4.2: Non-right Triangles - Law of Cosines

By the end of this section, you will be able to:

- Use the Law of Cosines to solve oblique triangles.
- Solve applied problems using the Law of Cosines.
- Use Heron’s formula to ﬁnd the area of a triangle.

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?

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.

## 8.2 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?

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 A B C is placed in the coordinate plane with vertex A A at the origin, side c c drawn along the *x*-axis, and vertex C C located at some point ( x , y ) ( 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.

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 , α , β , and γ , γ , and opposite corresponding sides a , b , a , b , and c , c , respectively, the Law of Cosines is given as three equations.

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.

### How To

**Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.**

- Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
- Apply the Law of Cosines to find the length of the unknown side or angle.
- Apply the Law of Sines or Cosines to find the measure of a second angle.
- Compute the measure of the remaining angle.

### Example 1

#### Finding the Unknown Side and Angles of a SAS Triangle

Find the unknown side and angles of the triangle in 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 , b , as we know the measurement of the opposite angle β . β .

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

## 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 A B C is placed in the coordinate plane with vertex A A at the origin, side c c drawn along the *x*-axis, and vertex C C located at some point ( x , y ) ( x , y ) in the plane, as illustrated in [link]. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.

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.

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 [link], with angles α , β , α , β , and γ , γ , and opposite corresponding sides a , b , a , b , and c , c , respectively, the Law of Cosines is given as three equations.

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.

**Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.**

- Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
- Apply the Law of Cosines to find the length of the unknown side or angle.
- Apply the Law of Sines or Cosines to find the measure of a second angle.
- Compute the measure of the remaining angle.

Find the unknown side and angles of the triangle in [link].

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 , b , as we know the measurement of the opposite angle β . β .

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

The complete set of angles and sides is

Find the missing side and angles of the given triangle: α = 30° , b = 12 , c = 24. α = 30° , b = 12 , c = 24.

## Non-right Triangles: Law of Sines

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in (Figure) that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles .

**Figure 1.**

### Using the Law of Sines to Solve Oblique Triangles

In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

**ASA (angle-side-angle)**We know the measurements of two angles and the included side. See (Figure).

**Figure 2.**

**Figure 3.**

**Figure 4.**

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in (Figure).

**Figure 5.**

Using the right triangle relationships, we know thatandSolving both equations forgives two different expressions for

We then set the expressions equal to each other.

Similarly, we can compare the other ratios.

Collectively, these relationships are called the **Law of Sines**.

Note the standard way of labeling triangles: angle(alpha) is opposite sideangle(beta) is opposite sideand angle(gamma) is opposite sideSee (Figure).

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.

**Figure 6.**

### Law of Sines

Given a triangle with angles and opposite sides labeled as in (Figure), the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

To solve an oblique triangle, use any pair of applicable ratios.

### Solving for Two Unknown Sides and Angle of an AAS Triangle

Solve the triangle shown in (Figure) to the nearest tenth.

**Figure 7.**

The three angles must add up to 180 degrees. From this, we can determine that

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle and its corresponding sideWe can use the following proportion from the Law of Sines to find the length of

Similarly, to solve forwe set up another proportion.

Therefore, the complete set of angles and sides is

[/hidden-answer]### Try It

Solve the triangle shown in (Figure) to the nearest tenth.

**Figure 8.**

### Using The Law of Sines to Solve SSA Triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

### Possible Outcomes for SSA Triangles

Oblique triangles in the category SSA may have four different outcomes. (Figure) illustrates the solutions with the known sidesandand known angle

**Figure 9.**

### Solving an Oblique SSA Triangle

Solve the triangle in (Figure) for the missing side and find the missing angle measures to the nearest tenth.

**Figure 10.**

Use the Law of Sines to find angleand angleand then sideSolving forwe have the proportion

However, in the diagram, angleappears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement ofLet’s investigate further. Dropping a perpendicular fromand viewing the triangle from a right angle perspective, we have (Figure). It appears that there may be a second triangle that will fit the given criteria.

**Figure 11.**

The angle supplementary tois approximately equal to 49.9°, which means that(Remember that the sine function is positive in both the first and second quadrants.) Solving for we have

We can then use these measurements to solve the other triangle. Sinceis supplementary to the sum ofand we have

Now we need to findand

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in (Figure).

**Figure 12.**

However, we were looking for the values for the triangle with an obtuse angleWe can see them in the first triangle (a) in (Figure).[/hidden-answer]

### Try It

Givenandfind the missing side and angles. If there is more than one possible solution, show both.

[/hidden-answer]### Solving for the Unknown Sides and Angles of a SSA Triangle

In the triangle shown in (Figure), solve for the unknown side and angles. Round your answers to the nearest tenth.

**Figure 13.**

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angleand its corresponding sideand we know sideWe will use this proportion to solve for

To findapply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values forIt is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

In this case, if we subtractfrom 180°, we find that there may be a second possible solution. Thus,To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

which is impossible, and so

To find the remaining missing values, we calculateNow, only sideis needed. Use the Law of Sines to solve forby one of the proportions.

The complete set of solutions for the given triangle is

[/hidden-answer]### Try It

Givenfind the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

### Finding the Triangles That Meet the Given Criteria

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

Using the given information, we can solve for the angle opposite the side of length 10. See (Figure).

**Figure 14.**

We can stop here without finding the value ofBecause the range of the sine function isit is impossible for the sine value to be 1.915. In fact, inputtingin a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.[/hidden-answer]

### Try It

Determine the number of triangles possible given

### Finding the Area of an Oblique Triangle Using the Sine Function

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given aswhereis base andis height. For oblique triangles, we must findbefore we can use the area formula. Observing the two triangles in (Figure), one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric propertyto write an equation for area in oblique triangles. In the acute triangle, we haveorHowever, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the baseto form a right triangle. The angle used in calculation isor

**Figure 15.**

### Area of an Oblique Triangle

The formula for the area of an oblique triangle is given by

This is equivalent to one-half of the product of two sides and the sine of their included angle.

### Finding the Area of an Oblique Triangle

Find the area of a triangle with sidesand angleRound the area to the nearest integer.

Using the formula, we have

[/hidden-answer]### Try It

Find the area of the triangle givenRound the area to the nearest tenth.

about

### Solving Applied Problems Using the Law of Sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

### Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in (Figure). Round the altitude to the nearest tenth of a mile.

**Figure 16.**

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side and then use right triangle relationships to find the height of the aircraft,

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

The distance from one station to the aircraft is about 14.98 miles.

Now that we knowwe can use right triangle relationships to solve for

The aircraft is at an altitude of approximately 3.9 miles.

The diagram shown in (Figure) represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, pointis 62°, and the distance between the viewing points of the two end zones is 145 yards.

**Figure 17.**

Access these online resources for additional instruction and practice with trigonometric applications.

### Key Equations

Law of Sines | |

Area for oblique triangles |

### Key Concepts

- The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
- According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
- There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See (Figure).
- The ambiguous case arises when an oblique triangle can have different outcomes.
- There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See (Figure) and (Figure).
- The Law of Sines can be used to solve triangles with given criteria. See (Figure).
- The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See (Figure).
- There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See (Figure).

### Section Exercises

#### Verbal

Describe the altitude of a triangle.

The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.

Compare right triangles and oblique triangles.

When can you use the Law of Sines to find a missing angle?

When the known values are the side opposite the missing angle and another side and its opposite angle.

In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?

What type of triangle results in an ambiguous case?

A triangle with two given sides and a non-included angle.

#### Algebraic

For the following exercises, assumeis opposite sideis opposite sideandis opposite sideSolve each triangle, if possible. Round each answer to the nearest tenth.

For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angleis opposite sideangleis opposite sideand angleis opposite side

Find sidewhen

Find side when

Find sidewhen

For the following exercises, assumeis opposite sideis opposite sideandis opposite sideDetermine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.

one triangle,

two triangles,or

two triangles,or

two triangles,or

For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.

Find anglewhen

Find anglewhen

or

Find anglewhen

For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth.

#### Graphical

For the following exercises, find the length of sideRound to the nearest tenth.

For the following exercises, find the measure of angleif possible. Round to the nearest tenth.

Notice thatis an obtuse angle.

For the following exercises, find the area of each triangle. Round each answer to the nearest tenth.

#### Extensions

Find the radius of the circle in (Figure). Round to the nearest tenth.

**Figure 18.**

Find the diameter of the circle in (Figure). Round to the nearest tenth.

**Figure 19.**

Findin (Figure). Round to the nearest tenth.

**Figure 20.**

Findin (Figure). Round to the nearest tenth.

**Figure 21.**

Solve both triangles in (Figure). Round each answer to the nearest tenth.

**Figure 22.**

Findin the parallelogram shown in (Figure).

**Figure 23.**

Solve the triangle in (Figure). (Hint: Draw a perpendicular fromtoRound each answer to the nearest tenth.

**Figure 24.**

Solve the triangle in (Figure). (Hint: Draw a perpendicular fromtoRound each answer to the nearest tenth.

**Figure 25.**

In (Figure),is not a parallelogram.is obtuse. Solve both triangles. Round each answer to the nearest tenth.

**Figure 26.**

#### Real-World Applications

A pole leans away from the sun at an angle ofto the vertical, as shown in (Figure). When the elevation of the sun isthe pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.

**Figure 27.**

To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in (Figure). Determine the distance of the boat from stationand the distance of the boat from shore. Round your answers to the nearest whole foot.

**Figure 28.**

(Figure) shows a satellite orbiting Earth. The satellite passes directly over two tracking stationsandwhich are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation atandare measured to beandrespectively. How far is the satellite from stationand how high is the satellite above the ground? Round answers to the nearest whole mile.

**Figure 29.**

The distance from the satellite to stationis approximately 1716 miles. The satellite is approximately 1706 miles above the ground.

A communications tower is located at the top of a steep hill, as shown in (Figure). The angle of inclination of the hill isA guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill isFind the length of the cable required for the guy wire to the nearest whole meter.

**Figure 30.**

The roof of a house is at aangle. An 8-foot solar panel is to be mounted on the roof and should be angledrelative to the horizontal for optimal results. (See (Figure)). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth.

**Figure 31.**

Similar to an angle of elevation, an *angle of depression* is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to beandas shown in (Figure). Find the distance of the plane from pointto the nearest tenth of a kilometer.

**Figure 32.**

A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in (Figure). Find the distance of the plane from pointto the nearest tenth of a kilometer.

**Figure 33.**

In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot.

In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot.

Pointsandare on opposite sides of a lake. Pointis 97 meters fromThe measure of angleis determined to be 101°, and the measure of angleis determined to be 53°. What is the distance fromtorounded to the nearest whole meter?

A man and a woman standingmiles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot.

Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile.

A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot.

Three cities,andare located so that cityis due east of cityIf cityis located 35° west of north from cityand is 100 miles from cityand 70 miles from cityhow far is cityfrom cityRound the distance to the nearest tenth of a mile.

Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet.

Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in (Figure).

**Figure 34.**

The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°.

A yield sign measures 30 inches on all three sides. What is the area of the sign?

Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in (Figure).

**Figure 35.**

## Law of Cosines

The **Law of Cosines** is used to find the remaining parts of an oblique (non-right) triangle when either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are known. In either of these cases, it is impossible to use the Law of Sines because we cannot set up a solvable proportion.

The Law of Cosines states:

*c 2 = a 2 + b 2 &minus 2 a b &thinsp &thinsp cos C .*

This resembles the Pythagorean Theorem except for the third term and if C is a right angle the third term equals 0 because the cosine of 90 ° is 0 and we get the Pythagorean Theorem. So, the Pythagorean Theorem is a special case of the Law of Cosines.

The Law of Cosines can also be stated as

*b 2 = a 2 + c 2 &minus 2 a c &thinsp &thinsp cos B or*

*a 2 = b 2 + c 2 &minus 2 b c &thinsp &thinsp cos A .*

**Example 1:** **Two Sides and the Included Angle-SAS**

Given a = 11 , b = 5 and m &ang C = 20 ° . Find the remaining side and angles.

c 2 = a 2 + b 2 &minus 2 a b &thinsp &thinsp cos C

c = a 2 + b 2 &minus 2 a b &thinsp &thinsp cos C

&thinsp &thinsp = 11 2 + 5 2 &minus 2 ( 11 ) ( 5 ) ( cos 20 ° )

To find the remaining angles, it is easiest to now use the Law of Sines.

Note that angle A is opposite to the longest side and the triangle is not a right triangle. So, when you take the inverse you need to consider the obtuse angle whose sine is 11 sin ( 20 ° ) 6.53 &asymp 0.5761 .

**Example 2:** **Three Sides-SSS**

Given a = 8 , b = 19 and c = 14 . Find the measures of the angles.

It is best to find the angle opposite the longest side first. In this case, that is side b .

cos B = b 2 &minus a 2 &minus c 2 &minus 2 a c = 19 2 &minus 8 2 &minus 14 2 &minus 2 ( 8 ) ( 14 ) &asymp &minus 0.45089

Since cos B is negative, we know that B is an obtuse angle.

Since B is an obtuse angle and a triangle has at most one obtuse angle, we know that angle A and angle C are both acute.

## Law of cosines - SSS example

If your task is to find the angles of a triangle given all three sides, all you need to do is to use the transformed cosine rule formulas:

Let&aposs calculate one of the angles. Assume we have a = 4 in, b = 5 in and c = 6 in. We&aposll use the first equation to find α:

You may calculate the second angle from the second equation in an analogical way, and the third angle you can find knowing that the sum of the angles in a triangle is equal to 180° (π/2).

If you want to save some time, type the side lengths into our law of sines calculator - our tool is a safe bet! Just follow these simple steps:

**Choose the option depending on given values**. We need to pick the second option - *SSS (3 sides)*.

**Enter the known values**. Type the sides: a = 4 in, b = 5 in and c = 6 in.

**The calculator displays the result!** In our case the angles are equal to α = 41.41°, β = 55.77° and γ = 82.82°.

After such an explanation, we&aposre sure that you understand what the law of cosine is and when to use it. Give this tool a try, solve some exercises, and remember that practice makes permanent!

## 4.2: Non-right Triangles - Law of Cosines

So far, we've only dealt with right triangles, but trigonometry can be easily applied to **non-right triangles** because any non-right triangle can be divided by an **altitude** * into two right triangles.

Roll over or tap the triangle to see what that means →

Remember that an **altitude** is a line segment that has one endpoint at a vertex of a triangle intersects the opposite side (or an extension of it outside the triangle) at a right angle. See triangles.

### Customary labeling of non-right triangles

This labeling scheme is commonly used for non-right triangles. **Capital letters** are **angles** and the corresponding **lower-case letters** go with the **side** opposite the angle: side **a** (with length of **a** units) is across from angle **A** (with a measure of **A** degrees or radians), and so on.

### Derivation of the law of sines

### Derivation

In mathematics, when we *derive* a formula, we basically *invent* it using one or more simpler, already-known principles. A derivation is essentially a proof that the formula we're creating works and can be relied upon.

Consider the triangle below. if we find the sines of angle **A** and angle **C** using their corresponding right triangles, we notice that they both contain the altitude, x.

We can rearrange those by solving each for **x** (multiply by **c** on both sides of the left equation, and by **a** on both sides of the right):

$x = ccdot sin(A) ext < and > x = acdot sin(C)$

Now the transitive property says that if both **c·sin(A)** and **a·sin(C)** are equal to **x**, then they must be equal to each other:

We usually divide both sides by **ac** to get the easy-to-remember expression of the law of sines:

We could do the same derivation with the other two altitudes, drawn from angles **A** and **C** to come up with similar relations for the other angle pairs. We call these together the law of sines. It's in the green box below.

The law of sines can be used to find the measure of an angle or a side of a non-right triangle if we know:

#### Law of Sines

Although three fractions are related here with two equal signs, we only use them in pairs in practice.

### Example 1

Find all of the missing measurements of this triangle:

**Solution** : The missing angle is easy, it's just

Now set up one of the law of sines proportions and solve for the missing piece, in this case the length of the lower side:

Then do the same for the other missing side. It's best to use the original known angle and side so that round-off errors or mistakes don't add up.

#### Pro tip:

I've skipped a couple of steps in the algebra above. Remember that when the variable for which you're trying to solve is in a denominator, your first task is clear: *Get it out of the denominator*, usually by multiplying by that variable on both sides. Then the rest of the algebra to isolate that variable from everything else attached to it is fairly clear.

### Example 2

Find all of the missing measurements of this triangle:

**Solution** : This time we have to find our first missing angle using the LOS, then we can use the sum of the angles of a triangle to find the third. First, set up one law of sines proportion. We'll be solving for a missing angle, so we'll have to calculate an inverse sine:

Now it's easy to calculate the third angle:

Then apply the law of sines again for the missing side. We have two choices, we can solve

Either gives the same answer,

### Caution : Law of sines ambiguity

We have to be careful about the law of sines, because it can give ambiguous solutions when we use the inverse sine function [f(x) = sin -1 (x)]. Here's how it works. For a given triangle in which only one side is known, two possible triangles can be formed. Take a look:

The magenta sides mark out an obtuse triangle (small one on the left) and an acute triangle (outer triangle) using the same combination of two sides, where the third (bottom) side can be one of two lengths because it isn't initially known. The two possible angles are always related: their sum is 180˚.

How do we handle this? Let's do an example.

First, we solve for angle **A** using the LOS:

We can rearrange and use the inverse sine function to get the angle:

Now if there is an ambiguity, its measure will be 180˚ minus the angle we determined:

Now if that angle, added to the original angle (30˚) is less than 180˚, such a triangle *can* exist, and we have an ambiguous case. Here are the two possible triangles in this example:

In such a case, we have to know a little more about our triangle. Is it acute or obtuse? Is a given diagram drawn to scale (as these are)? If so, that might be enough to resolve the ambiguity.

In some cases, the quantity (180˚ minus our calculated angle) added back to the calculated angle will be greater than 180˚, and such a triangle cannot exist, so the solution is unique.

Make sure to be aware of this as you work problems. Check for ambiguities unless you're clear about the results from other clues in the problem.

### The law of cosines

The law of cosines is used when we know:

- the lengths of two sides of a triangle and the measure of the angle between them, OR
- the lengths of all three sides of a triangle, but no angle measures.

The law of cosines is computationally a little more complicated to use than the law of sines, but fortunately, it only needs to be used once. After the law of cosines is applied to a triangle, the resulting information will always make it possible to use the law of sines to calculate further properties of the triangle.

### Derivation of the law of cosines

Consider another non-right triangle, labeled as shown with side lengths **x** and **y**. We can derive a useful law containing only the cosine function.

First use the Pythagorean theorem to derive two equations for each of the right triangles:

Notice that each contains and **x 2** , so we can eliminate **x 2** between the two using the transitive property:

Then expand the binomial **(b - y) 2** to get the equation below, and note that the **y 2** cancel:

Now we still have a y hanging around, but we can get rid of it using the cosine solution, notice that

$cos(A) = frac

Substituting c·cos(A) for y, we get

which is the law of cosines

The law of cosines can be used to find the measure of an angle or a side of a non-right triangle if we know:

We could again do the same derivation using the other two altitudes of our triangle, to yield three versions of the law of cosines for any triangle. They are listed in the box below.

#### Law of Cosines

The **Law of Cosines** is just the Pythagorean relationship with a correction factor, e.g. **-2bc·cos(A)**, to account for the fact that the triangle is not a right triangle. We can write three versions of the LOC, one for every angle/opposite side pair:

$ egin

#### Pro tip

You'll only need to use the law of cosines once if you need it at all. After finding the missing angle or side with the LOC, the law of sines will work for the rest.

### Example 3

Find all of the missing measurements of this triangle:

**Solution** : Set up the law of cosines using the only set of angles and sides for which it is possible in this case:

Now using the new side, find one of the missing angles using the law of sines:

And then the third angle is

In general, try to use the law of sines first. It's easier and less prone to errors (but be careful of the ambiguity when solving for non-acute angles). But in this case, that wasn't possible, so the law of cosines was necessary.

### Example 4

Find all of the missing measurements of this triangle:

**Solution** : Set up the law of cosines to solve for either one of the angles:

Rearrange to solve for A. You'll need an inverse cosine to get the angle.

Use the law of sines to find a second angle

Finally, just calculate the third angle:

### Practice problems

Find the measures of all missing sides and angles of these triangles:

*(Hint: Always use the LOS first if you can. It's simpler. When that fails, use the LOC, but then you can usually use other means to fill in the rest of the measurements using the LOS or the Pythagorean theorem.)*

## 4.2: Non-right Triangles - Law of Cosines

HSC Standard Maths Resources

**Browse:** 1. Home » 4. Trigonometry » 4.2 Non-right Triangle

Non-right triangles are triangles that do not have 90-degrees inside them. For this section, make sure you are across sine rule and cosine rule.

The sine rule is useful as you can use it to find either a missing angle or length of an edge of triangle if you have enough information to use the sine rule equation.

**Get access to 20 Mock Exams with over 700 exam-style questions for HSC Standard Maths.**

**Click here to check them out!!**

Consider the following triangles. Find x in each of the cases:

**Cosine Rule**

Consider the following situations:

All edges of the triangle are given but no angles are provided or two edges are given and the angle formed by them are given, and the third edge is to be calculated. In situations like these, the cosine rule is the quickest way to get to the answer.

Though there are three equations below, the basic idea is the same that you put the edges and angle enclosed by those edges on right side of equation and the edge opposite the angle on the left side.

## Solving Applied Problems Using the Law of Sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

### Example 6

#### Problem 1

##### Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16 . Round the altitude to the nearest tenth of a mile.

**Figure 16**

##### Solution

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a , a , and then use right triangle relationships to find the height of the aircraft, h . h .

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

The distance from one station to the aircraft is about 14.98 miles.

Now that we know a , a , we can use right triangle relationships to solve for h . h .