# 1.14: Right Triangles, Bearings, and other Applications

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

**Practice**Right Triangles, Bearings, and other Applications

While on a camping trip with your friends, you take an orienteering trip. You end up on a course which results in you hiking \begin{align*}30^\circ\end{align*}

Read on, and when you have completed this Concept, you'll be able to make this calculation.

### Watch This

### Guidance

We can also use right triangles to find distances using angles given as bearings. In navigation, a bearing is the direction from one object to another. In air navigation, bearings are given as angles rotated clockwise from the north. The graph below shows an angle of 70 degrees:

It is important to keep in mind that angles in navigation problems are measured this way, and not the same way angles are measured in trigonometry. Further, angles in navigation and surveying may also be given in terms of north, east, south, and west. For example, \begin{align*}N70^\circ E\end{align*}

#### Example A

A ship travels on a \begin{align*}N50^\circ E\end{align*}

**Solution:** The angle between \begin{align*}d\end{align*}

\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{10}{d}\\
\cos 40^\circ & = \frac{10}{d}\\
d \cos 40^\circ & = 10\\
d & = \frac{10}{\cos 40^\circ} \approx 13.05 \ nm\end{align*}

#### Example B

An airplane flies on a course of \begin{align*}S30^\circ E\end{align*}, for 150 km. How far south is the plane from where it originated?

**Solution:** We can construct a triangle using the known information, and then use the cosine function to solve the problem:

\begin{align*}\cos 30^\circ & = \frac{adjacent}{hypotenuse} = \frac{y}{150}\\ \cos 30^\circ & = \frac{y}{150}\\ 150 \cos 30^\circ & = y\\ y & = 150 \cos 30^\circ \approx 130 km\end{align*}

#### Example C

Jean travels to school each day by walking 200 meters due East, and then turning left and walking 100 meters due North. If she had walked in a straight line, what would the angle between her home and the school be if the beginning of the angle is taken from due East? What would be two different ways to describe the direction to take walking there in a straight line, using what we've learned in this Concept?

**Solution:** From the triangle given above, we can use the tangent function to determine the angle if she had walked in a straight line.

\begin{align*}\tan \theta & = \frac{opposite}{adjacent} = \frac{100}{200}\\ \tan \theta & = \frac{100}{200}\\ \theta & = 26.57^\circ \end{align*}

One way of describing her straight line path is how far north of east she is: \begin{align*}E26.57^\circ N\end{align*}

Also, since we know the angle between the East and the North directions is \begin{align*}90^\circ\end{align*}, her motion can also be described by how far east of north she is: \begin{align*}N63.43^\circ E\end{align*}

### Vocabulary

**Bearing:** In navigation, a ** bearing** is the direction taken to get from one place to another.

### Guided Practice

1. Plot a course of \begin{align*}S30^\circ W\end{align*} on a rectangular coordinate system.

2. Scott is boating on a course of \begin{align*}N15^\circ E\end{align*}. What course would he need to take to return to where he came from?

3. Adam hikes on a course of \begin{align*}N47^\circ E\end{align*} for 7 km. How far East is Adam from where he started?

**Solutions:**

1.

2. The opposite direction would return him to his starting point. This would be \begin{align*}S15^\circ W\end{align*}.

3.

We can find the length of the triangle above (which is how far he traveled East) by using the sine function:

\begin{align*} \sin 47^\circ = \frac{x}{7}\\ x = 7 \sin 47^\circ\\ x = (7)(.7313)\\ x =5.1191 \end{align*}

He is 5.1191 km East of where he started.

### Concept Problem Solution

From your knowledge of how to construct a triangle using bearings, you can draw the following:

This shows that the opposite side of the triangle is what's not known. Therefore, you can use the sine function to solve the problem:

\begin{align*} \sin 30^\circ = \frac{opposite}{5}\\ opposite = 5 \sin 30^\circ\\ opposite = (5)(.5) = 2.5\\ \end{align*}

You are 2.5 miles South of where you started.

### Practice

- Plot a course of \begin{align*}N40^\circ E\end{align*} on a rectangular coordinate system.
- Plot a course of \begin{align*}E30^\circ N\end{align*} on a rectangular coordinate system.
- Plot a course of \begin{align*}S70^\circ W\end{align*} on a rectangular coordinate system.
- Plot a course of \begin{align*}W85^\circ S\end{align*} on a rectangular coordinate system.
- Plot a course of \begin{align*}N42^\circ W\end{align*} on a rectangular coordinate system.
- You are on a course of \begin{align*}E35^\circ N\end{align*}. What course would you need to take to return to where you came from?
- You are on a course of \begin{align*}W56^\circ S\end{align*}. What course would you need to take to return to where you came from?
- You are on a course of \begin{align*}N72^\circ W\end{align*}. What course would you need to take to return to where you came from?
- You are on a course of \begin{align*}S10^\circ E\end{align*}. What course would you need to take to return to where you came from?
- You are on a course of \begin{align*}W65^\circ N\end{align*}. What course would you need to take to return to where you came from?
- You are on a course of \begin{align*}N47^\circ E\end{align*} for 5 km. How far East are you from where you started?
- You are on a course of \begin{align*}S32^\circ E\end{align*} for 8 km. How far East are you from where you started?
- You are on a course of \begin{align*}N15^\circ W\end{align*} for 10 km. How far West are you from where you started?
- You are on a course of \begin{align*}S3^\circ W\end{align*} for 12 km. How far West are you from where you started?
- You are on a course of \begin{align*}S67^\circ E\end{align*} for 6 km. How far East are you from where you started?

### Image Attributions

Here you'll learn how to relate right triangle relationships to course bearings.