# Bearing - Word Problems

Most of the bearing word problems involving trigonometry and angles can be reduced to that of finding relationships between angles and measurements of sides of a triangle. In this case, finding the right basic trigonometric functions to relate the angles and measurements are crucial for setting up and solving the problem correctly.

Trigonometry of angles and sides can be used on a daily basis in the workplace like carpentering, construction working and engineering etc.

## Basic Definitions

Before jumping to the word problems, here are the basic definitions you need to be familiar with.

The

angle of elevationof an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (also known as theline of sight). \(_\square\)Illustration of the above definition:

The angle of elevation in the above diagram is \(\alpha^\circ\).

If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the

angle of depression. \(_\square\)Illustration of the above definition:

The angle of depression in the above diagram is \(\beta^\circ\).

Here is a problem on angle of elevation.

## Word Problems

It is often useful to draw a diagram and remember how the basic trigonometric functions relate the angles and measurements of sides in a right triangle. Finding the right trigonometric function to relate the angles and measurements is crucial to to solve the problems. We will demonstrate the principle of setting up and solving trigonometric word problems by working through several examples.

A surveyor in a helicopter at an elevation of 1000 meters measures the angle of depression to the far edge of an island as \( 24^\circ \) and the angle of depression to the near edge as \( 31^\circ \). How wide is the island, to the nearest meter?

Let the horizontal distance between the helicopter and the island be \( d \), and the width of the island be \( w \). Then \( \tan 24^\circ = \frac{1000}{d+w} \) and \( \tan 31^\circ = \frac{1000}{d} \), implying \( d = \frac{1000}{\tan 31^\circ} \). Substituting this into \( \tan 24^\circ = \frac{1000}{d+w} \) gives

\[\begin{align} \tan 24^\circ &= \frac{1000}{\frac{1000}{\tan 31^\circ}+w} \\ \tan 24^\circ \left(\frac{1000}{\tan 31^\circ}+w \right) &= 1000 \\ w &= \frac{1000\left( 1 - \frac{\tan 24^\circ}{ \tan 31^\circ }\right)}{\tan 24^\circ} \\ w & \approx 581.75 . \end{align}\]

Thus, the width of the island is 582 meters. \(_\square\)

In alternative you can calculate the difference between cot(24°) and cot(31°) [equal to 0.581757292] and multiply it for 1000 mt.

Andrew was flying a kite on a hill, but he dumped his kite into the pond below. If the length of the string of his kite is 150 meters and the angle of depression from his position to the kite is \(30^\circ\), then how high is the hill where he is standing?

Let's first draw the diagram for explicit understanding of the problem.

So it is a right triangle with base angle \(30^\circ\), hypotenuse 150 meters, and the side opposite to the given angle \(h\) which is the same as the height of the hill. We use the sine ratio to find the height: \[\sin 30^\circ=\dfrac{h}{150} \\ \Rightarrow \dfrac 12 =\dfrac{h}{150} \\ \Rightarrow h=75 \text{ (m)}. \]

Hence the hill is 75 meters above the lake. \(_\square\)

The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 mt from its base is 45 degrees. If the angle of elevation of the top of the complete pillar at the same point is to be 60 degrees, then by how much the height of the incomplete pillar is to be increased?

Lets draw the figure of the problem's situation. Here we go:

Let \(BC\) be the height of the incomplete and \(BD\) be the height of the complete pillar. So we are given that \(BC=100\)m, \(\angle BAC = 45^\circ\) and \(\angle BAD = 60^\circ\). And we assume that the length of \(CD\) be \(x\) meter.

So in \(\delta ABC\), we know that \[\begin{array}{} \tan 45^\circ & =\frac{BC}{AB} \\ \Rightarrow AB & =BC \\ \Rightarrow AB & =100 \text{m}\end{array}\]

Similarly in \(\delta ABD\), we know that \[\begin{array}{} \tan 60^\circ & =\frac{BD}{AB} \\ \Rightarrow \sqrt 3 & = \frac{BD}{100} \\ \Rightarrow BD & =100 \sqrt 3 \text{m}\end{array}\]

And from the above figure, \[BD=BC+CD \\ \Rightarrow 100\sqrt3 =100+x \\ \Rightarrow x=100 \left( \sqrt3 -1 \right) \text{m} \]

Hence the height of the incomplete pillar is to be increased by \(100 \left( \sqrt3 -1 \right) \text{m} \) to complete the pillar.

A private plane flies for 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point?

The bearings tell us the angles from "due north" in a clockwise direction. Since 130 – 40 = 90, the two bearings give us a right triangle. From the times and rates, we have \[\begin{align} 1.3 × 110 &= 143\\ 1.5 × 110 &= 165. \end{align}\]

Now lets give the geometrical shape to our problem and set up a triangle.

Using the pythagorean theorem we get

\[\begin{array}{} m^2 & =143^2 + 165^2 \\

m^2 & =20449 + 27225 \\ m^2 & =47674 \\

m & = 218.34 \end{array} \]Hence the plane is approximately 218 miles away at the end of the time.\(\square\)

Try the following bearing word problems.

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB subtends an angle of \( 60^\circ\) at the foot of the tower and the angle of elevation of the top of the tower from A or B is \( 30^\circ\). Find the height of the tower.

## Example Problems

This section is meant to enhance the problem solving skills of bearing word problems. Here an example is illustrated followed by some problems for you to attempt.

A six-meter-long ladder leans against a building. If the ladder makes an angle of \(60^\circ\) with the ground,

(1) how far up the wall does the ladder reach, and

(2) how far from the wall is the base of the ladder?

To understand the situation, let's draw its diagram.

(1) From the above diagram, we get the following equation to obtain the value of \(h:\)

\[\begin{align} \sin 60^\circ & =\dfrac h6 \\ \Rightarrow h & =6 \times \sin 60^\circ \\ & =3 \sqrt 3. \end{align} \]

So the ladder can reach \(3\sqrt3 \) meters up the wall.

(2) Similarly, we can get the value of \(b\) as follows:

\[\begin{align} \cos 60^\circ &=\dfrac b6 \\ \Rightarrow b&=6 \times \cos 60^\circ \\ &=3. \end{align}\]

Hence the base of the ladder is 3 meters from the wall. \(_\square\)

Here are the problems to gain a strong grab over the bearing concept.

Aeroplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and 60° with respect to the horizontal, respectively, as shown in the figure.

The airspeed of A is \(100 \sqrt{3} \) meters per second. Initially, an observer in A sees B directly ahead and at the same altitude at a (horizontal) distance of 500 m.

Assuming they take no evasive action and collide, after how many seconds does this happen?

The angle of elevation of a cloud from a point \(h\) metres above a lake is \(\theta\). The angle of depression of its reflection in the lake is \(45^{\circ}\), then find the height of the cloud.

Linda measures the angle of elevation from a point on the ground to the top of the tree and find it to be 35 degrees. She then walks 20 meters towards the tree and finds the angle of elevation from this new point to the top of the tree to be 45 degrees. Find the height of the tree (in meters).

Give your answer to three significant figures.

## See Also

**Cite as:**Bearing - Word Problems.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/bearing-word-problems-easy/