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11. Applications of Trigonometry
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VCE 11 General 2023

11.03 2D Applications of Trigonometry

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Introduction

There are many practical applications to trigonometry, as well as Pythagoras' theorem. Sometimes problems may require a combination of Pythagoras' theorem and trigonometry in order to find the solution. In this chapter, several common applications will be discussed.

Angles of elevation and depression

In worded problems in trigonometry, the angle of elevation or depression is often used to describe a given angle. The angle of elevation is an angle measured upwards from a horizontal line of sight. The angle of depression is an angle measured downwards from a horizontal line of sight.

The image shows the different line of sight of an observer. Ask your teacher for more information.

Pythagoras' theorem: a^2+b^2=c^2, where c is the hypotenuse.

Angle of elevation: the angle from the observer's horizontal line of sight looking up at an object.

Angle of depression: the angle from the observer's horizontal line of sight looking down at an object.

Examples

Example 1

A man standing at point C, is looking at the top of a tree at point A. Identify the angle of elevation in the figure given.

A right angled triangle A B C showing angles theta, alpha, and sigma. Ask your teacher for more information.
A
\alpha
B
\theta
C
\sigma
Worked Solution
Create a strategy

Find angle between line of sight and horizontal.

Apply the idea

By tilting your head to look at point A from the horizontal line CB, we have line AC as your line of sight and the angle \alpha as the angle of elevation.

The correct option is A.

Example 2

From the top of a rocky ledge 188\text{ m} high, the angle of depression to a boat is 13\degree. If the boat is d\text{ m} from the foot of the cliff find d correct to two decimal places.

Worked Solution
Create a strategy

Sketch the situation to identify the appropriate trigonometric ratio.

Apply the idea

Here's the diagram showing the situation:

A diagram showing the angle of depression and location of the boat. Ask your teacher for more information.

With respect to the given angle 13 \degree, the opposite side is 188, and the adjacent side is d, so we can use the tangent ratio.

\displaystyle \tan \theta\displaystyle =\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan {13\degree}\displaystyle =\displaystyle \frac{188}{d}Substitute the values
\displaystyle d \times \tan {13\degree}\displaystyle =\displaystyle 188Multiply both sides by d
\displaystyle d\displaystyle =\displaystyle \frac{188}{\tan {13\degree}}Divide both sides by \tan {13\degree}
\displaystyle =\displaystyle 814.32 \text{ m}Evaluate using a calculator

Example 3

Consider the following diagram.

A diagram showing a rectangle with width of 63 and length of y. Ask your teacher for more information.
a

Find y, correct to two decimal places.

Worked Solution
Create a strategy

Determine the sides using the reference angle, then use the appropriate ratio.

Apply the idea

With respect to the given angle 39 \degree, the opposite side is 63, and the adjacent side is y, so we can use the tangent ratio.

\displaystyle \tan \theta\displaystyle =\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan {39\degree}\displaystyle =\displaystyle \frac{63}{y}Substitute the values
\displaystyle y \times \tan {39\degree}\displaystyle =\displaystyle 63Multiply both sides by y
\displaystyle y\displaystyle =\displaystyle \frac{63}{\tan {39\degree}}Divide both sides by \tan {39\degree}
\displaystyle \approx\displaystyle 77.80Evaluate using a calculator
b

Find w, correct to two decimal places.

Worked Solution
Create a strategy

Use the same method we used to find y in part (a).

Apply the idea

With respect to the given angle 67 \degree, the opposite side is 63, and the adjacent side is w, so we can use the tangent ratio.

\displaystyle \tan \theta\displaystyle =\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan {67\degree}\displaystyle =\displaystyle \frac{63}{w}Substitute the values
\displaystyle w \times \tan {67\degree}\displaystyle =\displaystyle 63Multiply both sides by w
\displaystyle w\displaystyle =\displaystyle \frac{63}{\tan {67\degree}}Divide both sides by \tan {67\degree}
\displaystyle \approx\displaystyle 26.74Evaluate using a calculator
c

Find x, correct to two decimal places.

Worked Solution
Create a strategy

Subtract w from y.

Apply the idea

We found in part (a) the value, y=77.80 and in part (b), we have the value, w=26.74.

\displaystyle x\displaystyle =\displaystyle 77.80-26.74Substitute the values
\displaystyle =\displaystyle 51.1Evaluate
Idea summary

Pythagoras' theorem: a^2+b^2=c^2, where c is the hypotenuse.

Angle of elevation: the angle from the observer's horizontal line of sight looking up at an object.

Angle of depression: the angle from the observer's horizontal line of sight looking down at an object.

Three-figure (true) bearings

In surveying and air navigation, bearings are used to help identify the location of an object.

A compass showing the directions north, east, south, and west and smaller divisions. Ask your teacher for more information.

The four main directions of a compass are known as cardinal directions. They are north (\text{N}), east (\text{E}), south (\text{S}), and west (\text{W}).

A three-figure bearing are:

  • Measured from north (\text{N})

  • Measured in a clockwise direction

  • Written using three figures (digits)

\text{T} is often but not always used to indicate a true bearing. If the angle measure is less than 100\degree it would be written as 040\degree or 040\degree \text{T}.

To use true bearing to describe the location of a plane at point B from the airport at point A:

  1. Place the centre of a compass on the starting point, in this case the airport

  2. Starting at North, rotate clockwise until we get to the line AB

  3. Write angle as the true bearing of point B

A compass with A as the centre and line A B at an angle of 127 degrees clockwise from the north direction.

The true bearing of B from A is 127\degree or 127\degree \text{T}

A compass with O as the centre and line O P at an angle of 55 degrees clockwise from the north direction.

The diagram describes the bearing of P from O. Rotating clockwise from North, we get an angle of 55\degree.

Since this measure is less than three digits, we put a 0 in front of the two digit number to write the three-figure bearing of P as 055\degree or 055\degree \text{T}.

Consider the true bearing of O from P. Since angle of elevation is equal to angle of depression and we are starting at P the true bearing would be 180+55=235\degree.

Examples

Example 4

What is the true bearing of Southwest?

Worked Solution
Create a strategy

Starting at North rotate in a clockwise direction towards Southwest.

Apply the idea
A compass showing the directions north, east, south, and west and in between directions. Ask your teacher for more information.

Southwest is located exactly halfway between South and West. Rotating starting from North going to South is a turn of 180\degree and from South going to halfway between South and West is an additional turn of 45\degree.

\displaystyle \text{True bearing}\displaystyle =\displaystyle 180+45Add the two angles
\displaystyle =\displaystyle 225\degree \text{T}Evaluate
Idea summary

True bearings are:

  • Measured from north (\text{N})

  • Measured in a clockwise direction

  • Written using three figures

  • Usually written with a \text{T} at the end

Compass bearings

A compass bearing describes the location of a point using:

  • The starting direction of either north or south;

  • The acute angle needed to rotate

  • The direction to rotate towards; either east or west

The image describes how to write an angle using the starting and rotation direction. Ask your teacher for more information.
A compass with A at the centre and an angle of 53 degrees between the south direction and line AB measured anti clockwise.

To describe the position of point B from A we would say: "Starting at South, I then rotate 53\degree towards East."

The compass bearing of B from A is S\,53\degree E

Exploration

The bearing needed or used completely depends on which position comes first. Have a look at the applet below. It quickly shows you how the angle changes depending on if we are measuring the bearing of A from B or B from A.

Loading interactive...

If A is between north and east, the compass bearing of A from B is measured clockwise from the north.

If A is between south and east, the compass bearing of A from B is measured anticlockwise from the south.

If A is between south and west, the compass bearing of A from B is measured clockwise from the south.

If A is between north and west, the compass bearing of A from B is measured anticlockwise from the north.

Examples

Example 5

Consider the point A.

A compass with O at the centre and line O A at an angle of 28 degrees clockwise from the east direction.

What is the compass bearing of point A from O?

Worked Solution
Create a strategy

Consider whether point A is closer to the North or South, and then calculate the number of degrees to be rotated East or West.

Apply the idea

In this case, the bearing is in the bottom-right corner, so it is closer to South than North. So our compass bearing will be from the South.

From point O facing South, we need to rotate towards East in order to be facing point A. So the angle would be: 90-28=62\degree.

So our compass would be: \text{S} \, 62\degree \text{E}

Example 6

In the figure below, point B is due East of point A. We want to find the position of point A relative to point C.

Three compasses with C, B, and A as the centres and are connected to make a triangle. Ask your teacher for more information.

What is the compass bearing of point A from point C?

Worked Solution
Create a strategy

Consider whether point A is closer to the North or South of point C, and then calculate the number of degrees to be rotated East or West from C.

Apply the idea

A is in the north west direction from C so it is closer to north than south at C.

From point C facing North, we need to rotate towards West in order to be facing point A. So the angle would be: 90-18=72\degree.

So our compass bearing would be: \text{N} \, 72\degree \text{W}.

Idea summary
The image describes how to write an angle using the starting and rotation direction. Ask your teacher for more information.

A compass bearing describes the location of a point using:

  • The starting direction of either north or south

  • The acute angle needed to rotate

  • The direction to rotate, east or west.

Applications of bearings

The cardinal directions \text{N},\,\text{S},\,\text{E}, and \text{W} form 90\degree angles to one another. This means that right-angled triangles can be used to represent distances and bearings measured from one location to another. Therefore trigonometry and Pythagoras' theorem can both be applied to solving navigation problems.

Examples

Example 7

The position of a ship S is given to be 20 kilometres from P, on a true bearing of 049\degree \text{T}.

The position of the ship can also be given by its \left(x,\, y \right) coordinates.

A compass with P at the centre and line P S with a true bearing of 49 degrees. Ask your teacher for more information.
a

If the ship's x-coordinate is x, find x to one decimal place.

Worked Solution
Create a strategy

Use the sine ratio.

Apply the idea
\displaystyle \sin \theta\displaystyle =\displaystyle \dfrac{\text{Opposite}}{\text{Hypotenuese}}Use the sine ratio
\displaystyle \sin 49\degree\displaystyle =\displaystyle \dfrac{x}{20}Substitute values and x
\displaystyle x\displaystyle =\displaystyle 20 \times \sin 49\degreeMultiply both sides by 20
\displaystyle =\displaystyle 15.1 \text{ km}Evaluate
b

If the ship's y-coordinate is y, find y to one decimal place.

Worked Solution
Create a strategy

Use the cosine ratio.

Apply the idea
\displaystyle \cos \theta\displaystyle =\displaystyle \dfrac{\text{Adjacent}}{\text{Hypotenuese}}Use the cosine ratio
\displaystyle \cos 49\degree\displaystyle =\displaystyle \dfrac{y}{20}Substitute values and y
\displaystyle y\displaystyle =\displaystyle 20 \times \cos 49\degreeMultiply both sides by 20
\displaystyle =\displaystyle 13.1 \text{ km}Evaluate

Example 8

In remote locations, photographers must keep track of their position from their base. One morning a photographer sets out from base, represented by point B, to the edge of an ice shelf at point S on a bearing of 055\degree. She then walked on a bearing of 145\degree to point P, which is 916 metres due east of base.

Diagram showing the triangle P S B with line BP measures 916 metres. Ask your teacher for more information.
a

From the information provided, which angle measures 90\degree?

A
\angle PBS
B
\angle SPB
C
\angle BSP
Worked Solution
Create a strategy

Derive additional angles from the provided angles.

Apply the idea

Remember that co-interior angles sum to 180\degree. By extending the side BS and cutting the angle 145\degree, we can see corresponding angles formed, which is 55\degree. In solving this, we have 145\degree - 55\degree = 90\degree. We can see that co-interior angles formed, so by:

\displaystyle \angle BSP\displaystyle =\displaystyle 180 - 90Subtract 90 from 180
\displaystyle =\displaystyle 90\degreeEvaluate

So the correct answer is C.

b

If the distance between B and S is d metres, find d to one decimal place.

Worked Solution
Create a strategy

Determine the sides using the reference angle, then use the appropriate ratio.

Apply the idea

In solving the other angles:

\displaystyle \angle PBS\displaystyle =\displaystyle 90-55Subtract 55 from 90
\displaystyle =\displaystyle 35\degreeEvaluate

We found \angle BSP = 90\degree in part (a). So, we subtract the \angle BSP and \angle PBS from 180\degree:

\displaystyle \angle SPB\displaystyle =\displaystyle 180-90-35Subtract the angles
\displaystyle =\displaystyle 55\degreeEvaluate

With respect to the \angle SPB=55 \degree, the opposite side is d, and the hypotenuse side is 916, so we can use the sine ratio.

\displaystyle \sin \theta\displaystyle =\displaystyle \dfrac{\text{Opposite}}{\text{Hypotenuese}}Use the sine ratio
\displaystyle \sin 55\degree\displaystyle =\displaystyle \dfrac{d}{916}Substitute values and d
\displaystyle d\displaystyle =\displaystyle 916 \sin 55\degreeMultiply both sides by 916
\displaystyle =\displaystyle 750.3 \text{ m}Evaluate
c

If the distance between S and P is e metres, find e to one decimal place.

Worked Solution
Create a strategy

Use the same method we used to find d in part (b).

Apply the idea

With respect to the \angle PBS = 35 \degree found in part (b), the opposite side is e, and the hypotenuse side is 916, so we can use the sine ratio.

\displaystyle \sin \theta\displaystyle =\displaystyle \dfrac{\text{Opposite}}{\text{Hypotenuese}}Use the sine ratio
\displaystyle \sin 35\degree\displaystyle =\displaystyle \dfrac{e}{916}Substitute values and e
\displaystyle e\displaystyle =\displaystyle 916 \sin 35\degreeMultiply both sides by 916
\displaystyle =\displaystyle 525.4 \text{ m}Evaluate
d

If the photographer were to walk back to her base from point P, what is the total distance she would have travelled? Round your answer to 1 decimal place.

Worked Solution
Create a strategy

Add the length of the three sides.

Apply the idea

In part (b), we found, d=750.3 and in part (c), we have, e=525.4.

\displaystyle \text{Total distance}\displaystyle =\displaystyle 525.4+750.3+916Add all sides
\displaystyle =\displaystyle 2191.7\text{ m}Evaluate

Example 9

A boat travels \text{S}\, 14 \degree \text{E} for 12 \text{ km} and then changes direction to \text{S}\, 49 \degree \text{E} for another 16 \text{ km}.

3 compasses connected to make a triangle. Ask your teacher for more information.
a

Find x, the distance of the boat from its starting point. Give your answer to two decimal places.

Worked Solution
Create a strategy

Find the opposite angle and then use the cosine rule to find x.

Apply the idea

The angle opposite x, let's call it \theta, is made up of two angles. The smaller angle is alternate to the angle of 14\degree since the two arrows pointing north are parallel. So the small angle equals 14\degree. The larger angle is on a straight line with 49\degree , so they add up to 180\degree.

\displaystyle \text{Larger angle}\displaystyle =\displaystyle 180-49(Angles on a straight line)
\displaystyle =\displaystyle 131\degreeEvaluate
\displaystyle \theta\displaystyle =\displaystyle 131+14Add the two angles
\displaystyle =\displaystyle 145\degreeEvaluate

Now we can use the cosine rule to find x.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2-2ab \cos CUse the cosine rule
\displaystyle x^2\displaystyle =\displaystyle 12^2+16^2 -2 \times 12 \times 16 \times \cos 145\degreeSubstitute c=x,a=12,b=16, C =145\degree
\displaystyle =\displaystyle 714.554385Evaluate
\displaystyle x\displaystyle =\displaystyle \sqrt{714.554385}Take the square root of both sides
\displaystyle \approx\displaystyle 26.73Evaluate and round
b

Find the angle b as labelled in the diagram. Express your answer to the nearest degree.

Worked Solution
Create a strategy

Use the sine rule.

Apply the idea
\displaystyle \dfrac{\sin A}{a}\displaystyle =\displaystyle \dfrac{\sin C}{c}Use the sine rule
\displaystyle \dfrac{\sin b}{16}\displaystyle =\displaystyle \dfrac{\sin 145}{26.73}Substitute A=b, a=16, C=145, c=26.73
\displaystyle \sin b\displaystyle =\displaystyle \dfrac{\sin 145}{26.73} \times 16Multiply both sides by 16
\displaystyle \approx\displaystyle 0.34333Evaluate
\displaystyle b\displaystyle =\displaystyle \sin ^{-1}(0.34333)Take the inverse sine of both sides
\displaystyle \approx\displaystyle 20 \degreeEvaluate and round
c

Hence write down the bearing that the boat should travel on to return to the starting point.

Worked Solution
Create a strategy

Find the angle, a, by using the alternate angles on parallel lines.

Apply the idea

Angle a is alternate to the angle (b+14)\degree at the starting point, because both north lines are parallel. So these angles are equal.

\displaystyle a\displaystyle =\displaystyle b+14Add the two angles
\displaystyle =\displaystyle 20+14Substitute b=20
\displaystyle =\displaystyle 34\degreeEvaluate

So the compass bearing is \text{N} \, 34\degree \text{W}.

Idea summary

We can use trigonometry once to find a new value, and then use trigonometry again with our new value to find another new value. We can repeat this process as many times as is necessary to find the value that we are looking for.

Outcomes

U2.AoS4.2

Pythagoras’ theorem and the trigonometric ratios (sine, cosine and tangent) and their application including angles of elevation and depression and three figure bearings

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