11.03 2D Applications of Trigonometry

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.

 

Practice questions

Question 1

Question 2

Question 3

 

True bearings and compass bearings

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

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

 

Three-figure (true) bearings

A three-figure bearing is:

• Measured from north ($N$N)
• Measured in a clockwise direction
• Written using three figures (digits)

$T$T is often but not always used to indicate a three-figure bearing. If the angle measure is less than $100^\circ$100°, it would be written in the form $040^\circ$040° or $040^\circ$040°T.

To use three-figure bearings to describe the location of a plane at point $B$B from the airport at point $A$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 $AB$AB
3. Write angle as the three-figure bearing of point $B$B

The true bearing of $B$B from $A$A is $127^\circ$127° or $127^\circ$127°T.

The diagram below describes the bearing of $P$P from $O$O. Between these two points, there is an angle measure of $55^\circ$55°.

Since this measure is less than three digits, a $0$0 must be placed in front of the two digit number to write the three-figure bearing of $P$P as $055^\circ$055° or $055^\circ$055°T.

Consider the bearing of $O$O from $P$P. Since we are effectively rotating to point in the opposite direction, the three-figure bearing starting from $P$P would be $180+55=235^\circ$180+55=235°.

 

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 following sentence could describe the position of point $B$B from $A$A:

"Starting at South, rotate $53^\circ$53° towards East."

The compass bearing of $B$B from $A$A is $S$S$53$53$^\circ$°$E$E.

 

Worked examples

Example 1

Find the three-figure and compass bearings of point $P$P from $O$O.

Solution:

Three-figure bearing:

Starting at North rotate in a clockwise direction.

$360^\circ-47^\circ=313^\circ$360°−47°=313°

The three-figure bearing of $P$P from $O$O is $313$313$^\circ$°T.

Compass Bearing

Point $P$P is closest to North, so starting at North, rotate $47^\circ$47° towards West.

The compass bearing of $P$P from $O$O is $N$N$47$47$^\circ$°$W$W.

 

Which position first?

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

​

 

Practice questions

Question 4

Question 5

Question 6

 

Applying trigonometry to problems involving bearings

The cardinal directions N, S, E and W form $90^\circ$90° 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.

 

Worked example

Example 2

A slightly lost hiker walks $300$300 m east before turning south and walking another $800$800 m. What are the hiker's three-figure and compass bearings from the original position (rounded to one decimal place)?

Think: It will be easiest to first visualise the situation by drawing a diagram, such as this one:

For the three-figure bearing, the angle we want to find has been marked. It is equal to $90^\circ$90°$+$+the angle inside the triangle, which we can find by using a tangent ratio, since the $800$800 m and $300$300 m lengths are the opposite and adjacent sides of the triangle respectively.

From above, $\tan\theta=\frac{800}{300}$tanθ=800300​

Therefore, $\theta=\tan^{-1}\frac{800}{300}$θ=tan−1800300​

We can then use this to find the compass bearing.

Do: The three-figure bearing is

$90+\tan^{-1}\frac{800}{300}$90+tan−1800300​	$=$=	$90+\tan^{-1}\frac{8}{3}$90+tan−183​
	$=$=	$90+69.44395\ldots$90+69.44395…
	$=$=	$159.4$159.4 (one d.p.)

So the three-figure bearing is $159.4$159.4$^\circ$°.

To determine this as a compass bearing, we want to know the acute angle made with the North-South line. This is $180-159.4=20.6$180−159.4=20.6.

Looking back at the diagram, we can see that this bearing is closer to South, and on the East side, so the compass bearing is $S$S$20.6$20.6$^\circ$°$E$E.

 

Practice question

Question 7

Question 8

Question 9