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IB Mathematics AI SL Applications of right and non-right angled trigonometry MAI Study Notes- New Syllabus

IB Mathematics AI SL Applications of right and non-right angled trigonometry MAI Study Notes

LEARNING OBJECTIVE

  • Applications of right and non-right angled trigonometry, including Pythagoras’ theorem.

Key Concepts: 

  • Applications of Trigonometry & Pythagoras

MAI HL and SL Notes – All topics

APPLICATIONS IN 3D GEOMETRY

ANGLE OF ELEVATION
Suppose that an object is above the horizontal level of an observer. The angle of elevation \( \theta \) to the object is shown below:

 

These are the terms we must be familiar with when dealing with the angle of elevation.

Angle

When two straight rays or lines meet at a single point, an angle is created. Degrees are used to express angles.

Line of Sight

The line of sight is the line drawn from the observer’s eye to the object being viewed. It is an oblique line which is therefore not horizontal nor vertical. The line of sight, together with the horizontal line, forms the angle of elevation. 

Horizontal Line

The horizontal line is a straight line on a flat surface. In a coordinate system, a horizontal line has points with the same y-coordinate. The line of sight, together with the horizontal line, form the angle of elevation. 

Observer’s Eye

The observer’s eye is the position where the line of sight and the horizontal line meet.

ANGLE OF DEPRESSION

If the object is below the level of the observer, the angle of depression \( \theta \) to the object is shown below:

These are the terms we must be familiar with when dealing with the angle of depression.

Angle

When two straight rays or lines meet at a single point, an angle is created. Degrees are used to express angles.

Line of Sight

The line of sight is the line drawn from the observer’s eye to the object being viewed. It is an oblique line which is therefore not horizontal nor vertical. The line of sight, together with the horizontal line, forms the angle of depression. 

Horizontal Line

The horizontal line is a straight line on a flat surface. In a coordinate system, a horizontal line has points with the same y-coordinate. The line of sight, together with the horizontal line, forms the angle of depression. 

Observer’s Eye

The observer’s eye is the position where the line of sight and the horizontal line meet.

Example

 

An observer is situated at point \( A \).

(a) Find the distance \( AG \) and the angle of elevation of point \( G \).

(b) Find the distance \( AF \) and the angle of elevation of point \( F \).

▶️Answer/Explanation

Solution:

(a) We consider the triangle \( AGB \).

By Pythagoras’ theorem:
$ AG^2 = 4^2 + 3^2 \Rightarrow AG = 5$
The angle of elevation is \( \angle BAG \). Hence:
$ \tan \angle BAG = \frac{3}{4} \Rightarrow \angle BAG = 36.9^\circ $

(b) For point \( F \), we consider the vertical height \( FC \) and thus the triangle \( AFC \).

We first need the side \( AC \). By Pythagoras’ theorem in \( ABC \):
$AC^2 = 4^2 + 5^2 \Rightarrow AC = \sqrt{41} $

Now, by Pythagoras’ theorem in \( AFC \):
$ AF^2 = (\sqrt{41})^2 + 3^2 \Rightarrow AF = \sqrt{50} $
The angle of elevation is \( \angle CAF \). Hence:
$\tan \angle CAF = \frac{3}{\sqrt{41}} \Rightarrow \angle CAF = 25.1^\circ $

Example

An object \( P \) is above a hill. Two observers \( A \) and \( B \) are situated as in the diagram above.

The angle of elevation from \( A \) is \( 45^\circ \).
The angle of elevation from \( B \) is \( 30^\circ \).
The distance between \( A \) and \( B \) is \( 10 \, \text{m} \).

Find the vertical height \( h \) of the object \( P \) above the ground.

▶️Answer/Explanation

Solution:

Consider the triangle:

$ \tan 45^\circ = \frac{h}{x} \iff \frac{h}{x} = 1 \iff h = x $
$ \tan 30^\circ = \frac{h}{x+10} \iff \frac{h}{x+10} = \frac{1}{\sqrt{3}} \iff h\sqrt{3} = x + 10 $

Therefore:
$ h\sqrt{3} = h + 10 \iff h(\sqrt{3} – 1) = 10 \iff h = \frac{10}{\sqrt{3} – 1} \approx 13.7 \, \text{m} $

Notice: Another approach is to work in triangle \( ABP \) first, to find \( AP = 19.318 \), and then by \( \sin 45^\circ = \frac{h}{19.318} \), we find \( h \approx 13.7 \).

 NAVIGATION – BEARING

Navigating with Cardinal and Intercardinal Directions

When using a map for navigation, it’s important to understand the eight principal compass directions:

Cardinal directions: North (N), East (E), South (S), and West (W)
Intercardinal (or intermediate) directions: Northeast (NE), Southeast (SE), Southwest (SW), and Northwest (NW)

These directions are evenly spaced around a circle, with an angle of $45^\circ$ between each consecutive direction.

Angle Between Directions:

The angle between North and East is $90^\circ$, since they are two steps apart (each step being $45^\circ$).
The angle between North and Southeast is $135^\circ$, which spans three steps clockwise from North.

Visual summary of the direction pairs and their angles:

Direction 1Direction 2Angle Between
NorthEast\( 90^\circ \)
NorthSoutheast\( 135^\circ \)
NorthNortheast\( 45^\circ \)
EastSouth\( 90^\circ \)
WestSouthwest\( 45^\circ \)

 

Bearings

Suppose that a moving body goes from point \( A \) to point \( B \).

The bearing of the course \( AB \) is the clockwise angle between the North direction and \( AB \).

According to the diagram:

the bearing of the course \( AB \) is \( 50^\circ \).
the bearing of the course \( BA \) is \( 230^\circ \) 

Types of bearing:

Compass Bearings are used as methods of navigation in relation to the north direction by angles. They help locate objects or positions within a two-dimensional plane, such as a map or diagram.

 

Interpreted as N50ºE (read as ‘fifty degrees north-east’) and S70ºE. Initial side of an angle starts from either North or South which moves clockwise or counterclockwise depending on the nearest initial side.

True bearings are the last navigation method we will be discussing. Often used in navigation, angles are often given in three digits, clockwise from pure north. Let’s learn how to report true bearings.

Interpreted as , no need to express directions. Always starts clockwise from north.

Example

A car travels:
from point \( A \) to point \( B \) with bearing \( 50^\circ \),
then from point \( B \) to point \( C \) with bearing \( 150^\circ \),
then goes back to point \( A \) with bearing \( 270^\circ \).

The distance \( AC \) is \( 10 \, \text{km} \).
Draw a diagram to show the details and find the distances \( AB \) and \( AC \).

▶️Answer/Explanation

Solution:

According to the diagram:
$ \angle A = 40^\circ, \quad \angle B = 80^\circ, \quad \angle C = 60^\circ $

Then, by using the sine rule:
$ \frac{10}{\sin 80^\circ} = \frac{AB}{\sin 60^\circ} = \frac{BC}{\sin 40^\circ} $
we find \( AB = 8.74 \, \text{km} \) and \( BC = 6.53 \, \text{km} \).

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