IB Mathematics AA Applications of trigonometry Study Notes
IB Mathematics AA Applications of trigonometry Study Notes
IB Mathematics AA Applications of trigonometry Study Notes Offer a clear explanation of Applications of trigonometry , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Applications of trigonometry.
Applications of Right and Non-Right Angled Trigonometry
Applications of Right and Non-Right Angled Trigonometry
Trigonometry is used to calculate unknown sides or angles in triangles.
Right-angled trigonometry: Uses sine, cosine, and tangent ratios:
\( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
Pythagoras’s Theorem:
\( a^2 + b^2 = c^2 \) where \(c\) is the hypotenuse of a right-angled triangle.
Non-right angled trigonometry: Uses sine rule and cosine rule:
\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \) \( c^2 = a^2 + b^2 – 2ab \cos C \)
Example :
A ladder leans against a wall making a \(60^\circ\) angle with the ground. The ladder is 10 m long. How high up the wall does it reach?
▶️ Answer/Explanation
We use sine because we know hypotenuse and need the opposite side:
\( \sin 60^\circ = \frac{\text{height}}{10} \)
\(\sin 60^\circ = 0.866\)
\( \text{height} = 10 \times 0.866 = 8.66 \ \text{m} \)
Example :
In triangle ABC, \(A = 50^\circ\), \(B = 60^\circ\), and \(a = 8 \ \text{cm}\). Find \(b\).
▶️ Answer/Explanation
Using sine rule:
\( \frac{a}{\sin A} = \frac{b}{\sin B} \)
\( \frac{8}{\sin 50^\circ} = \frac{b}{\sin 60^\circ} \)
\(\sin 50^\circ = 0.766\), \(\sin 60^\circ = 0.866\)
\( \frac{8}{0.766} = \frac{b}{0.866} \) \( 10.44 = \frac{b}{0.866} \) \( b = 10.44 \times 0.866 = 9.04 \ \text{cm} \)
Angles of Elevation and Depression
Angles of Elevation and Depression
- Angle of elevation: The angle formed between the horizontal and the line of sight when looking up at an object.
- Angle of depression: The angle formed between the horizontal and the line of sight when looking down at an object.
These angles are often used to solve real-world problems involving right triangles and trigonometric ratios.
Example:
A person is standing 30 m away from a building. The angle of elevation to the top of the building is \(40^\circ\). Find the height of the building.
▶️ Answer/Explanation
Let \(h\) be the height of the building.
\( \tan 40^\circ = \frac{h}{30} \)
\(\tan 40^\circ = 0.8391\)
\( h = 30 \times 0.8391 = 25.17 \ \text{m} \)
Example:
A lighthouse is 50 m high. The angle of depression to a boat is \(30^\circ\). How far is the boat from the base of the lighthouse?
▶️ Answer/Explanation
Let \(d\) be the distance from the base of the lighthouse to the boat.
\( \tan 30^\circ = \frac{50}{d} \)
\(\tan 30^\circ = 0.5774\)
\( d = \frac{50}{0.5774} = 86.6 \ \text{m} \)
Navigation – Bearing
Bearing
A bearing is a way of describing direction or angle in navigation and surveying.It is measured clockwise from the North (0°) and is always expressed as a three-digit number. Bearings are measured from the North line at the starting point to the line joining the starting point to the destination point.
For example, East is 090°, South is 180°, West is 270°.
Notation and Representation
- Bearing angles are given in the format: 000° to 359°.
- A bearing of 045° means 45° clockwise from North (i.e., NE direction).
- A bearing of 225° means SW direction (180° + 45°).
- Always use three figures when writing bearings (e.g., 045°, not 45°).
Types of Bearings
- True bearing: Measured from true North.
- Magnetic bearing: Measured from magnetic North (compass bearing).
- Relative bearing: Measured from the heading of a vessel or aircraft.
How to Measure and Use Bearings
- Start at the North line at your point of reference.
- Measure clockwise the angle formed between the North and the line joining the two points.
- Apply trigonometry (sine rule, cosine rule) to calculate distances or angles in navigation problems involving bearings.
Example:
A ship sails 50 km on a bearing of 135° from point A. How far east and how far south has the ship travelled?
▶️Answer/Explanation
Draw and label a diagram.
The bearing of 135° means the ship is sailing SE (135° clockwise from North).
Resolve the 50 km into east and south components:
- East distance = \( 50 \sin 45^\circ = 50 \times 0.7071 = 35.36 \ \text{km} \)
- South distance = \( 50 \cos 45^\circ = 50 \times 0.7071 = 35.36 \ \text{km} \)