CIE IGCSE Mathematics (0580) Right-angled triangles Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Right-angled triangles Study Notes
LEARNING OBJECTIVE
- Trigonometric Ratios in Right-Angled Triangles
Key Concepts:
- Trigonometric Ratios
- Bearings
- Use of Pythagoras’ Theorem and Trigonometry
Trigonometric Ratios in Right-Angled Triangles
Trigonometric Ratios in Right-Angled Triangles
In any right-angled triangle, the sine, cosine, and tangent ratios relate the angles to the lengths of the sides. These are used to calculate missing sides or angles when one side and one angle (other than the right angle) are known.
Key Definitions:
- Sine (sin): \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
- Cosine (cos): \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- Tangent (tan): \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)
Important Triangle Terms:
- Hypotenuse: The longest side opposite the right angle
- Opposite: The side opposite the given angle
- Adjacent: The side next to the given angle (not the hypotenuse)
Use of Calculator:
- Make sure your calculator is in degree mode (not radians).
- Round final angle answers to 1 decimal place, and sides to 2 decimal places (unless instructed otherwise).
Example:
In a right triangle, angle \( \theta = 30^\circ \) and the hypotenuse is 10 cm. Find the length of the side opposite to the angle.
▶️ Answer/Explanation
Use sine: \( \sin(30^\circ) = \frac{\text{opposite}}{10} \)
\( 0.5 = \frac{\text{opposite}}{10} \Rightarrow \text{opposite} = 0.5 \times 10 = 5 \)
Answer: 5 cm
Example:
In a right triangle, the opposite side is 7 cm and the adjacent side is 24 cm. Find the angle \( \theta \).
▶️ Answer/Explanation
Use tangent: \( \tan(\theta) = \frac{7}{24} \)
\( \theta = \tan^{-1}\left(\frac{7}{24}\right) \approx \tan^{-1}(0.2917) \approx 16.3^\circ \)
Answer: \( 16.3^\circ \)
Bearings
Understanding Bearings
Bearings are a way of describing direction using angles measured clockwise from the north direction. They are commonly used in navigation, maps, and word problems involving direction.
A bearing is an angle, measured clockwise from the north direction. Below, the bearing of B from A is 025 degrees (note 3 figures are always given). The bearing of A from B is 205 degrees.
Key Facts about Bearings:
- Bearings are measured clockwise from the north direction.
- Bearings are always written as three-digit numbers (e.g. 045°, 120°, 270°).
- The north line is drawn vertically upwards from the point of measurement.
- Use a protractor to measure bearings on a diagram.
Types of Bearings:
- True Bearing:
- Measured from the north in a clockwise direction.
- Back Bearing:
- The bearing from one point back to another point (difference of 180°).
Important Tips:
- Draw a diagram to visualize the problem.
- Mark north lines clearly at the point where a bearing is to be measured.
- Use geometry and angles in parallel lines to help deduce unknown bearings (e.g. alternate and corresponding angles).
Example:
Ship A is directly east of Ship B. What is the bearing of A from B?
▶️ Answer/Explanation
From B, draw a north line. Since A is to the east, the angle clockwise from north to east is 90°.
Answer: The bearing of A from B is 090°.
Example:
Point C is on a bearing of 135° from point D. What is the bearing of D from C?
▶️ Answer/Explanation
The back bearing is found by adding or subtracting 180°:
\( 135^\circ + 180^\circ = 315^\circ \)
(No need to subtract since it’s less than 180°)
Answer: The bearing of D from C is 315°.
Example :
A ship sails 200 km on a bearing of 243.7°.
- a) How far south has it traveled?
- b) How far west has it traveled?
▶️ Answer/Explanation
Interpret the bearing.
- Bearing of 243.7° is measured clockwise from North.
- Angle between the ship’s path and the south direction is:
\( \angle SOA = 243.7^\circ – 180^\circ = 63.7^\circ \)
Use trigonometry to resolve the 200 km path into south and west components.
a) Southward travel (opposite to angle):
Let \( x \) be the distance south.
\( \sin(63.7^\circ) = \frac{x}{200} \Rightarrow x = 200 \times \sin(63.7^\circ) \approx 179.3 \, \text{km} \)
b) Westward travel (adjacent to angle):
Let \( y \) be the distance west.
\( \cos(63.7^\circ) = \frac{y}{200} \Rightarrow y = 200 \times \cos(63.7^\circ) \approx 88.6 \, \text{km} \)
Solving 2D Problems Using Pythagoras’ Theorem and Trigonometry
Solving 2D Problems Using Pythagoras’ Theorem and Trigonometry
In real-world and geometric problems, Pythagoras’ Theorem and trigonometric ratios (sine, cosine, and tangent) are used to find missing lengths or angles in right-angled triangles. These are common in two-dimensional applications such as ladders leaning against walls, ramps, and distances on maps.
Pythagoras’ Theorem in 2D
Use when you are given two sides of a right-angled triangle and want to find the third side.
\( \text{Hypotenuse}^2 = (\text{Opposite side})^2 + (\text{Adjacent side})^2 \) or \( c^2 = a^2 + b^2 \)
Trigonometry in 2D
Use trigonometry when you know:
- One angle (other than 90°)
- And one side of the triangle
Use SOH CAH TOA:
- Sine: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
- Cosine: \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- Tangent: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)
Key Tips:
- Always check that the triangle is right-angled before applying trigonometry or Pythagoras.
- Label the triangle clearly with hypotenuse, opposite, and adjacent based on the given angle.
- Make sure your calculator is in degree mode.
Example :
A ladder is placed 3 m away from a wall and reaches a height of 4 m up the wall. How long is the ladder?
▶️ Answer/Explanation
Use \( c^2 = a^2 + b^2 \)
\( c^2 = 3^2 + 4^2 = 9 + 16 = 25 \)
\( c = \sqrt{25} = 5 \)
Answer: The ladder is 5 m long.
Example :
A ramp is 8 m long and makes an angle of 30° with the horizontal. How high does it rise?
▶️ Answer/Explanation
We are given hypotenuse and need opposite → use sine:
\( \sin(30^\circ) = \frac{\text{opposite}}{8} \)
\( \text{opposite} = 8 \times \sin(30^\circ) = 8 \times 0.5 = 4 \)
Answer: The ramp rises 4 m.
Example :
A fire truck ladder is 10 m long and is set at an angle of 60° to the ground. The base of the ladder is placed 3 m away from the wall. How high up the wall does the ladder reach, and how far is the top of the ladder from the ground vertically?
▶️ Answer/Explanation
Use trigonometry to find the height (opposite side).
Given: Hypotenuse = 10 m, angle = 60°
\( \sin(60^\circ) = \frac{\text{height}}{10} \)
\( \text{height} = 10 \times \sin(60^\circ) = 10 \times 0.866 = 8.66 \)
Check the horizontal distance using cosine:
\( \cos(60^\circ) = \frac{\text{base}}{10} \Rightarrow \text{base} = 10 \times \cos(60^\circ) = 10 \times 0.5 = 5 \, \text{m} \)
But the base is placed at 3 m from the wall. This means the actual position contradicts the angle. So we re-calculate using the known adjacent side.
Use Pythagoras to find vertical height based on adjacent = 3 m and hypotenuse = 10 m:
\( h^2 = 10^2 – 3^2 = 100 – 9 = 91 \Rightarrow h = \sqrt{91} \approx 9.54 \, \text{m} \)
Answer: The ladder reaches approximately 9.54 m up the wall.
Example :
A ship sails 50 km on a bearing of 045°, then changes course and sails 80 km on a bearing of 135°. How far is the ship from its starting point?
▶️ Answer/Explanation
Draw a diagram (not shown here) with two vectors from a common point (the origin):
- First leg: 50 km at 045° (i.e. 45° from north-east)
- Second leg: 80 km at 135° (i.e. 135° clockwise from north, which lies in the south-east direction)
Use the cosine rule to find the distance between the starting point and final position:
Let angle between the two bearings = \( 135^\circ – 45^\circ = 90^\circ \)
Using the cosine rule:
\( c^2 = a^2 + b^2 – 2ab\cos(C) \)
\( c^2 = 50^2 + 80^2 – 2(50)(80)\cos(90^\circ) \)
\( c^2 = 2500 + 6400 – 0 \Rightarrow c = \sqrt{8900} \approx 94.34 \)
Answer: The ship is approximately 94.34 km from its starting point.
Perpendicular Distance from a Point to a Line
Perpendicular Distance from a Point to a Line
The shortest distance from a point to a straight line is the length of the line segment drawn from the point perpendicular to the line.
Any other line segment from the point to the line is longer.
Key Fact:
- The perpendicular from a point to a line makes a right angle (90°) with the line.
- It is the minimum possible distance from the point to the line.
Example:
A point \( P \) lies 4 cm above a line \( AB \). A segment is drawn from \( P \) to point \( C \) on the line, such that \( PC = 5 \) cm and \( \angle PCB = 90^\circ \). Which is the shortest distance from \( P \) to the line?
▶️ Answer/Explanation
The shortest distance is the perpendicular from point \( P \) to line \( AB \).
Given this is 4 cm, and other paths (like \( PC \)) are longer (5 cm), we conclude:
Shortest distance = 4 cm
Example:
In the coordinate plane, the point \( A(2, 3) \) lies above the line \( y = 1 \). Find the perpendicular distance from point \( A \) to the line.
▶️ Answer/Explanation
The vertical (perpendicular) distance between the y-values:
\( \text{Distance} = 3 – 1 = \boxed{2 \text{ units}} \)
Angles of Elevation and Depression
Angles of Elevation and Depression
Angles of elevation and depression are measured from the horizontal line of sight. These angles are common in real-world trigonometry problems involving heights and distances.
- Angle of Elevation: The angle between the horizontal and the line of sight looking up.
- Angle of Depression: The angle between the horizontal and the line of sight looking down.
We solve these problems using right-angled triangle trigonometry:
- \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
- \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
Example:
A man observes the top of a tower at an angle of elevation of \( 35^\circ \). He is standing 40 m away from the base of the tower. Find the height of the tower.
▶️ Answer/Explanation
We form a right-angled triangle:
- Opposite = height of tower
- Adjacent = 40 m
- \( \tan(35^\circ) = \frac{\text{height}}{40} \)
\( \text{height} = 40 \times \tan(35^\circ) \approx 40 \times 0.7002 = \boxed{28.01 \, \text{m}} \)
Example:
A girl stands on top of a 20 m high cliff and observes a boat at sea. The angle of depression is \( 28^\circ \). How far is the boat from the base of the cliff (horizontal distance)?
▶️ Answer/Explanation
We form a right-angled triangle:
- Opposite = 20 m (height)
- Angle = \( 28^\circ \)
- \( \tan(28^\circ) = \frac{20}{x} \Rightarrow x = \frac{20}{\tan(28^\circ)} \)
\( x \approx \frac{20}{0.5317} = \boxed{37.62 \, \text{m}} \)
Example:
From a point on the ground, the angle of elevation to the top of a vertical pole is \( 60^\circ \). If the pole is 10 m tall, find the horizontal distance from the point to the base of the pole.
▶️ Answer/Explanation
Let \( x \) be the horizontal distance.
\( \tan(60^\circ) = \frac{10}{x} \Rightarrow x = \frac{10}{\tan(60^\circ)} = \frac{10}{\sqrt{3}} \approx \boxed{5.77 \, \text{m}} \)