a. Four planes of symmetry does the pyramid have.
b. Volume of the pyramid is 30 cm³.
c. The angle between PM and the base is 45°.
d. The angle between PB and the base is 37.53 approximately.
e. The length of PB = 5.
Given that,
The diagram depicts a pyramid on a foundation with the dimensions AB=6 cm and
AD=5 cm. At point F, the diagonals AC and BD converge. FP= 3 cm
in height vertically.
We know that,
a. Four planes of symmetry does the pyramid have.
b. Area of the base = 6 × 5 = 30 cm²
Volume of the pyramid is × area of base × height in cubic units
= × 30 × 3
= 30 cm³
c. Construct FM, joint F and M
From ΔABD, AB = 6cm and AD = 5cm
BD² = AB² + AD² = 36+25 =61
BD = √61
Then BF=
Again BM = 2.5
From ΔBMF
FM² = BF² - BM² =
=
FM = =3
Now, from ΔFMP
tan∠FMP =
=
tan∠FMP = = 1 = tan45°
∠FMP = 45°
The angle between PM and the base is 45°.
d. From Δ PFB
tan∠PBF =
=
tan∠PBF =
=
∠PBF = tan⁻¹( )
∠PBF = 37.532236
The angle between PB and the base is 37.53 approximately.
e. From right angle triangle PFB
PB² = PF² + BF² = 3² + ( )² =
PB =
= 4.9244 = 5
The length of PB = 5.
Problem Description
The height of a shop entrance above the pavement is to be calculated. A wheelchair
ramp with a length of 3.17 meters is inclined at an angle of 5 degrees to the
horizontal.
Solution
1. The ramp length is the hypotenuse of a right-angled triangle, and the
height of the entrance is the opposite side to the given angle.
2. The sine function is used to relate the opposite side and the hypotenuse.
3. The height is calculated by multiplying the ramp length by the sine of the
angle.
Answer:
1. The relationship between the height, ramp length and angle is expressed
as:
sin(5∘)=ℎ3.17
2. Height, ℎ is calculated as:
ℎ=3.17×sin(5∘)
3. Using a calculator,
sin(5∘)≈0.08716
4. Therefore,
ℎ=3.17×0.08716≈0.276
5. The height, ℎ, of the entrance above the pavement is approximately 0.276
meters.
What you're solving for
The distance of the submarine S from lighthouses A and B.
What's given in the problem




The distance between lighthouses A and B is 25 km.
Lighthouse A is due West of B.
The bearing of S from A is 137∘.
The bearing of S from B is 170∘
Helpful information


The sum of angles in a triangle is 180.
The sine rule states that a/sinA = b/sinB = c/sinC for a triangle with sides
a,b,c and opposite angles A,B,C.
How to solve
Use the sine rule to find the distances of S from A and B.
Step 1 . Find the angle ∠SAB
o
o
o
The bearing of S from A is 137∘.
The angle between the North direction and the line AB is 180∘.
The angle ∠SAB is 180∘−137∘=43∘
Step 2 . Find the angle ∠SBA
o
o
o
The bearing of S from B is 170∘
The angle between the North direction and the line BA is 0∘.
The angle ∠SBA is 170∘−180∘=-10∘.
The angle ∠SBA is 180∘−170∘=10∘.
Step 3 . Find the angle ∠ASB
The sum of angles in a triangle is 180∘.
∠ASB=180∘−∠SAB−∠SB.
∠ASB=180∘−43∘−10∘=127∘
Step 4 . Use the sine rule to find the distance AS
AS/sin(∠SBA)=AB/sin(∠ASB).
AS/sin(10∘)=25/sin(127∘).
AS=25×sin(10∘)/sin(127∘).
AS≈5.42km.
Step 5 . Use the sine rule to find the distance BS
BS/sin(∠SAB)=AB/sin(∠ASB)
BS/sin(43∘)=25/sin(127∘).
𝐵𝑆=25×sin(43∘)/sin(127∘)
BS≈20.45 km.
Solution
The distance of S from A is approximately 5.42 km and
the distance of S from B is approximately 20.45 km.
Provide formulas, theorems with derivations and important key points, examples to have vast knowledge
about Trigonometry keeping the syllabus of IGCSE Maths Extended syllabus
=======================================================================================
Certainly! Here's a comprehensive guide to Trigonometry tailored for the IGCSE Mathematics
Extended syllabus, covering key formulas, theorems, derivations, and examples.
📘 IGCSE Trigonometry Syllabus Overview
The Cambridge IGCSE Mathematics Extended syllabus (Code 0580) includes the following
trigonometric topics:(ckmacademy.com)
1. Right-Angled Triangles: Pythagoras’ Theorem, trigonometric ratios (sine, cosine, tangent),
angles of elevation and depression.
2. Sine and Cosine Rules: For non-right-angled triangles.
3. Area of a Triangle: Using the formula 12absin⁡C\frac{1}{2}ab \sin C.
4. Trigonometric Graphs: Understanding sine, cosine, and tangent functions.
5. 3D Trigonometry: Angles between lines and planes.
6. Bearings: Three-figure bearings and their applications.(tychr.com)
📐 Key Formulas and Theorems
1. Pythagoras’ Theorem
In a right-angled triangle:(tychr.com)
where cc is the hypotenuse, and aa and bb are the other two sides.
2. Trigonometric Ratios
For a right-angled triangle:



sin⁡θ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}
cos⁡θ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}
tan⁡θ=oppositeadjacent\tan \theta = \frac{\text{opposite}}{\text{adjacent}}
Mnemonic: SOHCAHTOA(savemyexams.com)
3. Sine Rule
Useful for non-right-angled triangles when two angles and one side are known.
4. Cosine Rule
Useful for non-right-angled triangles when two sides and the included angle are known.
5. Area of a Triangle
where aa and bb are two sides, and CC is the included angle.
6. Exact Trigonometric Values
For standard angles:(tychr.com)



sin⁡30∘=12\sin 30^\circ = \frac{1}{2}, cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2},
tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}
sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2}, cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2},
tan⁡45∘=1\tan 45^\circ = 1
sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}, cos⁡60∘=12\cos 60^\circ = \frac{1}{2},
tan⁡60∘=3\tan 60^\circ = \sqrt{3}
� Important Theorems and Concepts
Angles of Elevation and Depression


Angle of Elevation: The angle formed by the line of sight above the horizontal.
Angle of Depression: The angle formed by the line of sight below the horizontal.
These concepts are crucial for solving real-world problems involving heights and distances.
Bearings



Bearings are measured clockwise from the north direction.
Expressed as three-figure angles (e.g., 045°).
Used in navigation and surveying to determine directions.(ckmacademy.com, tychr.com)
📊 Trigonometric Graphs



Sine Graph: Starts at (0,0), oscillates between -1 and 1, with a period of 360°.
Cosine Graph: Starts at (0,1), oscillates between -1 and 1, with a period of 360°.
Tangent Graph: Starts at (0,0), has vertical asymptotes at 90° and 270°, with a period of 180°.
� Example Problems
Example 1: Using the Sine Rule
Problem: In triangle ABC, A=30∘A = 30^\circ, B=45∘B = 45^\circ, and side a=10a = 10. Find side bb.
Solution:
1. Calculate angle CC:
C = 180^\circ - A - B = 180^\circ - 30^\circ - 45^\circ = 105^\circ
\]:contentReference[oaicite:127]{index=127}
2. Apply the Sine Rule:
\frac{a}{\sin A} = \frac{b}{\sin B} \]:contentReference[oaicite:132]{index=132} \[ \frac{10}{\sin 30^\circ}
= \frac{b}{\sin 45^\circ} \]:contentReference[oaicite:134]{index=134} \[ \frac{10}{0.5} = \frac{b}{0.707}
\]:contentReference[oaicite:136]{index=136} \[ 20 = \frac{b}{0.707}
🔄 Bearings: Full Explanation
🔹 What Is a Bearing?
A bearing is a way to describe direction using angles measured clockwise from North.




Bearings are always written as three-figure numbers.
The reference direction is always due North (0°).
Bearings go clockwise from North.
Examples:
o North: 000°
o East: 090°
o South: 180°
o West: 270°
o North-East: 045°
o South-West: 225°
🔹 Why Three Digits?
Bearings are always expressed with three digits to avoid confusion.
Direction Bearing
North
000°
North-East 045°
East
090°
South
180°
South-West 225°
West
270°
� Real-World Use of Bearings
Bearings are commonly used in:



Navigation (e.g., ships, planes)
Mapping (e.g., plotting locations)
Surveying (e.g., finding angles and distances on land)
✍� How to Interpret Bearings



You always start at the first point.
Face North, then turn clockwise the given number of degrees to face the correct direction.
To go from one point to another, draw a North line at the starting point, then measure the
angle to the other point clockwise.
📐 How Bearings Relate to Trigonometry
Bearings often involve non-right angled triangles, where you apply:




Sine Rule
Cosine Rule
Basic Trigonometric Ratios (SOHCAHTOA) for right-angled triangle bearings
Scale diagrams and distance/angle estimation
📌 Key Concepts and Tips
Concept
Explanation
Bearings are measured from North Always draw a vertical line representing North
Clockwise
Bearings are always measured clockwise
Use accurate diagrams
Diagrams help understand direction and angle
Opposite Bearings
A to B is the reverse of B to A: Add or subtract 180°
📘 Worked Examples
� Example 1: Interpreting Bearings
Problem: A ship sails on a bearing of 120°. What direction is it heading?
Solution:



Start from North.
Move clockwise 120°.
That places the ship south-east, between east (90°) and south (180°).
� Example 2: Converting Between Directions and Bearings
Problem: What is the bearing of South-West?
Solution:



South = 180°
West = 270°
South-West is halfway between them:
Bearing=180∘+270∘2=225∘\text{Bearing} = \frac{180^\circ + 270^\circ}{2} = 225^\circ
� Example 3: Finding a Reverse Bearing
Problem: If a plane flies from A to B on a bearing of 045°, what is the bearing of B to A?
Solution:

Add 180°:
045∘+180∘=225∘045^\circ + 180^\circ = 225^\circ

So, B to A is on a bearing of 225°.
� Example 4: Solving Triangle Using Bearings (Right-Angled)
Problem: A ship leaves a port and sails 10 km on a bearing of 060°, then 8 km on a bearing of 150°.
Find how far the ship is from the port.
Approach:
1. Draw diagram with bearings.
2. Create triangle with known sides and angles.
3. Use cosine rule or sine rule to find unknown sides.
� Example 5: Using Bearings and Sine Rule
Problem:
A ship is at point A. It sails to point B on a bearing of 040°, and the distance AB = 200 km. Then it
sails to point C on a bearing of 130°, and BC = 150 km. Find:
a) the angle at B
b) the distance AC
Solution:
1. Draw the triangle ABC with accurate bearings.
2. Find angle B:
o Difference in bearings = 130° - 40° = 90°
o So, ∠ABC = 90°
3. Use cosine rule or Pythagoras:
o Since ∠B = 90°, triangle is right-angled.
o
AC^2 = AB^2 + BC^2
]
o
AC^2 = 200^2 + 150^2 = 40000 + 22500 = 62500
\Rightarrow AC = \sqrt{62500} = 250 \text{ km}
]
🔍 Questions to Practice
1. Convert the following directions into bearings:
o North-East
o South-West
o North-West
2. A plane flies on a bearing of 310° from City A to City B. What is the bearing from City B to
City A?
3. Two ships leave a port. Ship A travels 10 km on a bearing of 045°, Ship B travels 8 km on a
bearing of 135°. Find the distance between them.
4. A man walks 3 km due east and then 4 km due south. What is the bearing from his starting
point to his final position?
how to draw the bearings diagram step-by-step for this example:
📘 Scenario Recap:
A ship sails from point A to point B on a bearing of 045°, and then from point B to point C on a
bearing of 135°.
Distance AB = 10 km, and BC = 8 km.
Find the distance AC (from the starting point to the final point).
✏� Step-by-Step Guide to Drawing the Diagram
✅ Step 1: Draw North Lines
1. Start with point A.
2. Draw a vertical upward line through point A. Label it North (N).
3. Do the same later for point B.
✅ Step 2: Draw Line AB at 045° from A





Bearings are measured clockwise from North.
Measure a 45° angle clockwise from the vertical North line at point A.
From point A, draw a line slanting up and right at 45°. Label the end of this line as point B.
Mark the angle between the North line and AB as 45°.
Label the distance AB = 10 km.
✅ Step 3: Draw North Line at Point B

Draw a vertical upward line at point B, also labeled North.
✅ Step 4: Draw Line BC at 135° from B





From point B, measure 135° clockwise from the North line.
Since 135° is between East (90°) and South (180°), draw a line slanting downward to the
right.
Label the end of this line as point C.
Label distance BC = 8 km.
The angle between the North line at B and line BC is 135°.
✅ Step 5: Complete Triangle ABC


Connect point A to point C to form triangle ABC.
You can now use trigonometry (e.g., cosine rule) to calculate AC, the direct distance from A to
C.
🔺 Geometry Summary
Now that you have a triangle:


∠ABC is the angle between the directions AB (bearing 045°) and BC (bearing 135°).
To find the interior angle ∠ABC:
∠ABC=135∘−45∘=90∘\angle ABC = 135^\circ - 45^\circ = 90^\circ

So, triangle ABC is a right-angled triangle at point B.
� Now Calculate AC Using Pythagoras' Theorem
Since ∠B = 90°:
AC2=AB2+BC2AC^2 = AB^2 + BC^2 AC2=102+82=100+64=164AC^2 = 10^2 + 8^2 = 100 + 64 = 164
AC=164≈12.81 kmAC = \sqrt{164} \approx 12.81 \, \text{km}
✅ So the distance from A to C is approximately 12.81 km.
📐 1. Pythagoras' Theorem
🔹 Theorem Statement:
In a right-angled triangle:
Hypotenuse2=(Opposite Side)2+(Adjacent Side)2\text{Hypotenuse}^2 = (\text{Opposite Side})^2 +
(\text{Adjacent Side})^2 c2=a2+b2c^2 = a^2 + b^2
✅ Note: The hypotenuse is the side opposite the right angle — always the longest side.
� Example 1 – Finding the Hypotenuse
Given:


One side = 6 cm
Another side = 8 cm
Find: The hypotenuse.
c2=62+82=36+64=100⇒c=100=10 cmc^2 = 6^2 + 8^2 = 36 + 64 = 100 \Rightarrow c = \sqrt{100} = 10 \text{ cm}
� Example 2 – Finding a Missing Leg
Given:


Hypotenuse = 13 cm
One side = 5 cm
Find: The other side.
c2=a2+b2⇒132=52+b2⇒169=25+b2⇒b2=144⇒b=144=12 cmc^2 = a^2 + b^2 \Rightarrow 13^2 = 5^2 + b^2
\Rightarrow 169 = 25 + b^2 \Rightarrow b^2 = 144 \Rightarrow b = \sqrt{144} = 12 \text{ cm}
📐 2. Trigonometric Ratios
Use these only in right-angled triangles.
Ratio
Sine (sin⁡\sin)
Formula
Relation
OppositeHypotenuse\frac{\text{Opposite}}{\text{Hypotenuse}} For vertical sides
Cosine (cos⁡\cos) AdjacentHypotenuse\frac{\text{Adjacent}}{\text{Hypotenuse}} For horizontal sides
Tangent (tan⁡\tan) OppositeAdjacent\frac{\text{Opposite}}{\text{Adjacent}}
🎓 Mnemonic: SOH-CAH-TOA
� Example 3 – Finding a Side Using Sine
Opposite to Adjacent
Triangle ABC, right-angled at C:


Angle A = 30°
Hypotenuse = 10 cm
Find: Opposite side to angle A.
sin⁡A=OppositeHypotenuse⇒sin⁡30∘=x10⇒0.5=x10⇒x=5 cm\sin A =
\frac{\text{Opposite}}{\text{Hypotenuse}} \Rightarrow \sin 30^\circ = \frac{x}{10} \Rightarrow 0.5 =
\frac{x}{10} \Rightarrow x = 5 \text{ cm}
� Example 4 – Finding an Angle Using Tangent
Given:


Opposite = 5 cm
Adjacent = 8 cm
Find: Angle θ
tan⁡θ=58⇒θ=tan⁡−1(58)=tan⁡−1(0.625)⇒θ≈32∘\tan \theta = \frac{5}{8} \Rightarrow \theta = \tan^{1}\left(\frac{5}{8}\right) = \tan^{-1}(0.625) \Rightarrow \theta \approx 32^\circ
📐 3. Angles of Elevation and Depression
🔹 Definitions:


Angle of Elevation: The angle above horizontal when looking up.
Angle of Depression: The angle below horizontal when looking down.
✅ These angles form right-angled triangles in real-life height/distance problems.
� Example 5 – Angle of Elevation
A person is standing 40 m from a tower. The angle of elevation to the top of the tower is 30°. Find the
height of the tower.
Use:
tan⁡(θ)=oppositeadjacent⇒tan⁡(30∘)=h40\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \Rightarrow
\tan(30^\circ) = \frac{h}{40} 0.577=h40⇒h=0.577×40=23.1 m0.577 = \frac{h}{40} \Rightarrow h = 0.577 \times
40 = 23.1 \text{ m}
� Example 6 – Angle of Depression
From a 25 m tall building, the angle of depression to a car on the road is 40°. How far is the car from
the base of the building?
tan⁡(40∘)=25x⇒x=25tan⁡(40∘)=250.8391≈29.8 m\tan(40^\circ) = \frac{25}{x} \Rightarrow x =
\frac{25}{\tan(40^\circ)} = \frac{25}{0.8391} \approx 29.8 \text{ m}
📐 4. Shortest Distance from a Point to a Line
🔹 Key Concept:


The shortest distance from a point to a line is the perpendicular line.
It’s the “height” when forming a right-angled triangle with the point and line.
� Example 7 – Perpendicular Distance
A tree 20 m tall casts a shadow 15 m long. Find the angle of elevation of the sun.
We form a right-angled triangle:


Opposite = 20
Adjacent = 15
tan⁡θ=2015=1.333⇒θ=tan⁡−1(1.333)≈53.1∘\tan \theta = \frac{20}{15} = 1.333 \Rightarrow \theta = \tan^{1}(1.333) \approx 53.1^\circ
� Summary Table
Concept
Use
Formula
Pythagoras' Theorem
Finding a side in right triangle c2=a2+b2c^2 = a^2 + b^2
Sine
Opposite and hypotenuse
sin⁡θ=OH\sin \theta = \frac{O}{H}
Cosine
Adjacent and hypotenuse
cos⁡θ=AH\cos \theta = \frac{A}{H}
Tangent
Opposite and adjacent
tan⁡θ=OA\tan \theta = \frac{O}{A}
Elevation
Above horizontal line
Use right triangle
Depression
Below horizontal line
Use right triangle
Perpendicular Distance Shortest path
tan(40∘)=x25⇒x=tan(40∘)25=0.839125≈29.8 m
Use trigonometry (often sine or cosine)
📌 OVERVIEW OF KEY CONCEPTS:
1.
2.
3.
4.
5.
Trigonometric Functions & Definitions
Graphing Trigonometric Functions
Interpreting Trigonometric Graphs
Solving Trigonometric Equations (0° to 360°)
Applications in Real-World Contexts
� Recap: What You Should Know
Topic
Key Points
Sine, Cosine, Tangent Ratios in right triangles and unit circle
Graphs
Understand amplitude, period, range, and shift
Solving equations
Use inverse functions and unit circle rules
Real-life applications Apply trig to geometry, physics, architecture, etc.