Trigonometry – GCSE Maths
Introduction
- Trigonometry is all about Triangles.
- It is a branch of mathematics that deals with the relationships between the angles and sides of triangles—especially right-angled triangles.
Basics of Trigonometry
Trigonometry is the study of the relationship between the angles and sides of triangles.
1. Why do we use it?
To find:
- How long a side is
- What an angle is
—when we have the values of some other parts of the triangle.
2. The Three main Functions:
In a right-angled triangle:
- sin (as: “sine”)
- cos (as: “cosine”)
- tan (as: “tangent”)
They are simply the ratios (fractions) of the given triangle’s sides.
3. All about Triangles:
Triangles are three-sided polygons with several important properties. Here are some key properties of triangles:-
Basic Properties-
- A triangle has three sides, three vertices, and three angles.
- The sum of the interior angles is always 180°.
- The sum of the exterior angles is always 360°.
Side Length Rule (Triangle Inequality Theorem)
There are mainly four types of Triangles that can be distinguished uniquely.
Let us understand about them in detail:
Angles Of Elevation & Depression
Definitions-
- Angle of Elevation: The angle formed between the horizontal line (eye level) and the line of sight when an observer looks upwards at an object.
- Angle of Depression: The angle formed between the horizontal line (eye level) and the line of sight when an observer looks downwards at an object.
Key Points-
- Both angles are measured from the horizontal (eye level).
- They are always between 0° and 90°.
- The angle of elevation and depression are congruent (equal) when the observer and object are at the same horizontal level (i.e., in symmetric positions).
Real Life Applications-
- Angle of Elevation: Used in measuring heights of buildings, mountains, or trees.
- Angle of Depression: Used in aviation (pilots landing planes), navigation, or determining distances between objects at different heights.
Step by Step Procedure-
- Step#1: Draw a Diagram
- Step#2: Identify known and unknown values
- Step#3: Choose the Right Trigonometric Ratio
- Step#4: Solve for the Unknown
- Step#5: Check for Angle of Depression
Solved Example:
Solved Example: Example: “A bird sits on a tree 10m high. A man 20 m away looks up at the bird.”
Solution:
Step#1: Draw a Diagram-
- Sketch the scenario based on the problem statement.
- Label-
- The observer’s eye level (horizontal line).
- The line of sight (angle of elevation or depression).
- The height (vertical side) and distance (horizontal side).
Step#2: Identify Known & Unknown Values-
- Given:
- Distance from observer to object (adjacent side).
- Height (opposite side).
- Angle (if given).
- Find:
- The missing side or angle.
Example:
- Given:
- Find: Angle of Elevation (θ).
Step#3: Choose the Right Trigonometric Ratio-
- SOH-CAH-TOA helps decide which ratio to use:
- Sine (sinθ) = Opposite / Hypotenuse
- Cosine (cosθ) = Adjacent / Hypotenuse
- Tangent (tanθ) = Opposite / Adjacent
In our example:
- We have opposite (height) = 10m and adjacent (distance) = 20m.
- Use tangent-
Step#4: Solve for the Unknown-
- If finding an angle, use inverse trig functions (tan⁻¹, sin⁻¹, cos⁻¹).
- If finding a side, rearrange the formula.
Example (continued):
- To find θ:
- θ = tan−1(0.5) ≈ 26.57°
Step#5: Check for Angle of Depression-
- If the problem involves looking downward, the steps are the same, but the angle is measured below the horizontal.
Key Fact:
- Angle of elevation from point A to B = Angle of depression from B to A (they are equal due to alternate angles).
Therefore,
Angle of Elevation = Angle of Depression
Hence,
Angle of depression ≈ 26.57°
Solved Example:
Example: A bird is perched on a 15-meter-high tree. It spots a worm on the ground 9 meters away from the base of the tree. What is the angle of depression from the bird to the worm?
Solution:
Step#1: Draw a Diagram-
- Sketch the scenario based on the problem statement.
- Label:
- The observer’s eye level (horizontal line).
- The line of sight (angle of elevation or depression).
- The height (vertical side) and distance (horizontal side).
Step#2: Identify Known & Unknown Values-
- Given:
- Distance from observer to object (adjacent side).
- Height (opposite side).
- Angle (if given).
- Find:
- The missing side or angle.
Example:
- Given:
- Distance (adjacent) = 9m
- Height (opposite) = 15m
- Find: Angle of depression(θ).
Step#3: Choose the Right Trigonometric Ratio-
- SOH-CAH-TOA helps decide which ratio to use:
- Sine (sinθ) = Opposite / Hypotenuse
- Cosine (cosθ) = Adjacent / Hypotenuse
- Tangent (tanθ) = Opposite / Adjacent
In our example:
- We have opposite (height) = 15m and adjacent (distance) = 9m.
- Use tangent-
Step#4: Solve for the Unknown-
- If finding an angle, use inverse trig functions (tan⁻¹, sin⁻¹, cos⁻¹).
- If finding a side, rearrange the formula.
Example (continued):
- To find θ:
- θ = tan−1(1.67) ≈ 59.3°
Step#5: Check for Angle of Elevation-
- If the problem involves looking downward, the steps are the same, but the angle is measured below the horizontal.
Key Fact:
- Angle of elevation from point A to B = Angle of depression from B to A (they are equal due to alternate angles).
Therefore,
Angle of Elevation = Angle of Depression
Hence,
Angle of depression ≈ 59.3°
Triangles Exact Values
Let us understand about some important ratios in brief:
- Opposite = side opposite the angle
- Adjacent = side next to the angle (not the hypotenuse)
- Hypotenuse = the longest side (opposite the 90° angle
- Tip: We have to summarize this table given above to solve each of the question accurately.
Solved Example:
Example: In a right triangle, the angle is 30° and the adjacent side is 6 units. Find the opposite side.
Solution:
Solved Example:
Example: In a right triangle, the angle is 30° and the opposite side is 9 units. Find the opposite side.
Solution:
Given:
- Angle = 30°
- Adjacent side = 6 units
We know that,
So, therefore we got an answer to our question that is:
Solved Example:
Problem: A shed roof makes an angle of 41° with the horizontal. Given that the width of the shed is 6 m and the length of its slope is 4 m. Calculate the height of the roof.
Solution:
Given:
- Angle (θ) = 41° (between the roof and the horizontal)
- Slope length (L) = 4 m (the hypotenuse of the right triangle formed by the roof)
- Width (W) = 6 m (total horizontal span of the shed)
The width of the shed (6 m) is the total span, but the roof slope only covers half of this (since it’s a symmetrical shed roof).
The height of the roof is approximately 2.624 meters.
Solved Example:
Problem: A zip wire runs between two poles 45m apart. The zip wire is at an angle of 10° to the horizontal. Calculate the length of the zip wire.
Solution:
Given:
- Angle (θ) = 10° (between the zip wire and the length)
- Width (W) = 25 m (Distance between two poles)
The width of the shed (6 m) is the total span, but the roof slope only covers half of this (since it’s a symmetrical shed roof).
Solved Example:
Problem: Triangle ABC is an isosceles. Calculate the height of the given triangle.
Given:
- Angle (θ) = 71° (between the two sides)
- Side length = 12 cm (Distance between two poles)
The width of the shed (6 m) is the total span, but the roof slope only covers half of this (since it’s a symmetrical shed roof).
Box Plots – GCSE Maths
Introduction
- Box plot is an important concept used in Statistics to represent quantities related to a given dataset graphically and helps us to provide conclusions about that dataset.
- Box plots include graphical representation of these five quantities – Median, Maximum Value, Smallest value, First quartile and the Third Quartile.
Box Plots
- Basically Box Plots are graphical representation of the following quantities which describe a dataset’s important features –
- Example-
- Total values are 15 the increasing order of these will be –
- Median – When the elements of the dataset are sorted( in increasing or decreasing order) then the middle element is called the Median.
- Median will be the middle element that is 4, we can also use the following formula to find median –
- Greatest value – The maximum value among of dataset’s elements.
- In the given example the greatest value is 9.
- Smallest value – The smallest value among the elements.
- In the given example the smallest value is 0.
- Quartile(1st Quartile) – Basically it is first 25% part of the data. The formula to find 1st quartile is –
- Quartile(3rd Quartile) – Basically it is first 75% part of the data. The formula to find 3rd quartile is –
In the given example- 
Examples including even and odd number of elements:
Problem: Draw the box plot for the following dataset –
Examples including even and odd number of elements:
Problem: Following is a dataset given that is the time taken by 11 students to complete an essay –
(b) Find out the upper and lower Quartiles
(c) Draw the box plot for the dataset
Solution:
(a) Median for the dataset is- 
Examples including even and odd number of elements:
Problem: Here is the data collected from a company of the experience of their 10 employees –
To draw the box plot for the given dataset containing an even number of elements first we find Median, Quartiles and Minimum and Maximum value –
- Final Answer-
Table of Content
Interior and Exterior Angles in Polygons - GCSE Maths
Introduction
- The Word Polygon is made up of two words –
- A closed shape made of line segments .
- To make a Polygon, minimum three line segments are required which end up making a Triangle.
- Basic Polygons are Triangle, Square and Rectangle.
- Polygons have vertices, angles and sides.
- An Angle is basically the distance between two rays starting at the same point.
- Polygons have two types of angles, they are – Interior and Exterior angles.
- Polygons are 2-Dimensional shapes and we can use them to make 3-Dimensional objects.
Importance of polygons:
- Polygons play a vital role in understanding geometric concepts like shapes, angles, area and perimeter.
- Polygons are present in our daily life, in buildings, houses, and the design of objects.
- Students learn about angle and length measurements, which are used to solve real-world problems and make maths meaningful.
Types of Polygons
Polygons are classifies into two types –
- Regular Polygons
- Irregular Polygons
Regular Polygons:
- Polygons with equal sides and equal angles.
Irregular Polygons:
- Polygons with unequal sides and unequal angles.
Some Important polygons are as follows-
Interior Angles in Polygons
- The Angles present inside the polygon are known as Interior Angles.
- The polygon with the minimum number of sides is a Triangle and the sum of the interior angles of a Triangle is 180 degree.
- Consider other Polygons divided into triangles –
- We can conclude that every polygon can be divided into triangles, with the number of triangles formed being two fewer than the number of sides of the polygon.
- Since each triangle has interior angles that add up to 180°, the sum of the interior angles of a polygon is given by: –
Where n = Number of sides of the Polygon
Example: The Pentagon has 5 sides so –
- If we want to find the interior angle of a regular polygon, the formula is-
Solved Example:
Problem: Find the missing interior angles in the following Polygon.
The Polygon shown in the diagram is a Hexagon.
In which – 
Solved Example:
Problem: Work out the size of the angle for the following value of n (Number of sides of Regular Polygon).
Solution:
Using formula- 
Put (n = 5)
A Polygon with 5 sides is called a Pentagon.
Divide the sum of angles by number of sides :
Exterior Angles in Polygons
- When we extend any side of a Polygon, then the resulting angle made is called an Exterior Angle.
- When the exterior angles are combined together they form a circle which represents a complete angle of 360 degrees.
- The angles shown below are Exterior angles.
- In the following diagram a Regular Pentagon is shown.
- We know exterior angles summed up together gives us 360 degrees.
- Hence, the relationship between an exterior angle and the sides of the regular polygon, pentagon is-
Solved Example:
Problem: Find the number of sides of the polygon shown in the image given below: 
Step#1: Find the angle of a regular pentagon-
The angle of a regular pentagon will be –
The two angles together make: 
As interior and exterior angles are supplementary –
Solved Example:
Problem: Find out the value of exterior angle x and interior angle y of polygon.
Step#1: Finding the Exterior Angle-
The polygon is an Octagon, and we can find the exterior angle by the formula –
As the Interior and Exterior angles are supplementary, thus-
Solved Example:
Problem: Find the values of the unknown angles.
Step#1: Find the Exterior Angle-
The interior angle of a rectangle is 90°, hence the exterior angle will be-
In triangle A and B, 
In triangle B,
Solved Example:
Problem: Find the value of the unknown angle.
In the diagram, interior angle is 85° and x is unknown.
As they are supplementary-
Conditional Probability – GCSE Maths
Introduction
- Conditional Probability is the probability of an event occurring given that another event has already occurred.
- Studying of Conditional Probability is important because it helps us understand how the probability of an event changes when we know that another event has occurred.
- This concept is essential in real-world situations where outcomes are not independent.
What is Conditional Probability?
- We know, if one event depends upon the outcome of another event, the two events are Dependent events.
- A Conditional Probability is the probability of a dependent event in which probability of the second outcome depends on what has already happened in the first outcome.
Example:
- If there is a bag with red and blue balls. Picking one ball out and don’t put it back, then take another one, the chance of getting a red or blue ball on the second draw depends on what happened first.
How to Calculate Conditional Probability using Tree Diagrams?
- A Tree Diagram can be used to solve Conditional Probability using dependent events
Steps to solve conditional Probability using Tree Diagram:
- Step #1: Draw the Branch and label the probabilities.
- Step #2: Add Dependent Branches
- Step #3: Apply the Condition
- Step #4: Find the Probability
Solved Example:
Problem: Ivan has a Bag with 3 red and 2 green marbles. He picks 2 marbles without replacement. What’s the probability the second marble is red given the first was green?
Solution:
Step #1: Draw the Branch and label the probabilities.
Possible outcomes for first pick:
- Total Marbles = 5
- Red (3 out of 5 marbles) = 3/5
- Green (2 out of 5 marbles) = 2/5
Step #2: Add Dependent Branches
If first was Red:
- Remaining marbles: 2 red, 2 green
- Next pick will be:
If first was Green:
- Remaining marbles: 3 red, 1 green
- Next pick will be:
Step #3: Apply the Condition
First marble was green, so only follow the green path.
Step #4: Find the Probability
On the Green path, the chance the second marble is red is 3/4
Final Answer: The probability the second marble is red is 3/4
Use Two-way Table to Calculate Conditional Probability
- A Two-way table is a table shows how often different combinations of two events happen together.
Steps to solve conditional Probability using Two-way table:
- Step #1: Create the Two-Way Table
- Step #2: Apply the Condition
- Step #3: Find the Probability
Solved Example:
Problem: If a random cheesy pizza slice is picked from
- 6 pepperonis (with cheese)
- 3 olive pizzas (with cheese)
What’s the probability it’s pepperoni with cheese?
Solution:
Step #1: Create the Two-Way Table
Total Pizza Slices = 9
Step #2: Apply the Condition
- The Condition is pepperoni with cheese slice.
Step #3: Find the Probability
Using the Table:
The probability of the cheesy slice is pepperoni is 6/9
Final Answer: The probability of the cheesy slice is pepperoni is 6/9
Solved Example:
Problem: A bag contains:
You randomly pick one ball, don’t put it back, then pick a second ball. What’s the probability the second ball is red, given that the first ball was blue?
Solution:
Step #1: Create the Two-Way Table
If the first ball was blue then,
- Total number of balls left: 10 – 1 = 9
Step #2: Apply the Condition
- The condition is that if the first ball picked is blue, then the second ball is red.
Step #3: Find the Probability
Using the Table:
The probability of the second ball is red if the first was blue 4/9
Final Answer: The probability of the second ball is red if the first was blue 4/9
Solved Example:
Problem: A standard deck has 52 cards. You draw 2 cards without replacement. What’s the probability the second card is red, given the first card was black?
Solution:
Step #1: Draw the Branch and label the probabilities.
- P(Black): 26/52 = 1/2
- P(Red): 26/52 = 1/2
Step #2: Add Dependent Branches
If first was Black:
- Remaining Cards: 26 red, 25 black
- Next pick will be:
If first was Red:
- Remaining Cards: 25 red, 26 black
- Next pick will be:
Step #3: Apply the Condition
First Card was black, so only follow the black path.
Step #4: Find the Probability
On the Black path, the chance the second card is red is 26/51
Final Answer: The probability the second card is red is 26/51
Venn Diagram – GCSE Maths
Introduction
- A Venn diagram is a simple method to compare and group items using overlapping circles.
- It is fundamental tool in mathematics, logic, and problem-solving.
- Venn diagrams make complex data simple by showing it visually.
What is Venn Diagram?
- A Venn diagram is a visual way to show relationships between different sets.
- It uses circles to represent sets, and the overlapping areas show what the sets have in common.
Example
Suppose in a class of 30 students:
- 18 like Math (M)
- 12 like Science (S)
- 7 like both Math and Science
Set Operations in Venn Diagrams
Venn diagrams visually represent different set operations.
- Curly brackets { } show a set of values.
- ∈ means ‘is an element of’.
Common Set Operations in Venn Diagrams:
Union of Set:
- The Union of set represents that all elements that belong to either A or B or both.
where,
Intersection of Set:
- The Intersection of set represents that only elements that belong to both A and B.
where,
Compliment of a Set:
- The Compliment of a set represents that all elements not in set A, but in the universal set.
where,
Difference of Set:
- The Difference of set represents that elements in A but not in B or elements in B but not in A.
where,
How to Calculate Probability Using Venn Diagram?
- Probability can be visualized and calculated using Venn diagrams with the help of common set operations:
Steps to Calculate Probability Using a Venn Diagram
- Step #1: Define the Sample Space
- Step #2: Define the Events
- Step #3: Draw the Venn Diagram
- Step #4: Calculate the Probabilities
Solved Example:
Problem: Roll a fair 6-sided die. Define two events
- Event A: Roll an even number
- Event B: Roll a number > 3
Find the Probability of P(A) and P(A and B).
Solution:
Step #1: Define the Sample Space
All possible outcomes:
S = {1,2,3,4,5,6}
- Total outcomes = 6
Step #2: Define the Events
- Event A: {2, 4, 6}
- Event B: {4, 5, 6}
Step #3: Draw the Venn Diagram
Step #4: Calculate the Probabilities
Probability of an event,
Solved Example:
Problem: Draw 1 card from a standard 52-card deck. Define two events:
- Event H: Draw a Heart(♥)
- Event K: Draw a King (♠K, ♥K, ♦K, ♣K)
Find the Probability of P(H), P(K) and P(not H).
Solution:
Step #1: Define the Sample Space
All possible outcomes:
- Total cards = 52
- Hearts = 13
- Kings = 4
Step #2: Define the Events
- Event H = 13 cards
- Event K = 4 cards
- H ∩ K (King of Hearts) = 1 card (♥K)
Step #3: Draw the Venn Diagram
Step #4: Calculate the Probabilities
Probability of an event,
Solved Example:
Problem: Toss two fair coins A and B. Define two events:
- Event A: At least one Head appears
- Event B: Both coins show the same face
Find the Probability of P(A) and P(B).
Solution:
Step #1: Define the Sample Space
All possible outcomes:
S = {HH,HT,TH,TT}
- Total outcomes = 4
Step #2: Define the Events
- Event A (At least one Head) = {HH, HT, TH}
- Event B (Same face) = {HH, TT}
- A ∩ B (Both A and B) = {HH}
Step #3: Draw the Venn Diagram
Step #4: Calculate the Probabilities
Probability of an event,
Find The Exact Value of The Trigonometric Function – GCSE Maths
Introduction
- In GCSE Maths, you’re often asked to find sin, cos, or tan of specific angles without a calculator. These specific values are called exact trigonometric values.
- Exact trigonometric values refer to the known and precise values of sine, cosine, and tangent for specific standard angles, without using a calculator.
- These values are written as fractions or square roots, not rounded decimals.
- We study exact trigonometric values to solve non-calculator GCSE exam questions accurately.
Example:
Table of Exact Trigonometric Values
- If we need to calculate exact values of sin, cos, or tan for special angles like 0°, 30°, 45°, 60°, or 90°, there’s no need for a calculator.
- We can use a simple trigonometric values table that shows all the exact answers using fractions and square roots.
Let’s draw and understand the full table step by step:
- First, we will find the value of sine, because using the sine values, we can also find the values of functions like cosine and tangent.
- So, to start with this, first we will take the sine value and its corresponding angle on one side. Now, for each angle, we will take a number from 0 to 4, find its square root, and then divide it by 2, like-
- After solving these values, we will get the sine values for each angle.
- The cos values are just the reverse order of sine values.
- Now, to find the value of tan, we will again use a method similar to sine. First, we will take the same numbers and find their square roots, but this time, instead of dividing by 2, we will divide by the reverse of these numbers.
- After solving these values, we will get the tan values for each angle and tan(90°) is undefined because there is division by zero, which is mathematically impossible.
- The final exact trigonometric table is:
Using Exact value with SOHCAHTOA
- SOHCAHTOA is also a way to remember how sine (sin), cosine (cos), and tangent (tan) relate to the sides of a right-angled triangle.
SOH-CAH-TOA Stands For –
Note: To Learn more about SOH-CAH-TOA, please click on the link: How to Use SOHCAHTOA
Steps to Find Missing Side or Angle by SOHCAHTOA (Using Exact Trig Values):
- Step #1: Identify the Trigonometric function from given values.
- Step #2: Plug the known values into the formula.
- Step #3: Solve it.
Solved Example:
Problem: Find the length of the opposite side if the angle θ = 30° and the hypotenuse = 8.
Solution:
Step #1: Identify the Trigonometric function.
Given
- Angle = 30°
- Hypotenuse = 8
We will use,
Step #2: Plug the known values into the formula.
Step #3: Solve it.
Now,
Final Answer: 4
Solved Example:
Problem: Find the adjacent side if the angle θ = 60∘ and the hypotenuse = 10.
Solution:
Step #1: Identify the Trigonometric function.
Given
- Angle = 60°
- Hypotenuse = 10
We will use,
Step #2: Plug the known values into the formula.
Step #3: Solve it.
Using exact trigonometric table,
Now,
Final Answer: The adjacent side = 5 units
Solved Example:
Problem: Find the angle θ if the opposite side = 5 and the adjacent side = 5.
Solution:
Step #1: Identify the Trigonometric function.
Given
- Opposite side = 5
- Adjacent side = 5
We will use,
Step #2: Plug the known values into the formula.
Step #3: Solve it.
Using exact trigonometric table,
We know:
Final Answer: The angle θ = 45°
Law of Sine and Cosine Rule – GCSE Maths
Introduction
- Laws of Sine and Cosine are trigonometric formulas used to solve triangles when certain information is given.
- They are especially useful for non-right triangles.
- These laws are fundamental in trigonometry and have applications in physics, engineering, and navigation.
What is the Sine Rule?
- The Sine Rule is a fundamental trigonometric formula that relates the sides of a triangle to the sines of their opposite angles.
- Mathematically,
For any triangle with sides a, b and c opposite angles A, B and C respectively, for finding missing side:
Alternatively, it can be written as for finding missing angle:
Where:
- a, b and c are the lengths of the sides of the triangle
- A, B and C are the angles opposite those sides
Solved Example
Problem: A = 40∘, B = 60∘ and side a = 10 cm. Find side b.
Solution:
Use the formula
Put the values:
Now calculate using a calculator:
Final Answer: b = 13.5 cm
What is the Cosine Rule?
- The Cosine Rule is also a trigonometric formula used to find a side or angle in a triangle.
- It works for any triangle whether it’s acute, obtuse, or right-angled.
- Mathematically,
For any triangle with sides a, b and c opposite angles A, B and C opposite those sides:
If you know all three sides, then we can find an angle using this rearranged version of the cosine rule:
Where:
- a, b and c are the lengths of the sides of the triangle.
- A, B and C are the angles opposite those sides.
Solved Example
Problem: Side a = 5cm, side b = 7 cm, angle C = 60∘. Find side c.
Solution:
Use the formula:
Put the values:
Final Answer: c = 6.24 cm
How to Find Missing Side and Angle?
- The Sine Rule or the Cosine Rule, both are used to find the missing side or missing angle depending on what information is given in the question.
Use the Sine Rule:
- If we know the 2 angles and one side, then we use it to find another side
- If we know the 2 sides and one non-included angle, then we use it to find the other angle.
Use the Cosine Rule:
- If we know the 2 sides and one included angle, then we use it to find third side.
- If we know all the three sides, then we use it to find any angle.
Steps to Find the Missing Side or Angle:
- Step#1: Identify the known values.
- Step#2: Write the formula based on the side or angle you’re finding.
- Step#3: Plug the values.
- Step#4: Solve for the missing value.
Solved Example
Problem: In Triangle ABC, Side a = 10cm, Side b = 14cm and Angle A = 45°. Find angle B.
Solution:
Step#1: Given:
- Side a = 10 cm
- Side b = 14 cm
- Angle A = 45°
Step#2: Use The Formula:
Step#3: Plug the values:
Step#4: Solve for the missing angle:
The Missing angle of B ≈ 81.6°
Final Answer: B ≈ 81.6°
Solved Example
Problem: In Triangle ABC, Side a = 7cm, Side b = 8cm and Side c = 9cm. Find angle C.
Solution:
Step#1: Given:
- Side a = 7cm
- Side b = 8cm
- Side c = 9cm
Step#2: Use The Formula:
Step#3: Plug the values:
Step#4: Solve for the missing angle:
The Missing angle of C ≈ 73.4°
Final Answer: C ≈ 73.4°
Solved Example
Problem: In Triangle ABC, Angle A = 50°, Angle B = 60° and Side a = 10cm. Find side b.
Solution:
Step#1: Given:
- Side a = 10cm
- Angle A = 50°
- Angle B = 60°
Step#2: Use The Formula:
Step#3: Plug the values:
Step#4: Solve for the missing Side:
The Missing side of b ≈ 11.31 cm
Final Answer: b ≈ 11.31 cm
Calculate Area Using Sine Rule – GCSE Maths
Introduction
- Law of Sines are trigonometric formulas used to solve any triangles when certain information is given.
- This law is used to find unknown sides or angles in non-right-angled triangles, it can also be applied to calculate the area of a triangle when certain information is given.
- It is a fundamental tool used to solve real-world problems involving triangles.
What is the Sine Rule?
- The Sine Rule states that, in any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all three sides.
- The Sine Law is expressed as:
For any triangle with sides a, b and c opposite angles A, B and C respectively, for finding missing side:
Sine Rule for Calculating the Area of a Triangle
- The Sine Rule is not just used for solving sides and angles, but it is also helpful to calculate the area of triangle especially when height is unknown.
- Mathematically,
- For any triangle with sides a, b, c and opposite angles A, B and C:
Where:
- a and b are two known sides.
- C is the angle between them (included angle).
It can also use as:
Solved Example
Problem: A triangle has sides a = 9cm, b = 6cm, and the included angle C = 62∘. Find its area.
Solution:
Using the formula:
Plug the values and solve:
Final Answer: 23.8383 cm2
How to Calculate Area of Triangle Using Sine Rule?
- We can calculate the area of any triangle using the sine rule, based area formula, especially when we know two sides and the included angle between them.
Steps to Calculate The Area of Triangle:
- Step#1: Identify the known values
- Step#2: Use the formula based on information.
- Step#3: Plug the values in the formula.
- Step#4: Calculate the area
Solved Example
Problem: In Triangle ABC, side AB = 11 cm, side AC = 8 cm and the angle between them ∠A = 50°.Find the area of triangle ABC.
Solution:
Step#1: Identify the known values:
Given:
- Side AB = 11 cm
- Side AC = 8 cm
- Included angle ∠A = 50°
Step#2: Use The Formula:
Step#3: Plug the values in the formula:
Step#4: Calculate the area:
Area of ABC triangle is 33.7cm2.
Final Answer: 33.7cm2
Solved Example
Problem: In Triangle, sides a = 10 cm, side c = 7 cm and the angle B = 40°. Find the area of triangle.
Solution:
Step#1: Identify the known values:
Given:
- Side A = 10 cm
- Side C = 7 cm
- Angle B = 40°
Step#2: Use The Formula:
Step#3: Plug the values in the formula:
Step#4: Calculate the area:
Area of triangle is 22.50 cm2.
Final Answer: 22.50 cm2
Solved Example
Problem: The area of a triangle is 30 cm2. One side a = 6 cm, and the included angle C = 50°. Find the other side b.
Solution:
Step#1: Identify the known values:
Given:
- a = 6 cm
- Area of Triangle = 30 cm2
- Angle c = 50°
Step#2: Use The Formula:
Rearrange it,
Step#3: Plug the values in the formula:
Step#4: Calculate the area:
Final answer is 13.05 cm
Final Answer: 13.05 cm
Decimal Recurring to Fraction - GCSE Maths
Introduction
- A Recurring Decimal is denoted with a dot over the number and is any decimal in which the digits repeat themselves.
Types of Recurring Decimals
Pure Recurring Decimals:
- Decimal where all the digits after the decimal point repeat indefinitely.
Examples:
- 0.333…
- 0.7474….
- 0.4545….
- 0.98549854….
Mixed Recurring Decimals:
- After the decimal point, some digits do not repeat, and a sequence of digits starts repeating indefinitely after the non-repeating part.
Examples:
- 0.23434….
- 0.165858….
- 0.2358989….
- 0.7852222….
How to Convert Recurring Decimals to Fractions (Type 1)
Type 1: Converting Pure Recurring Decimals to Fraction:
Step#1: Take your term as x.
Step#2: Multiply both sides by 10n (where n is the number of repeating digits).
- Multiply 10 for 1 Recurring Decimal.
Example:
For 0.333…, there is only 1 digit repeat, so multiply by
101 = 10
So, 10x = 3.333…
- Multiply 100 for 2 Recurring Decimal.
Example:
For 0.2929…, there is 2 digits repeat, so multiply by
102 = 100
So, 100x = 29.2929…
- Multiply 1000 for 3 Recurring Decimal.
Example:
For 0.816816…, there is 3 digits repeat, so multiply by
103 =1000
So, 1000x = 816.816816…
Step#3: Subtract original equation from new equation to eliminate the repeating part.
Step#4: Solve for x and simplify fraction, if possible.
How to Convert Recurring Decimals to Fractions (Type 2)
Type 2: Converting Mixed Recurring Decimals to Fraction:
Step#1: Take your term as x.
Step#2: Multiply both sides by 10 (where m is the number of non-repeating digits).
Step #3: Multiply both sides by 10n (where n is the number of repeating digits) to shift the Decimal.
Step#4: Subtract original equation from new equation to eliminate the repeating part.
Step#5: Solve for x and simplify fraction, if possible.
Solved Example
Problem: Convert 0.12323… into Fraction.
Solution:
Step #1: Let x = 0.12323…
Step #2: Multiply both side by 101 =10 to move the non-repeating part:
Step #3: Multiply both sides by 102 =100 to shift the Decimal:
Step #4: Subtract the original equation from this new equation:
Step #5: Solve for x:
Final Answer: 61/495
Why it is important to convert Recurring Decimals into Fraction?
- Converting recurring decimals into fractions is important because recurring decimals are approximations of fractions, but fractions provide an exact representation of the number.
Examples:
- A recipe might call for 1/3 cup of flour, which is more practical than 0.3333… cups.
- If a bank offers an interest rate of 0.3333… it’s easier to express it as 1/3 to simplify calculations.
- In Chemistry, Mole ratios in reactions often involve fractions (e.g., 0.1666… moles = 1/6 mole).
Three Additional Solved Examples
Solved Example 1
Problem: Convert 0.333… into Fraction.
Solution:
Step #1: Let x = 0.333…
Step #2: Multiply both side by 10:
Step #3: Subtract the original equation from this new equation:
Step #4: Solve for x:
Final Answer: 1/3
Solved Example 2
Problem: Convert 0.181818… into Fraction.
Solution:
Step #1: Let x = 0.1818…
Step #2: Multiply both side by 100:
Step #3: Subtract the original equation from this new equation:
Step #4: Solve for x:
Final Answer: 2/11
Solved Example 3
Problem: Convert 0.6333… into Fraction
Solution:
Step #1: Let x = 0.6333…
Step #2: Multiply both side by 101 =10 to move the non-repeating part:
Step #3: Multiply both sides by 102 =100 to shift the Decimal:
Step #4: Subtract the original equation from this new equation:
Step #5: Solve for x:
Final Answer: 19/30
Practice Questions and Answers on Decimal Recurring to Fraction
Question 1: Convert the Pure Recurring Decimal 0.121212… to a fraction.
Question 2: Convert the Pure Recurring Decimal 0.2222… to a fraction.
Question 3: Convert the Pure Recurring Decimal 0.090909… to a fraction.
Question 4: Convert the Pure Recurring Decimal 0.142142… to a fraction.
Question 5: Convert 0.479479… form of Pure Recurring decimal to a fraction.
Question 6: Convert the Mixed Recurring Decimal 2.333… to a fraction.
Question 7: Converting the Mixed Recurring Decimal 0.10909… to a fraction.
Question 8: Converting the Mixed Recurring Decimal 0.5666… to a fraction.
Question 9: Convert the Mixed Recurring Decimal 2.272727… to a fraction.
Question 10: Convert the Pure Recurring Decimal 0.5555… to a fraction.
Solutions
Question 1:
Solution:
Step#1: Let x = 0.1212…
Step#2: Multiply both sides by 100
100x = 12.1212…
Step#3: Subtract the original equation
100x − x = 12.1212… − 0.1212…
99x = 12
Step#4: Solve for x
x = 12 ÷ 99
= 4 ÷ 33
Answer: 4/33
Question 2:
Solution:
Step#1: Let x = 0.2222…
Step#2: Multiply both sides by 10
10x = 2.222…
Step#3: Subtract the original equation
10x − x = 2.222… − 0.222…
9x = 2
Step#4: Solve for x
x = 2 ÷ 9
Answer: 2/9
Question 3:
Solution:
Step#1: Let x = 0.090909…
Step#2: Multiply both sides by 100
100x = 9.0909…
Step#3: Subtract the original equation
100x − x = 9.0909… − 0.0909…
99x = 9
Step#4: Solve for x
x = 9 ÷ 99
= 1 ÷ 11
Answer: 1/11
Question 4:
Solution:
Step#1: Let x = 0.142142…
Step#2: Multiply both sides by 1000
1000x = 142.142142…
Step#3: Subtract the original equation
1000x − x = 142.142142… − 0.142142…
999x = 142
Step#4: Solve for x
x = 142 ÷ 999
Answer: 142/999
Question 5:
Solution:
Step#1: Let x = 0.479479…
Step#2: Multiply both sides by 1000
1000x = 479.479479…
Step#3: Subtract the original equation
1000x − x = 479.479479… − 0.479479…
999x = 479
Step#4: Solve for x
x = 479 ÷ 999
Answer: 479/999
Question 6:
Solution:
Step#1: Let x = 2.333…
Step#2: Multiply both sides by 10
10x = 23.333…
Step#3: Subtract the original equation
10x − x = 23.333… − 2.333…
9x = 21
Step#4: Solve for x
x = 21 ÷ 9 = 7 ÷ 3
Answer: 7/3
Question 7:
Solution:
Step#1: Let x = 0.10909…
Step#2: Multiply both sides by 10
10x = 1.0909…
Step#3: Multiply both sides again by 100
1000x = 109.0909…
Step#4: Subtract
1000x − 10x = 109.0909… − 1.0909…
990x = 108
Step#5: Solve for x
x = 108 ÷ 990
= 6 ÷ 55
Answer: 6/55
Question 8:
Solution:
Step#1: Let x = 0.5666…
Step#2: Multiply both sides by 10
10x = 5.666…
Step#3: Multiply both sides again by 10
100x = 56.666…
Step#4: Subtract
100x − 10x = 56.666… − 5.666…
90x = 51
Step#5: Solve for x
x = 51 ÷ 90 = 17 ÷ 30
Answer: 17/30
Question 9:
Solution:
Step#1: Let x = 2.2727…
Step#2: Multiply both sides by 100
100x = 227.2727…
Step#3: Subtract
100x − x = 227.2727… − 2.2727…
99x = 225
Step#4: Solve for x
x = 225 ÷ 99 = 25 ÷ 11
Answer: 25/11
Question 10:
Solution:
Step#1: Let x = 0.555…
Step#2: Multiply both sides by 10
10x = 5.555…
Step#3: Subtract the original equation
10x − x = 5.555… − 0.555…
9x = 5
Step#4: Solve for x
x = 5 ÷ 9
Answer: 5/9
Table of Content
- Introduction
- Types of Recurring Decimals
- How to convert Recurring Decimals to Fractions (Type 1)
- How to convert Recurring Decimals to Fractions (Type 2)
- Why it is important to convert Recurring Decimals into Fraction?
- Three Additional Solved Examples
- Practice Questions and Answers on Decimal Recurring to Fraction
Reverse Percentages – GCSE Maths
Reverse Percentage also known as “Reverse Percent“ is a mathematical operation that involves finding the original value or quantity from which a percentage was calculated.
What are Reverse Percentages?
- Reverse percentage is a mathematical concept which is used to find or determine the original value before the Percentage Increase or Decrease.
- Start with the final amount after a percentage change and work backward to find the original number.
- If you know the final value after a percentage change, reverse percentages help you find the original value before the change.
Formula used in Reverse Percentages:
Solved Example
Problem: A TV now costs £300 after a 25% increase. What was its original price?
Solution:
Step #1: Given:
- New value = £300
- Percentage increased = 25%
Step #2: Applying the formula:
Step #3: Put the values in formula:
Step #4: Simplify the denominator:
Step #5: The final value is:
The original price of TV was £240.
Final Answer: £240
How Reverse Percentage is Different From Original Percentage?
Original Percentages:
- It is use when we know the original number or value and want to calculate a percentage of it.
- If original price of shirt is £100 then after 20% increase, the new price will be £120.
Reverse Percentages:
- It is use when we know the final value after a percentage change and want to find the original value.
- After a 20 % increase, the new price of a shirt is £120 then the original price of a shirt was £100.
Steps to Solve Reverse Percentages
Step#1: Understand the Question
- Check the new value is increased or decreased after the change of original value:
- If the final value is after a percentage increase, the formula is:
- If the final value is after a percentage decrease, the formula is:
Step#2: Work out what percentage you now have.
Step#3: Solve the equation
- We know the original equivalent percentage for all the process is 100%.
- Use this to find the 1% of original price or value.
Step#4: Now multiply the 1% with 100% to get the original value or price.
Note: To solve this easily, we should also know the concept of Original Percentages.
Three Additional Solved Examples
Solved Example 1
Problem: A jacket costs £60 after a 20% discount. What was the original price?
Solution:
Step #1: Given:
- New value = £60
- Percentage increased = 20%
Step #2: Applying the formula:
Step #3: Put the values in formula:
Step #4: Simplify the denominator:
Step #5: The final value is:
The original price of jacket was £75.
Final Answer: £75
Solved Example 2
Problem: A product costs £120 including 20% VAT. What was the price before tax?
Solution:
Step #1: Given:
- New value = £120
- Percentage increased = 20%
Step #2: Applying the formula:
Step #3: Put the values in formula:
Step #4: Simplify the denominator:
Step #5: The final value is:
The original price of product was £100.
Final Answer: £100
Solved Example 3
Problem: A house increased in value by 15% and is now worth £230,000. What was its original price?
Solution:
Step #1: Given:
- New value = £230,000
- Percentage increased = 15%
Step #2: Applying the formula:
Step #3: Put the values in formula:
Step #4: Simplify the denominator:
Step #5: The final value is:
The original price of house was £200,000.
Final Answer: £200,000
Practice Questions and Answers on Reverse Percentages
Question 1: A shirt is on sale for £60 after a 20% discount. What was the original price?
Question 2: After a 15% increase, the price of a phone is £345. What was the original price?
Question 3: A house is now valued at £110,000 after a 10% decrease. What was the original price?
Question 4: The price of a book is £25 after a 30% increase. What was the original price?
Question 5: A laptop is now £850 after a 25% discount. What was the original price?
Question 6: A company’s revenue is £1,27,500after a 12.5% decrease due to economic downturn. What was the original revenue before the decrease?
Question 7: A car’s price increased by 18% and is now valued at £4,720. What was the original price before the increase?
Question 8: A company reduced its workforce by 22%, leaving 4,290 employees. How many employees did the company originally have?
Question 9: The price of gold increased by 27%, and the new price is £1,397 per ounce. What was the original price?
Question 10: A business made £315,800 in profit after a 19.5% loss compared to the previous year. What was the previous year’s profit?
Solutions
Question 1:
Solution:
Step#1: Given
• New value = £60
• Percentage decreased = 20%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 60 ÷ (1 − 20 ÷ 100)
Step#4: Simplify the denominator
= 1 − 20 ÷ 100 = 1 − 0.2
= 0.8
Step#5: The final value is
Original Value = 60 ÷ 0.8
= 60 × 10 ÷ 8
= 600 ÷ 8 = 75
The original price of the shirt was £75
Question 2:
Solution:
Step#1: Given
• New value = £345
• Percentage increased = 15%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 345 ÷ (1 + 15 ÷ 100)
Step#4: Simplify the denominator
= 1 + 15 ÷ 100 = 1 + 0.15
= 1.15
Step#5: The final value is
Original Value = 345 ÷ 1.15
= 345 × 100 ÷ 115
= 34500 ÷ 115 = 300
The original price of the phone was £300
Question 3:
Solution:
Step#1: Given
• New value = £110,000
• Percentage decreased = 10%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 110000 ÷ (1 − 10 ÷ 100)
Step#4: Simplify the denominator
= 1 − 10 ÷ 100 = 1 − 0.1
= 0.9
Step#5: The final value is
Original Value = 110000 ÷ 0.9
= 110000 × 10 ÷ 9
= 1100000 ÷ 9 = 122222.22
The original price of the house was £122,222.22
Question 4:
Solution:
Step#1: Given
• New value = £25
• Percentage increased = 30%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 25 ÷ (1 + 30 ÷ 100)
Step#4: Simplify the denominator
= 1 + 30 ÷ 100 = 1 + 0.3
= 1.3
Step#5: The final value is
Original Value = 25 ÷ 1.3
= 25 × 10 ÷ 13
= 250 ÷ 13 = 19.23
The original price of the book was £19.23
Question 5:
Solution:
Step#1: Given
• New value = £850
• Percentage decreased = 25%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 850 ÷ (1 − 25 ÷ 100)
Step#4: Simplify the denominator
= 1 − 25 ÷ 100 = 1 − 0.25
= 0.75
Step#5: The final value is
Original Value = 850 ÷ 0.75
= 850 × 100 ÷ 75
= 85000 ÷ 75 = 1133.33
The original price of the laptop was £1133.33
Question 6:
Solution:
Step#1: Given
• New value = £127,500
• Percentage decreased = 12.5%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 127500 ÷ (1 − 12.5 ÷ 100)
Step#4: Simplify the denominator
= 1 − 12.5 ÷ 100 = 1 − 0.125
= 0.875
Step#5: The final value is
Original Value = 127500 ÷ 0.875
= 127500 × 1000 ÷ 875
= 127500000 ÷ 875 = 145714.28
The original revenue of the company was £145,714.28
Question 7:
Solution:
Step#1: Given
• New value = £4720
• Percentage increased = 18%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 4720 ÷ (1 + 18 ÷ 100)
Step#4: Simplify the denominator
= 1 + 18 ÷ 100 = 1 + 0.18
= 1.18
Step#5: The final value is
Original Value = 4720 ÷ 1.18
= 4720 × 100 ÷ 118
= 472000 ÷ 118 = 4000
The original price of the car was £4000
Question 8:
Solution:
Step#1: Given
• New value = 4290
• Percentage decreased = 22%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 4290 ÷ (1 − 22 ÷ 100)
Step#4: Simplify the denominator
= 1 − 22 ÷ 100 = 1 − 0.22
= 0.78
Step#5: The final value is
Original Value = 4290 ÷ 0.78
= 4290 × 100 ÷ 78
= 429000 ÷ 78 = 5500
The original number of employees in the company was 5500
Question 9:
Solution:
Step#1: Given
• New value = £1397
• Percentage increased = 27%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 1397 ÷ (1 + 27 ÷ 100)
Step#4: Simplify the denominator
= 1 + 27 ÷ 100 = 1 + 0.27
= 1.27
Step#5: The final value is
Original Value = 1397 ÷ 1.27
= 1397 × 100 ÷ 127
= 139700 ÷ 127 = 1100
The original price of gold was £1100
Question 10:
Solution:
Step#1: Given
• New value = £315,800
• Percentage decreased = 19.5%
Step#2: Applying the formula
Original Value = Final Value ÷ (1 ± Percentage ÷ 100)
Step#3: Put the values in the formula
Original Value = 315800 ÷ (1 − 19.5 ÷ 100)
Step#4: Simplify the denominator
= 1 − 19.5 ÷ 100 = 1 − 0.195
= 0.805
Step#5: The final value is
Original Value = 315800 ÷ 0.805
= 315800 × 1000 ÷ 805
= 315800000 ÷ 805 = 392298.13
The original revenue of the company was £392,298.13