# IGCSE Mathematics Foundation: Pythagoras’ Theorem and Trigonometry

## Scheme of work: IGCSE Foundation: Year 11: Term 1: Pythagoras’ Theorem and Trigonometry

#### Prerequisite Knowledge

1. Finding the Area of Triangles and Perimeter of 2D Shapes
• Concept: Understanding how to calculate the area of triangles and the perimeter of various 2D shapes.
• Example: $\text{For a triangle with base } b \text{ and height } h, \text{ the area } A = \frac{1}{2} b \times h.$ $\text{For a rectangle with length } l \text{ and width } w, \text{ the perimeter } P = 2l + 2w.$
2. Ability to Calculate Square Numbers and Square Roots
• Concept: Understanding how to calculate the square of a number and find the square root of a number.
• Example: $\text{The square of 4 is } 4^2 = 16.$ $\text{The square root of 16 is } \sqrt{16} = 4.$

#### Success Criteria

1. Know, Understand and Use Pythagoras’ Theorem in Two Dimensions
• Objective: Apply Pythagoras’ theorem to find the length of a side in a right-angled triangle.
• Example: $\text{For a right-angled triangle with legs } a \text{ and } b \text{ and hypotenuse } c, \text{ use } a^2 + b^2 = c^2.$ $\text{If } a = 3 \text{ and } b = 4, \text{ then } c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$
2. Know, Understand and Use Sine, Cosine and Tangent of Acute Angles to Determine Lengths and Angles of a Right-Angled Triangle
• Objective: Use the sine, cosine, and tangent ratios to find unknown sides or angles in a right-angled triangle.
• Example (Finding a Side): $\text{For a right-angled triangle with an angle } \theta, \text{ opposite side } a, \text{ and hypotenuse } c, \text{ use } \sin(\theta) = \frac{a}{c}.$ $\text{If } \theta = 30^\circ \text{ and } c = 10, \text{ then } a = 10 \sin(30^\circ) = 10 \times 0.5 = 5.$
• Example (Finding an Angle): $\text{For a right-angled triangle with adjacent side } b \text{ and hypotenuse } c, \text{ use } \cos(\theta) = \frac{b}{c}.$ $\text{If } b = 8 \text{ and } c = 10, \text{ then } \theta = \cos^{-1}\left(\frac{8}{10}\right) = \cos^{-1}(0.8).$ $\theta \approx 36.87^\circ.$
3. Apply Trigonometric Methods to Solve Problems in Two Dimensions
• Objective: Use trigonometric ratios and Pythagoras’ theorem to solve real-world and mathematical problems involving right-angled triangles.
• Example: $\text{A ladder leans against a wall, forming a right-angled triangle with the ground. The ladder is 10 meters long, and the base of the ladder is 6 meters from the wall. Find the height the ladder reaches on the wall.}$ $\text{Using Pythagoras’ theorem:}$ $6^2 + h^2 = 10^2 \rightarrow 36 + h^2 = 100 \rightarrow h^2 = 64 \rightarrow h = \sqrt{64} = 8.$ $\text{The height the ladder reaches on the wall is 8 meters.}$

#### Key Concepts

1. Know, Understand and Use Pythagoras’ Theorem in Two Dimensions
• Concept: Pythagoras’ theorem relates the lengths of the sides of a right-angled triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
• Example: $a^2 + b^2 = c^2.$ $\text{For a right-angled triangle with legs } a = 3 \text{ and } b = 4, \text{ and hypotenuse } c:$ $3^2 + 4^2 = 9 + 16 = 25 \rightarrow c = \sqrt{25} = 5.$
2. Know, Understand and Use Sine, Cosine and Tangent of Acute Angles to Determine Lengths and Angles of a Right-Angled Triangle
• Concept: The sine, cosine, and tangent ratios are fundamental in trigonometry. They relate the angles of a right-angled triangle to the lengths of its sides.
• Example: $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}.$ $\text{For a right-angled triangle with } \theta = 30^\circ, \text{ opposite side } a = 5, \text{ and hypotenuse } c = 10:$ $\sin(30^\circ) = \frac{a}{c} \rightarrow \frac{5}{10} = 0.5.$
3. Apply Trigonometric Methods to Solve Problems in Two Dimensions
• Concept: Trigonometric methods, including the use of Pythagoras’ theorem and the sine, cosine, and tangent ratios, can solve various real-world and mathematical problems involving right-angled triangles.
• Example: $\text{A ladder leaning against a wall forms a right-angled triangle with the ground. If the ladder is 10 meters long and the base is 6 meters from the wall, find the height } h \text{ the ladder reaches.}$ $\text{Using Pythagoras’ theorem:}$ $6^2 + h^2 = 10^2 \rightarrow 36 + h^2 = 100 \rightarrow h^2 = 64 \rightarrow h = \sqrt{64} = 8.$

#### Common Misconceptions

1. Know, Understand and Use Pythagoras’ Theorem in Two Dimensions
• Common Mistake: Incorrectly identifying the hypotenuse or misapplying the Pythagorean theorem.
• Example: $\text{Incorrect: For a triangle with legs } a = 3, b = 4, \text{ and hypotenuse } c, \text{ calculating } a^2 + b^2 = c \text{ instead of } a^2 + b^2 = c^2.$ $\text{Correct: Identify the hypotenuse correctly and apply the theorem: } 3^2 + 4^2 = c^2 \rightarrow 9 + 16 = 25 \rightarrow c = \sqrt{25} = 5.$
2. Know, Understand and Use Sine, Cosine and Tangent of Acute Angles to Determine Lengths and Angles of a Right-Angled Triangle
• Common Mistake: Confusing the sine, cosine, and tangent ratios or using incorrect sides for the given angle.
• Example: $\text{Incorrect: Using } \sin(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \text{ instead of } \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}.$ $\text{Correct: For } \theta = 30^\circ, \text{ opposite side } a = 5, \text{ and hypotenuse } c = 10:$ $\sin(30^\circ) = \frac{a}{c} = \frac{5}{10} = 0.5.$
• Example (Finding an Angle): $\text{Incorrect: Using the wrong ratio to find an angle, such as } \tan(\theta) = \frac{\text{adjacent}}{\text{opposite}}.$ $\text{Correct: For adjacent side } b = 8 \text{ and hypotenuse } c = 10:$ $\cos(\theta) = \frac{b}{c} = \frac{8}{10} = 0.8 \rightarrow \theta = \cos^{-1}(0.8) \approx 36.87^\circ.$
3. Apply Trigonometric Methods to Solve Problems in Two Dimensions
• Common Mistake: Misapplying trigonometric ratios or Pythagoras’ theorem in word problems, leading to incorrect solutions.
• Example: $\text{Incorrect: For a ladder problem, using } a^2 + b^2 = h \text{ instead of } a^2 + b^2 = h^2.$ $\text{Correct: For a ladder leaning against a wall, forming a right-angled triangle with the ground, and the ladder length } l = 10 \text{ meters, base } b = 6 \text{ meters from the wall, and height } h:$ $6^2 + h^2 = 10^2 \rightarrow 36 + h^2 = 100 \rightarrow h^2 = 64 \rightarrow h = \sqrt{64} = 8 \text{ meters.}$

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