Mathematics (0580) | Right-angled triangles

Welcome to the World of Right-Angled Triangles!

Hello! This chapter is where mathematics really starts to feel powerful and practical. We are moving into Trigonometry, which is basically the study of how the sides and angles of triangles relate to each other.

You might ask, "Why do I need this?" Trigonometry lets engineers calculate the height of skyscrapers, navigators plot courses across oceans, and architects design stable roofs—all without having to climb or measure impossible distances! This whole chapter focuses on the simplest and most important triangle: the right-angled triangle.

Don't worry if this seems tricky at first; we will break everything down step-by-step using easy-to-remember tools!

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Section 1: The Essential Right-Angled Triangle Vocabulary

A right-angled triangle is any triangle that contains one angle of 90°.

Before we can use any formula, we must correctly label the sides. The labels always depend on which non-90° angle (\(\theta\), pronounced theta) you are focusing on:

Key Terms to Label

1. Hypotenuse (H):
• This is always the longest side of the triangle.
• It is always directly opposite the 90° angle.

2. Opposite (O):
• This is the side that is directly across from the angle (\(\theta\)) you are using in your calculation.

3. Adjacent (A):
• This is the side that is next to the angle (\(\theta\)) you are using.
• (Think: "Adjacent" means next to or joining.)

! Important Note: The Opposite and Adjacent sides swap places if you choose to use the other acute angle in the triangle!

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Section 2: Finding Sides with Pythagoras' Theorem (Review)

Before trigonometric ratios, remember how to find a side length if you already know the other two sides. This is done using Pythagoras' Theorem (Syllabus C7.1 & E7.1).

The Formula

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$$a^2 + b^2 = c^2$$ Where \(c\) must always be the Hypotenuse.

Step-by-Step Process for Pythagoras:

1. Finding the Hypotenuse: (The longest side, \(c\))
• Square the two shorter sides, add them together, then find the square root.
• Example: If \(a=3\) and \(b=4\), then \(c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).

2. Finding a Shorter Side: (Side \(a\) or \(b\))
• Square the hypotenuse, subtract the square of the known shorter side, then find the square root.
• Example: If \(c=10\) and \(a=6\), then \(b = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\).

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Section 3: Introducing Trigonometric Ratios (SOH CAH TOA)

When you have one side and one acute angle (not 90°) and you need to find another side or angle, you must use Trigonometry.

The three basic trigonometric ratios (or functions) are Sine, Cosine, and Tangent. They define the relationship between the angle and the ratios of the sides.

The Famous Memory Aid: SOH CAH TOA

This mnemonic tells you which sides to use for each ratio:

1. SOH (Sine):
• Sine = Opposite / Hypotenuse
$$ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} $$

2. CAH (Cosine):
• Cosine = Adjacent / Hypotenuse
$$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} $$

3. TOA (Tangent):
• Tangent = Opposite / Adjacent
$$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $$

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Section 4: Calculating Unknown Sides

When solving for an unknown side, you need two known pieces of information (one side length and one angle).

The 4-Step Method for Finding a Side

Step 1: Label the Triangle.
• Relative to the known angle (\(\theta\)), label the sides: O, A, and H.
• Ignore the side you neither know nor need to find.

Step 2: Choose the Correct Ratio.
• Look at the sides you have a value for (known side) and the side you want to find (unknown side, often \(x\)).
• Choose the ratio (SOH, CAH, or TOA) that links these two sides.

Step 3: Set up the Equation.
• Substitute the known angle and side lengths (using \(x\) for the unknown) into the chosen formula.

Step 4: Solve the Equation.
• Use algebra to rearrange and find \(x\).

Case A: Unknown Side is in the Numerator (on top)

This is the easiest case!

Example: You know the angle is 30°, the Adjacent side (A) is 10 cm, and you want to find the Opposite side (O), \(x\).

1. Label: O = \(x\), A = 10, H = (ignored).
2. Ratio: O and A means use TOA (\(\tan\)).
3. Equation: \( \tan(30^{\circ}) = \frac{x}{10} \)
4. Solve: Multiply both sides by 10.
\( x = 10 \times \tan(30^{\circ}) \)
\( x \approx 5.7735...\)

(Remember to round your final answer to 3 significant figures, unless instructed otherwise. So, \(x = 5.77\) cm.)

Case B: Unknown Side is in the Denominator (on the bottom)

This requires a bit more algebra!

Example: You know the angle is 40°, the Opposite side (O) is 12 m, and you want to find the Hypotenuse (H), \(x\).

1. Label: O = 12, A = (ignored), H = \(x\).
2. Ratio: O and H means use SOH (\(\sin\)).
3. Equation: \( \sin(40^{\circ}) = \frac{12}{x} \)
4. Solve:
• Step 4a: Multiply both sides by \(x\) to get it off the bottom:
\( x \times \sin(40^{\circ}) = 12 \)
• Step 4b: Divide by \(\sin(40^{\circ})\) to isolate \(x\):
\( x = \frac{12}{\sin(40^{\circ})} \)
\( x \approx 18.668...\)
\(x = 18.7\) m (3 s.f.)

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Section 5: Calculating Unknown Angles

If you know two side lengths but need to find an unknown angle (\(\theta\)), you must use the inverse trigonometric functions.

Inverse Functions (The "Undo" Buttons)

To find the angle, you must "undo" the sine, cosine, or tangent function. On your calculator, these are usually marked as \(\sin^{-1}\), \(\cos^{-1}\), and \(\tan^{-1}\) (often accessed by pressing "Shift" or "2nd F" before the function button).

1. If \( \sin(\theta) = \text{ratio} \), then \( \theta = \sin^{-1}(\text{ratio}) \)
2. If \( \cos(\theta) = \text{ratio} \), then \( \theta = \cos^{-1}(\text{ratio}) \)
3. If \( \tan(\theta) = \text{ratio} \), then \( \theta = \tan^{-1}(\text{ratio}) \)

The 4-Step Method for Finding an Angle

Step 1: Label the Triangle.
• Relative to the unknown angle (\(\theta\)), label the two known sides (O, A, or H).

Step 2: Choose the Correct Ratio.
• Choose SOH, CAH, or TOA that links the two known sides.

Step 3: Set up the Equation.
• Substitute the known side lengths into the ratio formula.

Step 4: Solve using the Inverse Function.
• Use \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\) to find the angle.

Example: You know the Opposite side (O) is 5 cm and the Adjacent side (A) is 8 cm. Find angle \(\theta\).

1. Label: O = 5, A = 8, H = (ignored).
2. Ratio: O and A means use TOA (\(\tan\)).
3. Equation: \( \tan(\theta) = \frac{5}{8} \)
4. Solve: \( \theta = \tan^{-1}(\frac{5}{8}) \)
\( \theta \approx 32.005...\)

(Remember to round angles to one decimal place, unless instructed otherwise. So, \(\theta = 32.0^{\circ}\).

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Section 6: Solving Practical Problems (2D Applications)

Trigonometry is most often used to solve real-world problems. This usually involves drawing a diagram and correctly identifying the right-angled triangle within the situation (Syllabus C7.2/E7.2).

1. Angles of Elevation and Depression (Extended Focus E7.2)

These terms describe the angles involved when looking up or looking down from a horizontal line.

• Angle of Elevation: The angle measured up from the horizontal line to a point above (e.g., looking up at the top of a tower).

• Angle of Depression: The angle measured down from the horizontal line to a point below (e.g., looking down from a cliff to a boat).

Analogy: Imagine you are standing on the ground, looking straight ahead (that's the horizontal).
• Elevation is tilting your head up.
• Depression is tilting your head down.

! Crucial Point: Because the line of sight and the ground are usually parallel, the angle of elevation from Point A to Point B is equal to the angle of depression from Point B to Point A (due to alternate angles in parallel lines).

2. Bearings and Distances (Syllabus C7.2/E7.2)

Trigonometry and Pythagoras are often combined with bearing questions (which use three figures measured clockwise from North).

• If a journey involves moving North/South and East/West, you can always draw a right-angled triangle where the North/South line and the East/West line meet at 90°.

• This allows you to use Pythagoras to find the final distance or use SOH CAH TOA to find the angle needed to convert your geometric angle back into a bearing.

Example: A ship sails 5 km East, then 12 km North.
• The path forms a right-angled triangle.
• Final distance from start (Hypotenuse) is found using Pythagoras: \( \sqrt{5^2 + 12^2} = 13 \) km.

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Section 7: Extended Topic: 3D Problems

In the Extended syllabus (E7.6), you must be able to apply Pythagoras and trigonometry to solve problems in three dimensions (3D).

The trick in 3D geometry is to identify the necessary right-angled triangle hidden inside the 3D shape (like a cuboid or pyramid).

Strategy for 3D Problems

1. Draw and label the 3D shape.
2. Find the "Hidden" Right Angle: Look for a 90° angle between a vertical edge and a line lying flat on the base, or between two lines that are perpendicular in the base itself.
3. Work in Stages: You often need to use Pythagoras' theorem *twice* or use Pythagoras first to find a necessary base length, and then use trigonometry (SOH CAH TOA) to find the final angle or height.

Example: Finding the space diagonal of a cuboid.
• First, use Pythagoras on the floor (2D) to find the diagonal across the base.
• Second, use Pythagoras again, using that base diagonal and the vertical height of the cuboid, to find the final space diagonal.