WaywardHouse

18  Trigonometry

Angles and the ratios they fix

You’re building a wheelchair ramp. The code says the angle can’t exceed 4.8°. You know how high the step is. How long does the ramp need to be?

You’re standing 30 metres from a building, looking up at the roof. You want to know how tall the building is, but you can’t climb it. You can measure the angle you’re looking up at. Is that enough?

Your phone screen is 15.6 cm tall and 7.2 cm wide. What’s the diagonal? Easy — Pythagoras. But what’s the angle in the corner? That’s where trig comes in.

Each of these situations has a right triangle in it somewhere. You know some sides and angles, and you want to find others. Trigonometry gives you three ratios that connect them.

18.1 What the notation is saying

Start with a right triangle. One angle is exactly 90°. Pick one of the other angles and call it \(\theta\) (the Greek letter theta — just a name for “the angle you’re working with”).

Now name the three sides relative to \(\theta\):

  • The hypotenuse is the longest side, always opposite the right angle.
  • The opposite side is the one directly across the triangle from \(\theta\) — the side \(\theta\) “faces”.
  • The adjacent side is the one next to \(\theta\) that isn’t the hypotenuse — the side \(\theta\) “leans against”.

The three trig ratios are:

\[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\]

Read it: “the sine of theta equals opposite over hypotenuse.”

\[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]

Read it: “the cosine of theta equals adjacent over hypotenuse.”

\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]

Read it: “the tangent of theta equals opposite over adjacent.”

The memory aid is SOH-CAH-TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. It’s a mnemonic, not mathematics — use it until you don’t need it any more.

18.2 The method

18.2.1 Finding an unknown side

You know an angle and one side. You want another side. Pick the ratio that connects what you know to what you want.

Example. A ramp rises 0.5 m. The angle at the base is 4°. How long is the ramp surface?

The 0.5 m is the side opposite the 4° angle. The ramp surface is the hypotenuse. The ratio connecting opposite and hypotenuse is sine:

\[\sin(4°) = \frac{0.5}{\text{hyp}}\]

Rearrange step by step:

\[\sin(4°) \times \text{hyp} = 0.5 \quad \text{[multiply both sides by hyp]}\]

\[\text{hyp} = \frac{0.5}{\sin(4°)} \quad \text{[divide both sides by } \sin(4°)\text{]}\]

\[\text{hyp} \approx \frac{0.5}{0.0698} \approx 7.2 \text{ m}\]

The ramp needs to be about 7.2 m long.

18.2.2 Finding an unknown angle

You know two sides. You want the angle between them. Use the inverse trig function.

\(\sin^{-1}\) reads as “inverse sine” or “arcsine”. It asks: what angle has this sine value? If \(\sin(\theta) = 0.5\), then \(\theta = \sin^{-1}(0.5) = 30°\). The \(^{-1}\) notation means “undo the sine” — it does not mean “one over sine.”

Similarly, \(\cos^{-1}\) and \(\tan^{-1}\) undo the cosine and tangent respectively. All three inverse functions work the same way: give them a ratio, get back an angle.

Example. Your phone screen is 15.6 cm tall and 7.2 cm wide. What angle does the diagonal make with the long side?

The two known sides are 15.6 cm (adjacent to the angle at the bottom-left corner) and 7.2 cm (opposite). The ratio connecting opposite and adjacent is tangent:

\[\tan(\theta) = \frac{7.2}{15.6} \approx 0.4615\]

Apply inverse tangent to both sides:

\[\theta = \tan^{-1}(0.4615) \approx 24.8°\]

The diagonal meets the long side at about 25°.

Why this works

The ratios are fixed by the angle alone — not by how big the triangle is. Any right triangle with a 30° angle has the same ratio of opposite to hypotenuse: exactly 0.5. Scale the triangle up by ten times and both sides get ten times longer, but the ratio stays the same.

This is why the ratios are useful. They depend only on the angle, so you can look up \(\sin(30°) = 0.5\) once and use it for any triangle containing a 30° angle, whatever its size.

18.3 Worked examples

Example 1 — Finding the height of a building (geography/surveying)

A surveyor stands 40 m from the base of a radio mast. She measures the angle of elevation to the top as 62°. How tall is the mast?

The 40 m is the side adjacent to the 62° angle. The mast height is the opposite side. The ratio connecting adjacent and opposite is tangent:

\[\tan(62°) = \frac{\text{height}}{40}\]

Multiply both sides by 40:

\[\text{height} = 40 \times \tan(62°)\]

\(\tan(62°) \approx 1.8807\), so:

\[\text{height} = 40 \times 1.8807 \approx 75.2 \text{ m}\]

The mast is approximately 75 m tall. Check: \(\tan^{-1}(75.2/40) = \tan^{-1}(1.880) \approx 62°\). Yes.

Example 2 — Rope from a fence post (geometry/practical)

A rope is tied from the top of a fence post to a peg in the ground. The post is vertical and the rope makes an angle of 35° with the post. The rope has a tension of 80 N — the force acting along the rope. What is the component of that tension pulling horizontally away from the post?

The 80 N tension acts along the rope — that’s the hypotenuse of a right triangle. The angle at the top of the post is 35°. The horizontal pull is the side opposite that angle:

\[\sin(35°) = \frac{\text{horizontal component}}{80}\]

\[\text{horizontal component} = 80 \times \sin(35°)\]

\(\sin(35°) \approx 0.5736\), so:

\[\text{horizontal component} = 80 \times 0.5736 \approx 45.9 \text{ N}\]

The peg must hold against about 46 N of horizontal pull.

Example 3 — Finding the ramp angle for a target slope (science/engineering)

A physiotherapy exercise ramp needs a slope of 1 in 8 — for every 8 cm of horizontal run, it rises 1 cm. What angle does this correspond to?

The slope ratio gives you opposite over adjacent: \(\frac{1}{8} = 0.125\). That’s the tangent of the angle:

\[\tan(\theta) = \frac{1}{8} = 0.125\]

\[\theta = \tan^{-1}(0.125) \approx 7.1°\]

The ramp makes an angle of about 7° with the floor.

Example 4 — Rotating a point in 2D (computing/game development)

A game object is at position \((5, 0)\) — five units to the right of the origin, on the horizontal axis. The game rotates everything 40° anticlockwise. Where does the object end up?

The object starts at distance 5 from the origin (its “hypotenuse” from the rotation centre). After a 40° rotation, the new \(x\)-coordinate is the adjacent side of the 40° angle, and the new \(y\)-coordinate is the opposite side:

\[x' = 5 \times \cos(40°) \approx 5 \times 0.7660 \approx 3.83\]

\[y' = 5 \times \sin(40°) \approx 5 \times 0.6428 \approx 3.21\]

The object moves to approximately \((3.83,\ 3.21)\). This is the core of every 2D rotation in a game engine.

18.4 Where this goes

The three ratios in this chapter are functions — you put in an angle and get out a number. Volume 3, Chapter 4 (Functions and relations) gives that idea its full treatment: a function is a rule that maps one value to another, and sine, cosine, and tangent are the first examples most people meet of functions that are not straight lines. Once you see trig through the function lens you can ask new questions: what does the graph look like? Where is it increasing? Does it repeat?

It repeats — and that turns out to matter enormously. Differential calculus (Volume 5) uses sine and cosine as its central worked examples because they are the simplest functions whose rate of change is also a trig function. The derivative of \(\sin\) is \(\cos\); the derivative of \(\cos\) is \(-\sin\). That self-referential property is why trig is indispensable in physics and engineering, and it’s the first real glimpse of how calculus and trigonometry are connected at a deep level.

Where this shows up

  • A structural engineer resolves a cable force into horizontal and vertical components before checking whether a joint will hold.
  • A physicist writes the equation of a pendulum’s motion using sine — the angle oscillates like a sine wave over time.
  • A navigator uses bearing angles and the sine rule to triangulate a ship’s position from two known landmarks.
  • A game developer rotates sprites, aims projectiles, and swings camera views using \(\sin\) and \(\cos\) applied to coordinates.
  • A sound engineer represents audio waveforms as sums of sine waves — the starting point of Fourier analysis.

18.5 Exercises

These are puzzles. Each one has a clean answer. The work is in identifying which sides and which ratio to use before you reach for the calculator.

1. A ski slope is 800 m long and the vertical drop is 240 m. What angle does the slope make with the horizontal? Give your answer to one decimal place.

2. A ladder 6 m long leans against a wall. It makes an angle of 72° with the ground. How high up the wall does the top of the ladder reach? Give your answer to two decimal places.

3. A phone screen has a diagonal of 16.5 cm and the screen is 8.0 cm wide. What angle does the diagonal make with the bottom edge of the screen?

4. A solar panel is mounted on a roof. To maximise energy collection in Alberta, the panel should face the sun at an angle of elevation of 53°. The vertical rise from the base of the panel to its top edge is 1.2 m. How long is the panel surface?

5. A zip line descends from a platform 18 m high. The cable makes an angle of 35° with the horizontal. How long is the cable, and how far from the base of the platform does it end?

6. A ladder 5 m long leans against a wall. Its base is 1.8 m from the wall. (a) What angle does the ladder make with the ground? (b) How high up the wall does it reach?