Trigonometry

An interactive guide — 10 essential concepts

1. Similar Triangles

Similar triangles have the same shape but may differ in size. Two properties hold:

$$\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$$

Three ways to prove similarity: AA (angle-angle), SSS (all side ratios equal), SAS (two side ratios + included angle).

Interactive: Similar Triangle Scaler

Drag the slider to scale the second triangle. Ratios stay constant.

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Real-World Example: Measuring a Building’s Height with Shadows

A 2 m stick casts a 1.5 m shadow at the same time a building casts a 18 m shadow. The sun's rays form equal angles with the ground for both, creating two similar right triangles (AA similarity — both share a right angle and equal sun-angle).

$$\frac{\text{building height}}{18} = \frac{2}{1.5} \implies \text{height} = \frac{2 \times 18}{1.5} = 24 \text{ m}$$

Surveyors, architects, and engineers use this principle constantly — measuring inaccessible heights via proportional shadows or mirror reflections on the ground.

2. SOHCAHTOA

In any right triangle with a reference angle $\theta$, name the sides:

$$\sin\theta = \frac{O}{H} \qquad \cos\theta = \frac{A}{H} \qquad \tan\theta = \frac{O}{A}$$

Interactive: Right Triangle Calculator

Enter any angle (1–89°) and hypotenuse to compute the sides.

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Real-World Example: Wheelchair Ramp Slope

ADA guidelines require wheelchair ramps to have a maximum slope of 1:12 (rise:run). What is the ramp angle?

$$\tan\theta = \frac{\text{rise}}{\text{run}} = \frac{1}{12} \implies \theta = \arctan\!\left(\tfrac{1}{12}\right) \approx 4.76°$$

A building entrance 0.6 m above ground needs a ramp at least $0.6 \times 12 = 7.2$ m long horizontally. The actual ramp length (hypotenuse) is $\frac{0.6}{\sin 4.76°} \approx 7.22$ m. Civil engineers use SOHCAHTOA every day for grading roads, designing ramps, and calculating sight lines.

3. Sine Law & Cosine Law

For oblique triangles (no 90° angle):

Cosine Law (find a side): $$c^2 = a^2 + b^2 - 2ab\cos C$$ Cosine Law (find an angle): $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ Sine Law: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

When to use what:

Interactive: Oblique Triangle Solver

Enter three known values (at least one side). Leave unknowns as 0.

Real-World Example: Surveying a Land Parcel

A surveyor stands at point A and measures distances to two boundary markers: AB = 340 m and AC = 510 m. The angle between the two sight lines is 72°. What is the distance BC between the markers?

$$BC^2 = 340^2 + 510^2 - 2(340)(510)\cos 72°$$ $$BC^2 = 115600 + 260100 - 346800 \times 0.3090 \approx 268560$$ $$BC \approx 518.4 \text{ m}$$

Sine Law and Cosine Law are the backbone of triangulation — used in GPS, land surveying, marine navigation, and satellite ranging to compute distances and positions when direct measurement is impossible.

4. Special Triangles

Two triangles give exact trig values for 30°, 45°, and 60°:

45-45-90

Isosceles right triangle. Sides: $1, 1, \sqrt{2}$.

$\sin 45° = \dfrac{\sqrt{2}}{2}$
$\cos 45° = \dfrac{\sqrt{2}}{2}$
$\tan 45° = 1$

30-60-90

Half of an equilateral triangle. Sides: $1, \sqrt{3}, 2$.

$\sin 30° = \tfrac{1}{2},\; \cos 30° = \tfrac{\sqrt{3}}{2},\; \tan 30° = \tfrac{\sqrt{3}}{3}$
$\sin 60° = \tfrac{\sqrt{3}}{2},\; \cos 60° = \tfrac{1}{2},\; \tan 60° = \sqrt{3}$

Interactive: Special Triangle Explorer

Click a reference angle to see the triangle and exact ratios.

Real-World Example: Roof Pitch Calculation

A house has a roof with a 12/12 pitch (rises 12 inches for every 12 inches of run). The roof angle is $\arctan(12/12) = 45°$. For a house 10 m wide, each rafter spans half the width (5 m run), so the rafter length is:

$$\text{rafter} = \frac{5}{\cos 45°} = \frac{5}{\sqrt{2}/2} = 5\sqrt{2} \approx 7.07 \text{ m}$$

Carpenters and structural engineers memorise the special triangle ratios — a 30° roof (low slope, common in warm climates) versus a 60° roof (steep, sheds snow) have rafter lengths directly derived from these exact values.

5. Unit Circle & CAST Rule

The unit circle has radius 1, centred at the origin. A terminal arm at angle $\theta$ intersects it at the point:

$$(x, y) = (\cos\theta,\; \sin\theta)$$

Key points: $(1,0)$ at 0°, $(0,1)$ at 90°, $(-1,0)$ at 180°, $(0,-1)$ at 270°.

The CAST rule tells which ratios are positive in each quadrant:

Interactive: Unit Circle Explorer

Drag the slider or type an angle to see sin, cos, tan on the unit circle.

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Real-World Example: Ferris Wheel Rider Height

A Ferris wheel has radius 25 m and its centre is 27 m above the ground. A rider boards at the bottom (angle = 270° on the unit circle, or $-\pi/2$). After rotating through angle $\theta$ from the 3-o'clock position, the rider's height is:

$$h(\theta) = 27 + 25\sin\theta$$

At the top ($\theta = 90°$): $h = 27 + 25(1) = 52$ m. At the bottom ($\theta = 270°$): $h = 27 + 25(-1) = 2$ m. Theme park engineers use unit-circle relationships to model rider position, speed, and g-forces at every point of the rotation.

6. Ratios for Angles > 90°

To find exact trig values for angles beyond 90°:

  1. Locate the principal angle on the Cartesian grid.
  2. Find the reference angle — the acute angle between the terminal arm and the nearest x-axis.
  3. Use the CAST rule to assign the correct sign (+/−).
  4. Evaluate using the reference angle's special-triangle value.
Example: $\sin 150° = +\sin 30° = \tfrac{1}{2}$   (Q II, sin is positive, ref = 30°)
Example: $\cos 225° = -\cos 45° = -\tfrac{\sqrt{2}}{2}$   (Q III, cos is negative, ref = 45°)

Interactive: Reference Angle Finder

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Real-World Example: Force Components on an Inclined Conveyor

A 500 N crate sits on a conveyor belt inclined at 150° to the positive x-axis (measuring counter-clockwise). The horizontal and vertical force components are:

$$F_x = 500\cos 150° = 500 \times (-\cos 30°) = -250\sqrt{3} \approx -433 \text{ N}$$ $$F_y = 500\sin 150° = 500 \times \sin 30° = 250 \text{ N}$$

Mechanical engineers decompose forces at arbitrary angles daily — designing conveyor systems, crane booms, and bridge trusses. Knowing how to handle angles beyond 90° via reference angles is essential for resolving vectors in any quadrant.

7. Sine & Cosine Graphs

Thinking of $\sin$ and $\cos$ as functions of angle:

$$y = A\sin(Bx) + D \quad\text{or}\quad y = A\cos(Bx) + D$$ Amplitude $= |A|$   (half the distance between max and min)
Period $= \dfrac{360°}{|B|}$   (horizontal length of one cycle)
Vertical shift $= D$

For the base functions $y = \sin x$ and $y = \cos x$: amplitude = 1, period = 360°.

Interactive: Wave Plotter

Adjust amplitude, period multiplier, and vertical shift.

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Real-World Example: Modelling Tidal Patterns

A harbour's tide varies between 1.2 m (low) and 5.8 m (high) with a period of approximately 12.4 hours. If high tide occurs at $t = 0$:

$$\text{Amplitude} = \frac{5.8 - 1.2}{2} = 2.3 \text{ m}, \quad D = \frac{5.8 + 1.2}{2} = 3.5 \text{ m}$$ $$h(t) = 2.3\cos\!\left(\frac{2\pi}{12.4}\,t\right) + 3.5$$

Marine engineers, harbour pilots, and coastal planners use sinusoidal models to predict water levels for docking schedules, flood barriers, and coastal construction timing.

8. Radians

A radian is the angle subtended when the arc length equals the radius:

$$\theta \text{ (radians)} = \frac{\text{arc length}}{r} \qquad\qquad 2\pi \text{ rad} = 360° \qquad\qquad 1° = \frac{\pi}{180} \text{ rad}$$

Common conversions: 30° = $\pi/6$, 45° = $\pi/4$, 60° = $\pi/3$, 90° = $\pi/2$, 180° = $\pi$.

Interactive: Degree ↔ Radian Converter

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Interactive: Arc Length Visualiser

Drag to change the angle — watch the arc equal to the radius.

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Real-World Example: Angular Velocity of a Car Wheel

A car travels at 90 km/h. Its tyres have radius 0.32 m. The angular velocity $\omega$ (in rad/s) of each wheel is:

$$v = 90 \text{ km/h} = 25 \text{ m/s}$$ $$\omega = \frac{v}{r} = \frac{25}{0.32} \approx 78.1 \text{ rad/s} \approx 746 \text{ RPM}$$

Radians are the natural unit for rotational mechanics. Every rotating machine — engines, turbines, hard drives, centrifuges — is analysed using radians because the arc-length formula $s = r\theta$ and the velocity formula $v = r\omega$ only work directly in radians.

9. Trig Identities

Identities are equations true for all values of the variable.

Reciprocal Identities

$$\csc x = \frac{1}{\sin x} \qquad \sec x = \frac{1}{\cos x} \qquad \cot x = \frac{1}{\tan x}$$

Quotient Identities

$$\tan x = \frac{\sin x}{\cos x} \qquad \cot x = \frac{\cos x}{\sin x}$$

Pythagorean Identity

$$\sin^2 x + \cos^2 x = 1$$ Rearranged: $\sin^2 x = 1 - \cos^2 x$,   $\cos^2 x = 1 - \sin^2 x$

Interactive: Identity Verifier

Pick any angle — the identity holds true every time.

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Real-World Example: Simplifying an AC Power Formula

In electrical engineering, instantaneous power in an AC circuit is:

$$p(t) = V_0 I_0 \sin(\omega t)\sin(\omega t - \phi)$$

Using the product-to-sum identity (derived from the Pythagorean and angle-sum identities), this simplifies to:

$$p(t) = \frac{V_0 I_0}{2}\left[\cos\phi - \cos(2\omega t - \phi)\right]$$

The first term gives the average (real) power; the second is the oscillating component. Without trig identities, analysing AC circuits, signal processing, and electromagnetic wave interference would be impossibly complex.

10. Solving Trig Equations

Strategy:

  1. Isolate the trig function (factor if quadratic).
  2. Use special triangles / unit circle to find the reference angle.
  3. Use the CAST rule to find all principal angles in the given domain.
  4. If the domain is unrestricted, add $+ 2\pi k$ (or $+ 360° k$), $k \in \mathbb{Z}$.
Example: $\sin x = -\dfrac{\sqrt{2}}{2}$,   $0 \le x < 2\pi$

Reference angle $= \pi/4$. Sin is negative in Q III and Q IV.
$x_1 = \pi + \pi/4 = 5\pi/4$,   $x_2 = 2\pi - \pi/4 = 7\pi/4$.

Interactive: Equation Visualiser

Set a target sin value to see where the sine curve hits that value in $[0, 2\pi]$.

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Real-World Example: When Does a Pendulum Pass a Sensor?

A pendulum's angular displacement is modelled by $\theta(t) = 0.15 \sin(2\pi t)$ (radians), where $t$ is in seconds. A sensor triggers when $\theta = 0.075$ rad. When does it trigger in the first second?

$$0.15\sin(2\pi t) = 0.075 \implies \sin(2\pi t) = 0.5$$ $$2\pi t = \frac{\pi}{6} \;\text{or}\; \frac{5\pi}{6}$$ $$t = \frac{1}{12} \approx 0.083 \text{ s} \quad\text{or}\quad t = \frac{5}{12} \approx 0.417 \text{ s}$$

Robotics engineers solving for precise timing of periodic events — oscillating arms, vibrating components, rotating gears — solve trig equations like this constantly to synchronise sensors, actuators, and control loops.


Source material: JensenMath — Top 10 Things to Know About Trigonometry.