Learn: Trigonometry (SOH CAH TOA)

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The idea

SOH CAH TOA is the map of a right triangle — three ratios that connect an angle to its sides.

The three ratios:

1
SOH — $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ — sine: opposite over hypotenuse
2
CAH — $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ — cosine: adjacent over hypotenuse
3
TOA — $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ — tangent: opposite over adjacent
4
Pick the ratio — the one with the two sides you know and the side you need — then it is one clean multiplication or division

Quick memory tips

Opposite and adjacent are named from the angle θ — switch the angle and the names swap
The hypotenuse never moves: always the longest side, across from the right angle
O + H → SOH · A + H → CAH · O + A → TOA

The sides are named from the angle’s point of view: opposite faces θ, adjacent touches θ (but is not the hypotenuse), and the hypotenuse is always the longest side, across from the right angle.

Worked example

Every step explained — why we do it, and what happens.

1
$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$
Why: Name the sides from θ: the opposite side faces θ (length 3), the hypotenuse is the longest side, across from the right angle (length 5). SOH = Sine is Opposite over Hypotenuse.
What happens: sin θ = 3/5 — the first ratio of the 3-4-5 triangle.
2
$\cos\theta = \frac{4}{5}, \qquad \tan\theta = \frac{3}{4}$
Why: The side touching θ but not the hypotenuse is the adjacent (length 4). CAH and TOA are the same game with the other pairs of sides.
What happens: cos θ = 4/5 and tan θ = 3/4 — the full SOH CAH TOA set.
3
$\text{which ratio?}$
Ask which two sides the problem gives you: O and H → SOH, A and H → CAH, O and A → TOA. The ratio holding both known sides and the mystery side is the one to use.

Special Angles (Optional Toolbox)

Quick shortcuts (memorize or derive). Separate from the θ example above.

optional

These values come from two special right triangles: 30–60–90 and 45–45–90. If you see a 30° or 45° in a real problem, you can use these immediately.

30–60–90 triangle ratios

Side ratios: $1 : \sqrt{3} : 2$ (short : long : hypotenuse)

\sin 30^\circ = \tfrac{\text{opposite}}{\text{hypotenuse}} = \tfrac{1}{2}

\cos 30^\circ = \tfrac{\text{adjacent}}{\text{hypotenuse}} = \tfrac{\sqrt{3}}{2}

45–45–90 triangle ratios

Side ratios: $1 : 1 : \sqrt{2}$ (leg : leg : hypotenuse)

\sin 45^\circ = \tfrac{1}{\sqrt{2}} = \tfrac{\sqrt{2}}{2}

\tan 45^\circ = \tfrac{1}{1} = 1

Exact values:

30°

\sin 30^\circ = \tfrac{1}{2}

\cos 30^\circ = \tfrac{\sqrt{3}}{2}

\tan 30^\circ = \tfrac{1}{\sqrt{3}}

45°

\sin 45^\circ = \tfrac{\sqrt{2}}{2}

\cos 45^\circ = \tfrac{\sqrt{2}}{2}

\tan 45^\circ = 1

60°

\sin 60^\circ = \tfrac{\sqrt{3}}{2}

\cos 60^\circ = \tfrac{1}{2}

\tan 60^\circ = \sqrt{3}

Tip: $\tan\theta = \sin\theta / \cos\theta$ (when you already know sin and cos).

When a word problem says “gentle 30° incline”, that’s your signal to use $\sin 30^\circ = 1/2$ .

Real-world example

The community hall needs a wheelchair ramp: it must rise 1 m to the door, and the builders want a gentle 30° incline. How long must the ramp be? The ramp is the hypotenuse, and the 1 m rise is the side opposite the 30° angle.

1
$\sin 30^\circ = \frac{1}{L}$
Why: Recall sin 30° = 1/2 from the Special Angles toolbox above. The ramp is the hypotenuse (L), the rise is the opposite side, and we know the angle — SOH has exactly these three.
What happens: One equation, one unknown: sin 30° = 1 ÷ L.
2
$\frac{1}{2} = \frac{1}{L} \Rightarrow L = 2 \text{ m}$
Why: sin 30° = 1/2, so 1/2 = 1/L. Flip both sides: L = 2.
What happens: The ramp is 2 m — twice the rise, because sin 30° = 1/2.
3
$\text{Trap: } \tan 30^\circ = \frac{1}{b} \Rightarrow b = \sqrt{3} \approx 1.73 \text{ m}$
Why: The glitch: reaching for tan instead of sin. tan gives the base — how far the ramp’s foot sits from the wall — not the ramp itself.
What happens: 1.73 m is the base, not the ramp length.
4
$1^2 + (\sqrt{3})^2 = 1 + 3 = 4 = 2^2$
Why: The check: the three lengths must close a right triangle — Pythagoras: rise² + base² = ramp².
What happens: 1 + 3 = 4 = 2² ✓ — the 2 m ramp, 1 m rise and √3 m base fit perfectly.

Common glitch

Two classic glitches:

1.Swapping opposite and adjacent

Wrong:Opposite and adjacent are fixed labels on the triangle

Right:They are named from the angle you are using: opposite faces θ, adjacent touches θ. Switch the angle and the names swap.

2.Grabbing the wrong ratio

Wrong:Any ratio works — just divide the two numbers you see

Right:Pick the ratio with the side you know AND the side you need: O+H → SOH, A+H → CAH, O+A → TOA.

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Common glitch

Two classic glitches:

1.Swapping opposite and adjacent

Wrong:Opposite and adjacent are fixed labels on the triangle

Right:They are named from the angle you are using: opposite faces θ, adjacent touches θ. Switch the angle and the names swap.

2.Grabbing the wrong ratio

Wrong:Any ratio works — just divide the two numbers you see

Right:Pick the ratio with the side you know AND the side you need: O+H → SOH, A+H → CAH, O+A → TOA.

Learn: Trigonometry (SOH CAH TOA)

The idea

The three ratios:

Quick memory tips

Worked example

Real-world example

Common glitch

Try a related calculator

Learn: Trigonometry (SOH CAH TOA)

The idea

The three ratios:

Quick memory tips

Worked example

Real-world example

Common glitch

Try a related calculator