SOH-CAH-TOA: Sin, Cos, Tan from First Principles

Three ratios, one angle, every right-triangle problem. The visual lesson that turns sin/cos/tan from a memorization headache into a 5-second decision.

8 min TEKS 9A,9B Geometría

Why we need three ratios in the first place

Imagine you're standing on the ground looking up at the top of a building. You know how far you are from the base, and you can measure the angle you're looking up at. You want to know how tall the building is — without climbing up there with a tape measure.

That's the kind of problem the Texas Geometry CBE asks you to solve over and over again. To do it, you need a way to translate "angle + one side" into "the missing side". That translator has a name: trigonometry. And in a right triangle, it boils down to three ratios — sine, cosine, and tangent.

Big idea

For any right triangle, the ratio between two sides depends only on one of the non-right angles. Memorize three ratios — and you can find any side or angle from any pair of facts.

The anatomy of a right triangle

Before we name any ratios, let's name the parts. Pick one of the two non-right angles — call it θ (theta). Now look at the three sides relative to that angle:

θ adjacent (next to θ) opposite (across from θ) hypotenuse (longest, opposite the right angle)
Three sides, one labeled angle. Remember: opposite/adjacent always depend on which angle you're looking at.
Hypotenuse
The longest side, always across from the right angle. It never changes label — same regardless of which non-right angle you focus on.
Opposite
The side across from your chosen angle θ. Switch angles and this label switches sides.
Adjacent
The side next to θ that is not the hypotenuse. Switch angles and this also switches.
Watch out

"Opposite" and "adjacent" are not properties of a side — they're properties of a side relative to the angle you're working with. Pick the angle first, then label. The single most common mistake on the CBE is using the wrong pair.

The three ratios — SOH-CAH-TOA

Now the punchline. For a fixed angle θ, no matter how big or small you draw the right triangle, three ratios stay constant:

sin θ  =  opposite / hypotenuse cos θ  =  adjacent / hypotenuse tan θ  =  opposite / adjacent SOH — CAH — TOA

The acronym SOH-CAH-TOA just stitches the first letters together: Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj. Tape it to your forehead before the test.

Aha moment

Notice how each ratio uses exactly two sides. If a problem gives you any one ratio plus any one side, you can solve for the other side. That's the entire trick.

Walkthrough — finding a missing side

Let's apply this to a real CBE-style problem. From a point on level ground 150 feet from the base of a building, you measure the angle of elevation to the top of the building as 32°. How tall is the building?

32° 150 ft (adjacent) h = ? (opposite)
The 150 ft is adjacent to the 32° angle. The unknown height h is opposite. We never see the hypotenuse — so we want a ratio that uses only adjacent and opposite.

Step 1 — identify what you have and what you want. You have the adjacent side (150 ft) and the angle (32°). You want the opposite side (h).

Step 2 — pick the ratio that uses just those parts. tan is the only one that does: opposite / adjacent. So:

tan 32° = h / 150 h = 150 × tan 32° h = 150 × 0.625 = 93.75 ft

So the building is about 94 feet tall. The whole calculation took three lines once we picked tan correctly.

The decision tree

When you see a right-triangle word problem, ask: which two of {opposite, adjacent, hypotenuse} are involved?   Opposite + adjacent → tan. Opposite + hypotenuse → sin. Adjacent + hypotenuse → cos. Pick before you write anything down.

Try it yourself

Now that you know the method, here's the kind of problem you'll see on the CBE. Click an answer to check your work — the explanation appears below.

Practice question

Ladder against a wall

A 24-foot ladder leans against a wall, making a 65° angle with the level ground. To the nearest tenth of a foot, how high up the wall does the ladder reach?  (sin 65° ≈ 0.906, cos 65° ≈ 0.423)

Open the question →
Practice question

Wheelchair ramp angle

A wheelchair ramp must rise 18 inches over a horizontal run of 216 inches. To the nearest tenth of a degree, what is the angle of inclination?

Open the question →

Where this goes next

SOH-CAH-TOA handles every right-triangle problem on the CBE that gives you an angle and one side. There are two natural extensions you'll see in upcoming lessons:

  • Inverse trig (sin⁻¹, cos⁻¹, tan⁻¹) — used when you know two sides and need to find the angle. Same three ratios, run backwards.
  • Special right triangles (30-60-90 and 45-45-90) — exact ratios you can write down without a calculator. The CBE loves these.

For now, the win condition is the same as for your CBE: given any right-triangle setup, you can pick the right ratio in under five seconds. That's it. That's the lesson.