Three ratios. The whole skill is picking the one that uses what you know.
Choose the ratio
You know an angle and the side adjacent to it, and you want the side opposite. Which ratio do you use?
Tangent is opposite over adjacent — the only ratio that involves those two sides and no hypotenuse. A and B both need the hypotenuse, which you do not have. D needs two sides and gives no angle at all. Pick the ratio whose two letters match what you have and want.
Only the hypotenuse is fixed — it always faces the right angle. The other two labels swap the moment you look from the other acute angle, so mark θ before naming anything.
SOH CAH TOA
| Ratio | Is | Uses |
|---|---|---|
| sin θ | opposite / hypotenuse | SOH |
| cos θ | adjacent / hypotenuse | CAH |
| tan θ | opposite / adjacent | TOA |
Write down which two sides you have or want, then pick the ratio containing exactly those. Only one of the three will fit.
θ = 35°, adjacent = 40
find the opposite
tan 35° = x40
TOA
x = 40 tan 35°
≈ 28.0
θ = 28°, hypotenuse = 15
find the opposite
sin 28° = x15
SOH
x = 15 sin 28° ≈ 7.04
shorter than 15 ✓
When the unknown is on the bottom the algebra changes: from cos 40° = 9h you get h = 9 ÷ cos 40°, a division rather than a multiplication.
opposite 5, hypotenuse 13
find θ
sin θ = 513 ≈ 0.3846
the ratio
θ = sin⁻¹(0.3846)
the inverse key
≈ 22.6°
acute, as expected
The superscript −1 here means "the angle whose sine is this", not a reciprocal. Confusing the two produces an answer with no geometric meaning at all.
A sine greater than 1 is impossible and always means the ratio was written upside down. It is the fastest error-catcher in the topic.
Exact values to know. Read them off the 45-45-90 triangle with sides 1, 1, √2 and the 30-60-90 triangle with sides 1, √3, 2.
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 12 | √32 | 1√3 |
| 45° | 1√2 | 1√2 | 1 |
| 60° | √32 | 12 | √3 |
A ratio names two sides. Draw them, find the third with Pythagoras, then read off any ratio you like.
θ is acute and cos θ = 725. Find sin θ and tan θ.
adjacent 7, hypotenuse 25
CAH
opposite = √625 − 49 = √576 = 24
Pythagoras
sin θ = 2425
opposite over hypotenuse
tan θ = 247
opposite over adjacent; above 1 is fine
Answering tan θ = 725 uses the given sides in a new ratio — but 25 is the hypotenuse and tangent never uses it. Every ratio other than the one given needs the third side first.
Three things the basic examples did not need.
θ = 50° and the adjacent side is 12. Find the hypotenuse h.
cos 50° = 12h
CAH, unknown on the bottom
h = 12 ÷ cos 50° ≈ 12 ÷ 0.6428
divide, do not multiply
h ≈ 18.7
longer than 12 ✓
Multiplying gives 12 × 0.6428 ≈ 7.7, shorter than a leg — impossible for a hypotenuse.
The other acute angle: compare sin 35° and cos 55°.
35° + 55° = 90°
the two acute angles
opposite 35° = adjacent to 55°
the same side, seen from each angle
sin 35° = cos 55° ≈ 0.574
sin θ = cos(90° − θ)
Switching angles swaps opposite and adjacent, so sine and cosine trade places.
Opposite 9, adjacent 4. Find θ.
tan θ = 94 = 2.25
TOA
θ = tan⁻¹(2.25) ≈ 66.0°
the inverse key
A tangent above 1 is allowed and means θ is above 45°: the opposite side is the longer leg.
Expect a choice of ratio for the sides you have, a side or angle to compute, a sine or cosine of a special angle, and one ratio given with another asked. The fastest route is to mark θ, label opposite, adjacent and hypotenuse from it, and pick the ratio holding exactly your two sides; find a missing side with Pythagoras first. The 10-second check: a sine or cosine above 1 means the ratio is upside down, and a hypotenuse must come out longer than both legs.
Find the ratio
In a right triangle, the side opposite angle θ is 8 and the hypotenuse is 17. What is cos θ?
Cosine needs the adjacent side, which is not given — find it with Pythagoras: 17² − 8² = 289 − 64 = 225, so the adjacent side is 15. Then cos θ = 1517. A is sin θ, and D is tan θ. C is greater than 1, which no cosine can ever be.
Hint: cosine needs the adjacent side. It is not given, but two sides of a right triangle always determine the third.
