Trigonometric ratios

Stand at the base of a tall tree and look up at its top — the angle your eyes make with the ground and the tree's height and your distance from it are all locked together by a fixed relationship. That relationship is what trigonometric ratios capture. In a right-angled triangle, once you fix one acute angle, the ratio of any two sides relative to that angle is always the same number, no matter how big or small the triangle is drawn. This article names the three sides relative to an angle, defines the six trigonometric ratios as fractions of those sides, shows how Pythagoras' theorem supplies a missing side, and works through two fully solved examples so you can compute any of the six ratios from just two given pieces of information.

The core insight

In a right-angled triangle, each trigonometric ratio of a chosen acute angle is a fixed fraction formed from the opposite, adjacent and hypotenuse sides, and it depends only on the angle, never on the triangle's size.

The core idea

Every right-angled triangle has one right angle and two acute angles. Trigonometric ratios are always defined relative to one of the acute angles, so the first job before writing any ratio is to name the sides correctly for that angle.

  • Take a right triangle with the right angle at CC, and let AA be one of the acute angles. The side facing angle AA (not touching it) is the opposite side.
  • The side that touches angle AA but is not the hypotenuse is the adjacent side.
  • The side facing the right angle — always the longest side — is the hypotenuse.

Once the three sides are named, the three primary trigonometric ratios of angle AA are:

sin⁡A=oppositehypotenuse,cos⁡A=adjacenthypotenuse,tan⁡A=oppositeadjacent\sin A = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan A = \frac{\text{opposite}}{\text{adjacent}}

where: opposite, adjacent and hypotenuse are the side lengths measured relative to angle AA, and sin⁡A\sin A, cos⁡A\cos A, tan⁡A\tan A are pure numbers with no unit, since a ratio of two lengths always cancels the unit. Each primary ratio has a reciprocal, formed simply by flipping the fraction:

csc⁡A=1sin⁡A=hypotenuseopposite,sec⁡A=1cos⁡A=hypotenuseadjacent,cot⁡A=1tan⁡A=adjacentopposite\csc A = \frac{1}{\sin A} = \frac{\text{hypotenuse}}{\text{opposite}}, \qquad \sec A = \frac{1}{\cos A} = \frac{\text{hypotenuse}}{\text{adjacent}}, \qquad \cot A = \frac{1}{\tan A} = \frac{\text{adjacent}}{\text{opposite}}

where: csc⁡A\csc A, sec⁡A\sec A, cot⁡A\cot A are read as cosecant, secant and cotangent of angle AA, and each is undefined only when its denominator ratio is zero. A handy check links tangent to sine and cosine: dividing opposite/hypotenuse by adjacent/hypotenuse cancels the hypotenuse and leaves opposite/adjacent, so tan⁡A=sin⁡Acos⁡A\tan A = \dfrac{\sin A}{\cos A}. The classic memory aid for the three primary ratios is SOH-CAH-TOA — Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. If a triangle is given with only two of its three sides, Pythagoras' theorem, (hypotenuse)2=(opposite)2+(adjacent)2(\text{hypotenuse})^2 = (\text{opposite})^2 + (\text{adjacent})^2, must be used first to find the missing side before any ratio can be written down.

Figure
A right triangle with the right angle marked at C, angle A at the base, the side opposite A labelled opposite, the side next to A labelled adjacent and the slanted side labelled hypotenuse
A C B adjacent opposite hypotenuse
Derivation AFinding all six ratios from two given sides

The goal is to show, using nothing beyond the definitions above and Pythagoras' theorem, exactly how to obtain all six trigonometric ratios of an acute angle in a right triangle when only two side lengths are known, and to confirm along the way that the ratios never depend on the triangle's actual size.

Step 1. Identify which two sides are given relative to the chosen angle. In any right triangle, once the right angle is located, the hypotenuse is fixed as the side opposite it; the remaining two sides become "opposite" or "adjacent" only once a specific acute angle is chosen as the reference angle.

Step 2. If the given two sides are the opposite and adjacent sides (but not the hypotenuse), find the hypotenuse using (hypotenuse)2=(opposite)2+(adjacent)2(\text{hypotenuse})^2 = (\text{opposite})^2 + (\text{adjacent})^2, then take the square root. If instead the hypotenuse and one other side are given, rearrange the same equation to find the missing side.

Step 3. With all three sides now known, write the three primary ratios directly from their definitions: sin⁡A=opposite/hypotenuse\sin A = \text{opposite}/\text{hypotenuse}, cos⁡A=adjacent/hypotenuse\cos A = \text{adjacent}/\text{hypotenuse}, tan⁡A=opposite/adjacent\tan A = \text{opposite}/\text{adjacent}.

Step 4. Obtain the three reciprocal ratios simply by flipping each primary fraction upside down: csc⁡A=1/sin⁡A\csc A = 1/\sin A, sec⁡A=1/cos⁡A\sec A = 1/\cos A, cot⁡A=1/tan⁡A\cot A = 1/\tan A. Because every step used only the ratio of two lengths, scaling the whole triangle up or down leaves every ratio unchanged — this is why the same six numbers describe every right triangle containing that particular angle.

Worked example

In right triangle PQRPQR, right-angled at QQ, the side PQ=5PQ = 5 cm and QR=12QR = 12 cm. Find all six trigonometric ratios of angle RR, and then use tan⁡R\tan R to check the answer a second way.

Step 1 — name the sides relative to angle RR. The right angle is at QQ, so PRPR is the hypotenuse. Relative to angle RR, the side opposite RR is PQ=5PQ = 5 cm, and the side adjacent to RR is QR=12QR = 12 cm.

Step 2 — find the hypotenuse. PR2=PQ2+QR2=52+122=25+144=169PR^2 = PQ^2 + QR^2 = 5^2 + 12^2 = 25 + 144 = 169, so PR=169=13PR = \sqrt{169} = 13 cm.

Step 3 — write the three primary ratios. sin⁡R=PQPR=513\sin R = \dfrac{PQ}{PR} = \dfrac{5}{13}, cos⁡R=QRPR=1213\cos R = \dfrac{QR}{PR} = \dfrac{12}{13}, tan⁡R=PQQR=512\tan R = \dfrac{PQ}{QR} = \dfrac{5}{12}.

Step 4 — write the three reciprocal ratios. csc⁡R=135\csc R = \dfrac{13}{5}, sec⁡R=1312\sec R = \dfrac{13}{12}, cot⁡R=125\cot R = \dfrac{12}{5}.

Step 5 — check with the sine-over-cosine identity. sin⁡Rcos⁡R=5/1312/13=512\dfrac{\sin R}{\cos R} = \dfrac{5/13}{12/13} = \dfrac{5}{12}, which matches tan⁡R\tan R found directly in Step 3, confirming the calculation.

The six ratios of angle RR are: sin⁡R=513\sin R = \tfrac{5}{13}, cos⁡R=1213\cos R = \tfrac{12}{13}, tan⁡R=512\tan R = \tfrac{5}{12}, csc⁡R=135\csc R = \tfrac{13}{5}, sec⁡R=1312\sec R = \tfrac{13}{12}, cot⁡R=125\cot R = \tfrac{12}{5}.

Competitive stretch

If tan⁡A=34\tan A = \dfrac{3}{4}, find sin⁡A\sin A and cos⁡A\cos A without drawing a fresh triangle. Since tan⁡A=opposite/adjacent=3/4\tan A = \text{opposite}/\text{adjacent} = 3/4, take the opposite side as 3k3k and adjacent as 4k4k for some positive kk. Then the hypotenuse is (3k)2+(4k)2=25k2=5k\sqrt{(3k)^2 + (4k)^2} = \sqrt{25k^2} = 5k, giving sin⁡A=3k5k=35\sin A = \dfrac{3k}{5k} = \dfrac{3}{5} and cos⁡A=4k5k=45\cos A = \dfrac{4k}{5k} = \dfrac{4}{5} — the kk cancels, so the answer holds for every triangle with this angle, however large or small.

Common traps

  • Mixing up "opposite" and "adjacent" when the reference angle changes — the side opposite AA becomes the side adjacent to the other acute angle in the same triangle.
  • Assuming the longest-looking side in a rough sketch is the hypotenuse instead of checking that it faces the right angle.
  • Writing units on sin⁡A\sin A, cos⁡A\cos A, tan⁡A\tan A as if they were lengths — they are unit-free ratios.
  • Writing tan⁡A=sin⁡A×cos⁡A\tan A = \sin A \times \cos A instead of the correct tan⁡A=sin⁡Acos⁡A\tan A = \dfrac{\sin A}{\cos A}.

Ask the tutor: Want to practise finding all six trigonometric ratios starting from just one given ratio, or work through a triangle where the reference angle switches partway through the problem?

Why do sin A, cos A and tan A stay the same even if I draw the triangle bigger?

Because each ratio is formed by dividing one side length by another, and scaling a triangle up or down multiplies every side by the same factor, which cancels out in the division. In the competitive-stretch example, replacing the sides 3,4,53, 4, 5 with 3k,4k,5k3k, 4k, 5k for any kk still gives sin⁡A=3/5\sin A = 3/5 and cos⁡A=4/5\cos A = 4/5. The ratio depends only on the shape of the triangle, meaning only on the angle, never on how large you draw it.

How do I remember which side is opposite and which is adjacent?

Look at the angle you are given, then look at the two sides that are not the hypotenuse. The side that touches the angle (forms one of its two arms, besides the hypotenuse) is adjacent; the side that does not touch the angle, sitting across the triangle from it, is opposite. If you switch to the other acute angle in the same triangle, opposite and adjacent swap roles, but the hypotenuse never changes, since it always faces the right angle.

Do I always need Pythagoras theorem before finding trigonometric ratios?

Only when you are missing one of the three sides. If a question already gives you all three sides, or gives you one ratio directly (like tan⁡A=3/4\tan A = 3/4 in the competitive stretch), you do not need to apply Pythagoras separately — you can use the given ratio to set up sides as 3k3k and 4k4k and then find the hypotenuse from those. Pythagoras' theorem is the tool that fills in whichever one side is missing, nothing more.

Before this

Where this leads