Ratios in a Right Triangle

Idea

Take two right triangles with the same acute angle. One may be bigger, but it is just a scaled copy of the other.

Try it. Slide to grow the triangle below. Every side changes. Watch the three ratios under it: they do not move.

A right triangle

Try this. Compare the labeled lengths with sine, cosine and tangent. If size is adjustable, scale the triangle: lengths change together, but the ratios stay fixed.

Triangle size1

θ = 36.87° · sin θ = 0.6 · cos θ = 0.8 · tan θ = 0.75

That fixed number is what a trigonometric function gives you. Put in the angle, get back a ratio. The size of the triangle does not matter - only the angle does.

Rule

Pick one acute angle in the triangle and call it $\theta$ . Now name the three sides from that angle. The side across from $\theta$ is the opposite. The side next to $\theta$ that is not the longest is the adjacent. The longest side, across from the right angle, is the hypotenuse.

Three ratios come first.

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

The other three are the same fractions turned upside down. Each one is also a ratio of two sides, so you can read it straight off the triangle.

\cot\theta = \frac{\text{adjacent}}{\text{opposite}}

\sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}}

\csc\theta = \frac{\text{hypotenuse}}{\text{opposite}}

Written the other way, $\cot\theta = 1/\tan\theta$ , $\sec\theta = 1/\cos\theta$ and $\csc\theta = 1/\sin\theta$ .

How it is used

1. The names follow the angle

Turn the triangle and the "opposite" side moves with the angle you are using. Reading the sides from a fixed picture is the most common mistake here.

2. Which ratio answers which question

You know	You want	Use
opposite and hypotenuse	the angle	$\sin\theta$
adjacent and hypotenuse	the angle	$\cos\theta$
the two legs	the angle	$\tan\theta$
an angle and one side	another side	rearrange the matching ratio

The last three ratios rarely start a problem. Flip the first three when you need them.

3. Sine and cosine never pass 1

The hypotenuse is the longest side, so $\sin\theta$ and $\cos\theta$ are never larger than $1$ . If you get $1.4$ , you made an arithmetic mistake. For the same reason $\sec\theta$ and $\csc\theta$ are never smaller than $1$ .

4. The other angle swaps the pair

The two acute angles add to $90^\circ$ , so $\sin\theta = \cos(90^\circ - \theta)$ . Sine and cosine measure the same thing, seen from the other angle.

Worked example - the three main ratios

A right triangle has legs of $3$ and $4$ and hypotenuse $5$ . Find the sine, cosine and tangent of the angle $\theta$ opposite the side of length $3$ .

The side opposite $\theta$ is $3$ , the side adjacent is $4$ , and the hypotenuse is $5$ .

\sin\theta = \frac{3}{5}

\cos\theta = \frac{4}{5}

\tan\theta = \frac{3}{4}

Both sine and cosine came out below $1$ , as they must. Here is the triangle those numbers describe.

A right triangle

Try this. Compare the labeled lengths with sine, cosine and tangent. If size is adjustable, scale the triangle: lengths change together, but the ratios stay fixed.

θ = 36.87° · sin θ = 0.6 · cos θ = 0.8 · tan θ = 0.75

Worked example - all six ratios, from both angles

Triangle $ABC$ has its right angle at $B$ . The sides are $|AB| = 5$ , $|BC| = 12$ and $|AC| = 13$ . Write all six ratios for angle $A$ , and then for angle $C$ .

Start at angle $A$ . The side across from $A$ is $BC$ , so the opposite is $12$ . The side next to $A$ that is not the hypotenuse is $AB$ , so the adjacent is $5$ . The hypotenuse is $AC = 13$ .

\sin A = \frac{12}{13}

\cos A = \frac{5}{13}

\tan A = \frac{12}{5}

\cot A = \frac{5}{12}

\sec A = \frac{13}{5}

\csc A = \frac{13}{12}

Now move to angle $C$ . Nothing about the triangle changed. Only the angle you are reading from changed. The side across from $C$ is $AB$ , so the opposite is now $5$ , and the adjacent is now $12$ .

\sin C = \frac{5}{13}

\cos C = \frac{12}{13}

\tan C = \frac{5}{12}

\cot C = \frac{12}{5}

\sec C = \frac{13}{12}

\csc C = \frac{13}{5}

Compare the two lists. Every sine turned into a cosine, and every tangent turned into a cotangent. This happens because $A$ and $C$ add to $90^\circ$ .

In the picture below, $\alpha$ is angle $C$ and $\beta$ is angle $A$ . Switch between them and watch the two legs trade names while their lengths stay the same.

Read the ratios

Try this. Choose a ratio to highlight its numerator and denominator. Switch the reference angle: opposite and adjacent exchange roles.

Read a ratioRead from

sin θ = opposite / hypotenuse = 5 / 13 = 0.385. Blue: numerator. Green: denominator.

Explore

Pick a ratio. The triangle lights up the two sides that ratio reads, and the rule beside it updates.

Predict what happens to sine before you switch from $\alpha$ to $\beta$ . Switch. The lengths do not move, but every name changes, and so does every answer.