Trigonometric ratios

In a right triangle, fix one of the acute angles. The three sides relative to that angle have names , opposite, adjacent, and hypotenuse , and the six ratios formed by pairs of them are the trigonometric ratios.

Definitions

Consider a right triangle with one acute angle $\theta$ . Looking at the triangle from $\theta$ 's point of view:

the hypotenuse is the side opposite the right angle (always the longest);
the opposite side is the one across from $\theta$ ;
the adjacent side is the remaining one (between $\theta$ and the right angle).

The six ratios are: $\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}},$ $\csc\theta = \frac{1}{\sin\theta} = \frac{\text{hyp}}{\text{opp}}, \quad \sec\theta = \frac{1}{\cos\theta} = \frac{\text{hyp}}{\text{adj}}, \quad \cot\theta = \frac{1}{\tan\theta} = \frac{\text{adj}}{\text{opp}}.$

Also $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ .

Why these ratios depend only on the angle

By the AA similarity criterion (chapter 6), all right triangles sharing the same acute angle $\theta$ are similar. Their corresponding sides are in the same ratio. So $\dfrac{\text{opp}}{\text{hyp}}$ is the same for every such triangle , it depends only on $\theta$ , not on the triangle's size.

This is why $\sin 30^\circ$ , for instance, has a single fixed value: it's the ratio $\text{opp}/\text{hyp}$ for any right triangle with a $30^\circ$ angle, regardless of size.

Useful facts

All six ratios are positive for an acute angle $\theta \in (0^\circ, 90^\circ)$ .
$\sin\theta, \cos\theta \in (0, 1)$ for $\theta \in (0^\circ, 90^\circ)$ ; their reciprocals $\csc\theta, \sec\theta > 1$ .
$\tan\theta$ can be any positive real number for $\theta \in (0^\circ, 90^\circ)$ .
From Pythagoras: $\sin^2\theta + \cos^2\theta = 1$ , the famous Pythagorean identity, which we revisit in topic 3.

A useful memorisation device for the basic ratios is SOH-CAH-TOA: Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj.

Worked examples

Example 1. In a right triangle, $\sin\theta = 3/5$ . Find $\cos\theta$ and $\tan\theta$ .

Take opp $= 3$ and hyp $= 5$ . By Pythagoras, adj $= \sqrt{25 - 9} = 4$ .

$\cos\theta = 4/5$ and $\tan\theta = 3/4$ .

Example 2. Given a right triangle with sides $5, 12, 13$ (hypotenuse $13$ ) and angle $\theta$ opposite the side $5$ , find all six ratios.

$\sin\theta = 5/13$ , $\cos\theta = 12/13$ , $\tan\theta = 5/12$ .

$\csc\theta = 13/5$ , $\sec\theta = 13/12$ , $\cot\theta = 12/5$ .

Example 3. If $\tan\theta = 7/24$ , find $\sin\theta + \cos\theta$ .

opp $= 7$ , adj $= 24$ . Hyp $= \sqrt{49 + 576} = 25$ .

$\sin\theta = 7/25$ , $\cos\theta = 24/25$ . Sum $= 31/25$ .

Example 4. In $\triangle ABC$ right-angled at $C$ , $\sin A = 3/5$ . Find $\sin B, \cos B, \tan B$ .

In this triangle, $A + B = 90^\circ$ . $\sin A = 3/5$ means opp from $A$ is $3$ , hyp is $5$ , adj from $A$ is $4$ . From $B$ 's perspective: opp is $4$ , adj is $3$ , hyp is $5$ .

$\sin B = 4/5$ , $\cos B = 3/5$ , $\tan B = 4/3$ .

Example 5. Verify: in a right triangle with $\tan\theta = 4/3$ , $(1 - \tan^2\theta)/(1 + \tan^2\theta) = ?$ .

$\tan^2\theta = 16/9$ . Numerator: $1 - 16/9 = -7/9$ . Denominator: $1 + 16/9 = 25/9$ .

Ratio: $-7/25$ . (This equals $\cos 2\theta$ in general , but we don't need that for class X.)

Try it yourself

In a right triangle, $\cos\theta = 5/13$ . Find $\sin\theta, \tan\theta, \sec\theta$ .
If $\tan\theta = 4/3$ , find $\sin\theta, \cos\theta$ .
In $\triangle ABC$ , right-angled at $B$ , $AB = 24$ cm, $BC = 7$ cm. Find $\sin A, \cos A$ .
If $3 \sin\theta = 4 \cos\theta$ , find $\tan\theta$ .
Given $\sin\theta = a/b$ , find $\cos\theta$ and $\cot\theta$ in terms of $a, b$ .
In a right triangle, the hypotenuse is twice the shorter leg. Find $\sin$ and $\cos$ of the angle opposite the shorter leg.
Show $\sec^2\theta - \tan^2\theta = 1$ using a right triangle.
If $\cot\theta = 1/\sqrt{3}$ , find $\sin\theta, \cos\theta, \tan\theta$ .
Find the hypotenuse when $\sin\theta = 4/5$ and opp $= 12$ .
If $\sec\theta = 13/12$ , find $\sin\theta + \tan\theta$ .

Pitfalls / Insight

The terms opposite/adjacent are relative to the chosen angle. Switching to the other acute angle swaps them.
Hypotenuse is always longest , it's never the opposite or adjacent.
All ratios are positive for acute angles. Don't worry about signs in Class X.

Insight. A trigonometric ratio is a number , the ratio of two sides. It encodes the shape of the right triangle, not its size. That's why it depends only on the angle.