Trigonometry in right triangles

Three ratios that depend only on the angle

Take a right triangle and one of its acute angles, $\alpha$. Name the sides relative to that angle: the hypotenuse is opposite the right angle and is the longest side; the opposite leg faces $\alpha$; the adjacent leg touches $\alpha$ and is not the hypotenuse. Then

$$\sin\alpha = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\alpha = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\alpha = \frac{\text{opposite}}{\text{adjacent}}$$

These ratios depend only on the angle, not on the size of the triangle: any two right triangles with the same acute angle are similar, so their sides are in the same proportions. That is why a single number, $\sin 30° = \frac{1}{2}$, holds for every right triangle with a 30° angle.

Two consequences follow at once. Because the hypotenuse is the longest side, the sine and cosine of an acute angle lie strictly between 0 and 1, while the tangent can be any positive number. And $\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$, since dividing the two ratios cancels the hypotenuse.

Changing the point of view

The two acute angles of a right triangle add up to 90°. The leg opposite one is adjacent to the other, so

$$\sin(90° - \alpha) = \cos\alpha \qquad \cos(90° - \alpha) = \sin\alpha$$

When a question gives the "other" angle, either switch to it or switch the ratio; both work. The label "opposite" or "adjacent" always belongs to a particular angle, so name the angle first.

The exact values to know

The test does not allow a calculator, so questions use angles whose ratios are exact. They come from two triangles. Half a square, cut along its diagonal, is a 45°-45°-90° triangle with sides $1, 1, \sqrt{2}$. Half an equilateral triangle of side 2, cut along a height, is a 30°-60°-90° triangle with sides $1, \sqrt{3}, 2$: the short leg faces 30°, the long leg faces 60°.

Angle	Radians	sin	cos	tan
$30°$	$\frac{\pi}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
$45°$	$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$60°$	$\frac{\pi}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

If you forget a value, sketch the small triangle and read the ratio from it. A useful pattern: sine grows from 30° to 60° while cosine shrinks, and $\sin 30° = \cos 60°$, $\sin 60° = \cos 30°$, as the rule for complementary angles says.

Degrees and radians

A full turn is $360°$ or $2\pi$ radians, so $180° = \pi$. To convert degrees to radians multiply by $\frac{\pi}{180}$; to go back multiply by $\frac{180}{\pi}$. So $120° = \frac{2\pi}{3}$ and $\frac{5\pi}{4} = 225°$. Answer options may mix the two units; convert before comparing.

Solving a right triangle

"Solving" means finding every side and angle from what is given. Two situations cover almost everything:

• One side and one acute angle. Pick the ratio that links the known side with the side you want. If the hypotenuse $c$ is known, the leg opposite $\alpha$ is $c\sin\alpha$ and the adjacent leg is $c\cos\alpha$. If a leg is known, use the tangent between the two legs, or divide by the sine or cosine to get the hypotenuse.
• Two sides. Find the third with the Pythagorean theorem, and the angle from the ratio of the two sides you know. With a special ratio such as $\frac{1}{2}$ or $\sqrt{3}$, the angle is read from the table; this is what the inverse functions $\arcsin$, $\arccos$ and $\arctan$ do, so $\arctan\sqrt{3} = 60°$.

Where these triangles hide

Many questions do not say "right triangle". An angle of elevation or depression, the slope of a ramp, a ladder against a wall, a shadow on flat ground: each forms a right angle with the horizontal. In coordinate geometry, a line of slope $m$ makes an angle $\theta$ with the $x$-axis where $\tan\theta = m$: the slope is rise over run, opposite over adjacent. A line of slope $\sqrt{3}$ rises at 60°. The area of any triangle with two sides $a$, $b$ and the angle $\gamma$ between them is $\frac{1}{2}ab\sin\gamma$, because $b\sin\gamma$ is the height on side $a$.