Right Triangles and Trigonometry

The Pythagorean theorem

In a right triangle, the two legs and the hypotenuse (the side opposite the right angle) satisfy $a^2 + b^2 = c^2$, where $c$ is the hypotenuse. If you know any two sides, you can find the third:

• hypotenuse: $c = \sqrt{a^2 + b^2}$
• missing leg: $a = \sqrt{c^2 - b^2}$

Worked example: the hypotenuse

A right triangle has legs of $5$ and $12$, shown below. Find the hypotenuse.

Apply the Pythagorean theorem directly: $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$.

The ACT loves a few whole-number triples, so learn to spot them. When two sides match one of these (or a multiple of one), you can skip the calculation entirely:

Multiples count: legs of $9$ and $12$ are $3(3, 4)$, so the hypotenuse is $3(5) = 15$. A hypotenuse of $40$ with a leg of $24$ is $8(5, 3)$, so the other leg is $8(4) = 32$.

Sine, cosine, and tangent

For an acute angle $\theta$ in a right triangle, the three basic ratios compare sides, named relative to $\theta$:

• $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$
• $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
• $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$

The memory aid SOH-CAH-TOA holds all three. The hypotenuse is always the longest side, and "opposite" and "adjacent" change depending on which angle you are working from. Suppose a right triangle has legs of $9$ and $12$. The hypotenuse is $15$. For the acute angle whose adjacent leg is $9$, $\cos\theta = \frac{9}{15}$. A frequent slip is to use the opposite side for cosine, or to invert tangent, so name the sides out loud before you write the ratio.

When a problem gives only two sides, find the third with the Pythagorean theorem before computing a ratio. For example, with an opposite side of $8$ and a hypotenuse of $17$, the adjacent side is $\sqrt{17^2 - 8^2} = \sqrt{225} = 15$.

Two special right triangles

The $45$-$45$-$90$ triangle has two equal legs, and the hypotenuse is a leg times $\sqrt{2}$. A leg of $6$ gives a hypotenuse of $6\sqrt{2}$. (The same triangle is half of a square cut along its diagonal.)

The $30$-$60$-$90$ triangle has a hypotenuse that is twice the short leg, and the long leg is the short leg times $\sqrt{3}$. The short leg is opposite the $30^\circ$ angle, and the long leg is opposite the $60^\circ$ angle. With a hypotenuse of $20$, the short leg is $10$ and the long leg is $10\sqrt{3}$.

The classic mix-up is applying the $\sqrt{2}$ rule to a $30$-$60$-$90$ triangle, where the hypotenuse is simply doubled. Check which special triangle you have before picking a rule.

The unit circle

The unit circle assigns a point $(\cos\theta, \sin\theta)$ to every angle $\theta$. At the four axis angles, the coordinates are exact: $(1, 0)$ at $0^\circ$, $(0, 1)$ at $90^\circ$, $(-1, 0)$ at $180^\circ$, and $(0, -1)$ at $270^\circ$. The standard angles $30^\circ$, $45^\circ$, and $60^\circ$ produce exact values worth knowing without a calculator:

You can rebuild this table from the special triangles instead of memorizing it. The sine of $30^\circ$ is the short leg over the hypotenuse, $\frac{1}{2}$, and the cosine of $60^\circ$ is the same ratio because the sides swap roles for the other acute angle. Notice that $\sin$ and $\cos$ swap values at $30^\circ$ and $60^\circ$.

Worked example: tan(30°)

Find $\tan(30^\circ)$.

Tangent is sine divided by cosine at that angle. Since $\sin(30^\circ) = \frac{1}{2}$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$:

$\tan(30^\circ) = \frac{1}{2} \div \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Because $\tan\theta = \frac{\sin\theta}{\cos\theta}$, any tangent value comes from the unit-circle point. For the point $(x, y)$ on the circle, $\tan\theta = \frac{y}{x}$.

Where this appears as a model

Real objects with only some measurements given are right triangles in disguise. A $13$-foot ladder leaning against a wall with its base $5$ feet from the wall reaches a height of $12$ feet. The ladder, the wall, and the ground make a right triangle with the ladder as hypotenuse, so $\sqrt{13^2 - 5^2} = 12$. A diagonal across a rectangular field, a ramp, and a wire attached to a pole all work the same way.

This chapter covers the right-triangle and exact-value foundation. Trigonometric identities and triangles without a right angle build directly on the sine, cosine, and tangent relationships taught here, and they are developed in their own chapter later in this Part.

Strategy checklist

1. Draw the triangle and label the sides relative to the angle in question: opposite, adjacent, hypotenuse.
2. Look for a triple ($3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$, or a multiple) before squaring anything.
3. If the angles are $45^\circ$ or $30^\circ$ and $60^\circ$, use the special-triangle ratios instead of a calculator.
4. Rewrite tangent as $\frac{\sin\theta}{\cos\theta}$ when only the unit-circle point is known.
5. Sanity-check: the hypotenuse is the longest side, and sine and cosine of an acute angle are between $0$ and $1$.