Skills

Right-Triangle Trigonometry

In a right triangle, name the legs from the angle θ : opposite and adjacent.

Practice 1: Procedural and Symbolic Fluency

Part 1 · Hook

Why this matters

A surveyor cannot climb a 30-meter tree to measure it, and a builder cannot guess how steep a ramp is. Both stand back, measure one angle and one length, and let a ratio of sides do the rest. Those ratios, sine, cosine and tangent, are the starting point for every periodic function in this unit.

Part 2 · Prerequisite check

Quick check before you start

1. A line rises 6 units while it runs 8 units to the right. What is its slope?

$\frac{3}{4}$
$\frac{4}{3}$
$14$
$48$

Show the answer

Slope is rise over run: $\frac{6}{8} = \frac{3}{4}$ .

Correct: $\frac{3}{4}$ :
$\frac{4}{3}$ :
$14$ :
$48$ :

2. Solve $x^2 = 6^2 + 8^2$ for $x > 0$ .

$10$
$14$
$100$
$\sqrt{14}$

Show the answer

$x^2 = 36 + 64 = 100$ , so $x = 10$ .

Correct: $10$ :
$14$ :
$100$ :
$\sqrt{14}$ :

Part 3 · See it

See it first

A right triangle with the right angle at the lower right corner and the angle theta at the lower left corner. The longest side, across from the right angle, is the hypotenuse. The vertical side across from theta is the side opposite theta. The horizontal side that touches theta and is not the hypotenuse is the side adjacent to theta. To the right: sine of theta is opposite over hypotenuse, cosine of theta is adjacent over hypotenuse, tangent of theta is opposite over adjacent (SOH CAH TOA). — The names of the two legs depend on which acute angle you stand at; the hypotenuse is always across from the right angle. LevlPrep original diagram.

Part 4 · Step by step

How it works, step by step

Stand at an acute angle $\theta$ of a right trianglethe leg across from $\theta$ is the opposite side, the leg touching $\theta$ is the adjacent side, and the side across from the right angle is the hypotenuse
Right triangles with the same angle $\theta$ are similarthe ratios of their sides are the same, whatever the triangle's size, so each ratio depends only on $\theta$
Name the three ratios $\sin\theta$ $= \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta$ $= \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta$ $= \frac{\text{opposite}}{\text{adjacent}}$ (SOH CAH TOA)
You know one side and the anglepick the ratio that links the known side to the unknown one, write the equation, and solve it

Part 5 · Key ideas

Key ideas

The hypotenuse is the longest side, so $\sin\theta$ and $\cos\theta$ are between $0$ and $1$ for an acute angle.
$\tan\theta$ $= \frac{\text{opposite}}{\text{adjacent}}$ is rise over run: the slope of the hypotenuse when the adjacent side lies flat.
To find a side: write the ratio, put in the known values, then multiply (unknown on top) or divide (unknown on the bottom).

Part 6 · Misconception

A common mistake

The wrong idea: The opposite side is always the vertical side and the adjacent side is always the horizontal one.

What actually happens: Opposite and adjacent are named from the angle you use. Switch to the other acute angle and the two legs swap names; only the hypotenuse stays the same.

Part 7 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. A right triangle has legs of length $8$ and $15$ and hypotenuse $17$ . The angle $B$ is across from the leg of length $8$ . What is $\sin B$ ?

$\frac{8}{17}$
$\frac{15}{17}$
$\frac{8}{15}$
$\frac{17}{8}$

Show the answer

Sine is opposite over hypotenuse. From $B$ the opposite side is $8$ , so $\sin B = \frac{8}{17}$ .

Correct: $\frac{8}{17}$ : Opposite $8$ over hypotenuse $17$ .
$\frac{15}{17}$ : $\frac{15}{17}$ is adjacent over hypotenuse, which is $\cos B$ .
$\frac{8}{15}$ : $\frac{8}{15}$ is opposite over adjacent, which is $\tan B$ .
$\frac{17}{8}$ : This is the ratio upside down; the hypotenuse goes on the bottom.

2. In the same triangle (legs $8$ and $15$ , hypotenuse $17$ ), the other acute angle is $C$ , across from the leg of length $15$ . What is $\tan C$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From $C$ : opposite $= 15$ , adjacent $= 8$ , so $\tan C = \frac{15}{8}$ .

Answer: $\frac{15}{8}$ ≈ 1.875

3. A kite string 40 meters long makes an angle of $55^\circ$ with the level ground. Use $\sin 55^\circ \approx 0.8192$ to find the kite's height above the hand holding the string, in meters.

No calculator. Type the exact value (a fraction, π, e or a root is fine), in meters.

Show the answer

$\sin 55^\circ = \frac{h}{40}$ , so $h = 40(0.8192) = 32.768$ meters.

Answer: 32.768 meters

4. A right triangle has hypotenuse $8$ and an acute angle $\theta$ . Write the length of the side adjacent to $\theta$ in terms of $\theta$ .

Type an expression for $\text{adjacent} =$ .

Show the answer

$\cos\theta = \frac{\text{adjacent}}{8}$ , so adjacent $= 8\cos\theta$ .

Answer: $\text{adjacent} =$ $8\,\cos(\theta)$

5. A wheelchair ramp rises 2 feet over a horizontal distance of 24 feet. Let $\theta$ be the angle the ramp makes with the ground. What is $\tan\theta$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The rise is opposite $\theta$ and the horizontal run is adjacent: $\tan\theta = \frac{2}{24} = \frac{1}{12}$ , the slope of the ramp.

Answer: $\frac{1}{12}$ ≈ 0.083

Part 8 · Summary

Summary

In a right triangle, name the legs from the angle

\theta

: opposite and adjacent. Then

\sin\theta

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

\cos\theta

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

and

\tan\theta

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

. The ratios depend only on the angle, so one angle and one side are enough to find the other sides.

Part 9 · Up next

What comes next

Part 10 · Connections

Connections

Next: Circles, Arcs and Special Triangles

Question set for this topic