Right Triangle Trigonometry
Right Triangle Trigonometry
Right triangle trigonometry defines the sine, cosine, and tangent of an acute angle as ratios of sides in a right triangle. SOH-CAH-TOA — sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, tangent equals opposite over adjacent — is the mnemonic that has helped generations of students remember these ratios.
These three ratios, along with their reciprocals (cosecant, secant, cotangent), let you calculate unknown sides and angles in any right triangle, making them indispensable in surveying, navigation, construction, and physics.
This lesson teaches you to set up and solve right-triangle problems using the six trigonometric ratios.
The six trigonometric functions are defined as ratios of sides in a right triangle.
For an acute angle $\theta$ in a right triangle with sides labeled opposite ($O$), adjacent ($A$), and hypotenuse ($H$):
| Function | Ratio | Reciprocal |
|---|---|---|
| $\sin\theta$ | $\dfrac{O}{H}$ | $\csc\theta = \dfrac{H}{O}$ |
| $\cos\theta$ | $\dfrac{A}{H}$ | $\sec\theta = \dfrac{H}{A}$ |
| $\tan\theta$ | $\dfrac{O}{A}$ | $\cot\theta = \dfrac{A}{O}$ |
Sine = Opposite / Hypotenuse • Cosine = Adjacent / Hypotenuse • Tangent = Opposite / Adjacent
Special Right Triangles
Sides are in ratio $1 : 1 : \sqrt{2}$.
| $\sin$ | $\cos$ | $\tan$ | |
|---|---|---|---|
| $45°$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ |
Sides are in ratio $1 : \sqrt{3} : 2$.
| $\sin$ | $\cos$ | $\tan$ | |
|---|---|---|---|
| $30°$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$ |
| $60°$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
In a right triangle, the side opposite angle $\theta$ is 5 and the hypotenuse is 13. Find all six trig functions.
Adjacent $= \sqrt{13^2 - 5^2} = \sqrt{144} = 12$.
| $\sin\theta = \frac{5}{13}$ | $\csc\theta = \frac{13}{5}$ |
| $\cos\theta = \frac{12}{13}$ | $\sec\theta = \frac{13}{12}$ |
| $\tan\theta = \frac{5}{12}$ | $\cot\theta = \frac{12}{5}$ |
Find the exact value of $\sin 60° + \cos 30°$.
$$\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$$
Solving Right Triangles
A ladder 20 ft long leans against a wall making a $65°$ angle with the ground. How high up the wall does it reach?
$$\sin 65° = \frac{h}{20} \implies h = 20\sin 65° \approx 20(0.9063) \approx 18.1 \text{ ft}$$
From a point 50 m from the base of a building, the angle of elevation to the top is $40°$. Find the height.
$$\tan 40° = \frac{h}{50} \implies h = 50\tan 40° \approx 50(0.8391) \approx 42.0 \text{ m}$$
Practice Problems
Show Answer Key
1. adj $= 15$; $\sin\theta = \frac{8}{17}$, $\cos\theta = \frac{15}{17}$, $\tan\theta = \frac{8}{15}$
2. $1 + \frac{1}{2} = \frac{3}{2}$
3. $\frac{1}{2} \cdot 1 = \frac{1}{2}$
4. $15\sin 35° \approx 8.60$
5. hyp $= 25$; $\sin\theta = \frac{24}{25}$
6. $h = 100\sin 55° \approx 81.9$ m
7. $1$ (Pythagorean identity)
8. $\sec 60° = 2$; $\csc 45° = \sqrt{2}$
9. $\tan\theta = \frac{3}{12} = 0.25$; $\theta \approx 14.0°$
10. $\frac{\sqrt{3}/2}{1/2} = \sqrt{3} = \tan 60°$ ✓
11. $\tan\theta = \frac{9}{12} = 0.75$; $\theta \approx 36.9°$
12. $h = 25\tan 50° \approx 29.8$ ft