Right Triangle Trigonometry
Discover the ratios that unlock triangle measurements
If trigonometry sounds intimidating to you, know that you are not alone. For many people, the word conjures up images of confusing symbols and mysterious Greek letters. But here is the thing: trigonometry is really just about one simple idea: ratios. And ratios are something you already understand. When you say “the building is twice as tall as that tree,” you are comparing two measurements. When a recipe says “use 3 parts flour to 1 part sugar,” you are using ratios. Trigonometry just applies this familiar idea to triangles in a precise, powerful way.
The ancient Greeks, Egyptians, and Babylonians figured out something remarkable: in any right triangle, if you know one of the acute angles, the ratios between the sides are always the same, no matter how big or small the triangle is. A tiny right triangle with a 30-degree angle has the same ratios as a massive one with that same angle. This discovery allows us to measure things we cannot reach directly: the height of a mountain, the distance to a star, the depth of a canyon. All from standing safely on the ground with just an angle and a single measurement.
Core Concepts
Setting Up a Right Triangle
Before we dive into the trigonometric ratios, we need to establish some vocabulary. In a right triangle, we have three sides and three angles. One angle is always 90 degrees (the right angle). The other two angles are acute (less than 90 degrees), and together they add up to 90 degrees.
Here is how we label the sides relative to a particular acute angle, which we often call $\theta$ (the Greek letter “theta”):
- Hypotenuse: The longest side, always opposite the right angle. This one never changes regardless of which acute angle you are looking at.
- Opposite: The side directly across from the angle $\theta$. If you were standing at the angle and looking straight ahead, this is the side you would be facing.
- Adjacent: The side next to the angle $\theta$ that is not the hypotenuse. It forms one of the “arms” of the angle along with the hypotenuse.
The key insight here is that “opposite” and “adjacent” depend on which angle you are focused on. If you switch your attention to the other acute angle, the opposite and adjacent sides swap roles. The hypotenuse, however, always stays the same.
The Three Main Ratios: Sine, Cosine, and Tangent
Now for the heart of trigonometry. For any acute angle $\theta$ in a right triangle, we define three fundamental ratios:
$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$$
$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$
$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$$
That is it. These three ratios are the foundation of trigonometry. Everything else builds on them.
What makes these ratios special? Here is the amazing part: these ratios depend only on the angle, not on the size of the triangle. Every right triangle with a 30-degree angle has the same sine, cosine, and tangent for that angle, whether the triangle fits on a postage stamp or covers a football field. This is because all right triangles with the same acute angle are similar (same shape, different size), and similar triangles have proportional sides.
SOH-CAH-TOA: Your Memory Lifeline
Generations of students have remembered these ratios using the mnemonic SOH-CAH-TOA:
- SOH: Sine = Opposite / Hypotenuse
- CAH: Cosine = Adjacent / Hypotenuse
- TOA: Tangent = Opposite / Adjacent
Say it out loud a few times: “soh-cah-toh-ah.” It sounds a bit like a Native American word, which might help it stick. Some people remember it as “Some Old Horse Caught Another Horse Taking Oats Away” or other creative sentences, but the simple SOH-CAH-TOA works well for most people.
Finding Trig Ratios from a Triangle
When you have a right triangle with known side lengths, finding the trig ratios is straightforward. You just need to identify which sides are opposite, adjacent, and the hypotenuse relative to the angle you care about, then plug into the definitions.
Let us say you have a right triangle where one leg is 3 units, the other leg is 4 units, and the hypotenuse is 5 units. If angle $A$ is opposite the side of length 3, then:
- Opposite to $A$ = 3
- Adjacent to $A$ = 4
- Hypotenuse = 5
So: $\sin A = \frac{3}{5}$, $\cos A = \frac{4}{5}$, $\tan A = \frac{3}{4}$
Notice that these are just fractions (or ratios, hence the name “trigonometric ratios”). There is nothing mysterious about them. You are simply comparing side lengths.
Using Trig Ratios to Find Missing Sides
This is where trigonometry becomes truly useful. If you know one side of a right triangle and one acute angle, you can find the other sides.
The process is:
- Identify what you know (an angle and a side)
- Identify what you want to find
- Determine which trig ratio connects the known side to the unknown side
- Set up the equation and solve
For instance, if you know angle $\theta = 30°$ and the hypotenuse is 10, and you want the opposite side:
$$\sin 30° = \frac{\text{opposite}}{10}$$
Since $\sin 30° = 0.5$:
$$0.5 = \frac{\text{opposite}}{10}$$
$$\text{opposite} = 10 \times 0.5 = 5$$
Using Trig Ratios to Find Missing Angles
You can also work backwards. If you know two sides of a right triangle, you can find the angles using inverse trig functions.
The inverse functions are written as $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ (also called arcsin, arccos, and arctan). They “undo” the original functions:
- If $\sin \theta = x$, then $\theta = \sin^{-1}(x)$
- If $\cos \theta = x$, then $\theta = \cos^{-1}(x)$
- If $\tan \theta = x$, then $\theta = \tan^{-1}(x)$
For example, if the opposite side is 4 and the adjacent side is 4, then:
$$\tan \theta = \frac{4}{4} = 1$$
$$\theta = \tan^{-1}(1) = 45°$$
Your calculator has buttons for these inverse functions (sometimes labeled “sin$^{-1}$” or “arcsin” or “asin”). Make sure your calculator is in degree mode if you want your answer in degrees.
The Reciprocal Ratios: Cosecant, Secant, and Cotangent
In addition to the three main ratios, there are three more that are simply the reciprocals (read: flipped versions) of sine, cosine, and tangent:
$$\csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Opposite}}$$
$$\sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}}$$
$$\cot \theta = \frac{1}{\tan \theta} = \frac{\text{Adjacent}}{\text{Opposite}}$$
Why do we need these? Honestly, for basic right triangle problems, you do not strictly need them. But they become very useful in more advanced trigonometry, calculus, and when simplifying certain expressions. Think of them as convenient shortcuts that save you from writing “$\frac{1}{\sin \theta}$” all the time.
A helpful memory trick: notice that each reciprocal function pairs with the function that does not share its starting letter. Cosecant goes with sine. Secant goes with cosine. Cotangent goes with tangent. The “co” prefix creates a kind of partnership.
Special Right Triangles: 45-45-90
Some triangles show up so often that their trig ratios are worth memorizing. The first is the 45-45-90 triangle, also called an isosceles right triangle.
Start with a square with side length 1. Cut it diagonally, and you get two 45-45-90 triangles. Each has:
- Two legs of length 1
- A hypotenuse of length $\sqrt{2}$ (by the Pythagorean theorem: $1^2 + 1^2 = 2$)
From this, we can calculate the exact trig values for 45 degrees:
$$\sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\tan 45° = \frac{1}{1} = 1$$
Notice that sine and cosine are equal at 45 degrees. This makes sense because in a 45-45-90 triangle, the two legs are equal, so the opposite and adjacent sides are the same length.
Special Right Triangles: 30-60-90
The other special triangle is the 30-60-90 triangle. You can create one by cutting an equilateral triangle in half.
Start with an equilateral triangle with side length 2. Draw the altitude from one vertex to the opposite side. This altitude bisects both the vertex angle (creating 30-degree angles) and the opposite side. You now have a 30-60-90 triangle with:
- Short leg (opposite 30 degrees) = 1
- Long leg (opposite 60 degrees) = $\sqrt{3}$
- Hypotenuse (opposite 90 degrees) = 2
From this, we get exact values:
For 30 degrees: $$\sin 30° = \frac{1}{2}$$ $$\cos 30° = \frac{\sqrt{3}}{2}$$ $$\tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$
For 60 degrees: $$\sin 60° = \frac{\sqrt{3}}{2}$$ $$\cos 60° = \frac{1}{2}$$ $$\tan 60° = \frac{\sqrt{3}}{1} = \sqrt{3}$$
Notice a beautiful pattern: $\sin 30° = \cos 60°$ and $\cos 30° = \sin 60°$. This is not a coincidence! The sine of any angle equals the cosine of its complement (and vice versa). Since 30 and 60 are complementary angles (they add to 90), their sines and cosines swap.
Exact Values vs. Calculator Approximations
When working with special angles like 30, 45, and 60 degrees, you have a choice: use exact values (with square roots) or decimal approximations.
Exact values like $\frac{\sqrt{3}}{2}$ are mathematically precise. They are preferred in theoretical work, proofs, and when you need to simplify expressions further.
Calculator approximations like 0.866 (which is $\frac{\sqrt{3}}{2}$ rounded) are useful for practical calculations, like determining the actual height of a building.
Both are correct in their contexts. If a problem asks for “exact” values, use the radical form. If it asks you to “find” or “calculate” a measurement, decimals are usually fine. When in doubt, give the exact value and then say “which is approximately…” followed by the decimal.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| $\sin \theta$ | Opposite / Hypotenuse | $\sin 30° = \frac{1}{2}$ |
| $\cos \theta$ | Adjacent / Hypotenuse | $\cos 60° = \frac{1}{2}$ |
| $\tan \theta$ | Opposite / Adjacent | $\tan 45° = 1$ |
| $\csc \theta$ | Hypotenuse / Opposite (reciprocal of sin) | $\csc 30° = 2$ |
| $\sec \theta$ | Hypotenuse / Adjacent (reciprocal of cos) | $\sec 60° = 2$ |
| $\cot \theta$ | Adjacent / Opposite (reciprocal of tan) | $\cot 45° = 1$ |
| Hypotenuse | Longest side, opposite the right angle | In a 3-4-5 triangle, it is 5 |
| Opposite | Side across from the angle of interest | Depends on which angle you choose |
| Adjacent | Side next to the angle (not the hypotenuse) | Depends on which angle you choose |
| $\theta$ | Greek letter “theta,” commonly used for angles | $\theta = 45°$ |
| $\sin^{-1}$, arcsin | Inverse sine function | $\sin^{-1}(0.5) = 30°$ |
| $\cos^{-1}$, arccos | Inverse cosine function | $\cos^{-1}(0.5) = 60°$ |
| $\tan^{-1}$, arctan | Inverse tangent function | $\tan^{-1}(1) = 45°$ |
Examples
In a right triangle, the legs measure 3 and 4 units, and the hypotenuse measures 5 units. Find $\sin A$, $\cos A$, and $\tan A$, where angle $A$ is opposite the side of length 3.
Solution:
First, identify the sides relative to angle $A$:
- Opposite to $A$ = 3 (the side across from $A$)
- Adjacent to $A$ = 4 (the leg next to $A$)
- Hypotenuse = 5 (always the longest side, opposite the right angle)
Now apply the definitions:
$$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5}$$
$$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5}$$
$$\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}$$
These can also be written as decimals: $\sin A = 0.6$, $\cos A = 0.8$, $\tan A = 0.75$.
The 3-4-5 triangle is one of the most common Pythagorean triples, so you will see these ratios often.
Find $x$ if $\sin 30° = \frac{x}{10}$.
Solution:
We know that $\sin 30° = \frac{1}{2}$ (this is one of the special values worth memorizing).
So we have: $$\frac{1}{2} = \frac{x}{10}$$
Multiply both sides by 10: $$x = 10 \times \frac{1}{2} = 5$$
Interpretation: In a right triangle where the hypotenuse is 10 and one angle is 30 degrees, the side opposite that 30-degree angle is 5.
This makes sense! In a 30-60-90 triangle, the side opposite 30 degrees is always half the hypotenuse.
You stand 50 feet away from the base of a tree. You measure the angle from the ground to the top of the tree and find it is 35 degrees. How tall is the tree?
Solution:
Draw the situation: you, the tree, and the ground form a right triangle. The right angle is where the tree meets the ground. The angle at your position is 35 degrees.
- The 50 feet (your distance to the tree) is the side adjacent to the 35-degree angle.
- The height of the tree is the side opposite the 35-degree angle.
- We want to find the opposite side.
Which trig ratio connects opposite and adjacent? That is tangent!
$$\tan 35° = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height}}{50}$$
Use a calculator to find $\tan 35° \approx 0.7002$:
$$0.7002 = \frac{\text{height}}{50}$$
$$\text{height} = 50 \times 0.7002 \approx 35.01 \text{ feet}$$
The tree is approximately 35 feet tall.
Note: This is called an “angle of elevation” problem because you are looking up from a horizontal line. The angle between the horizontal and your line of sight to the top of the tree is the angle of elevation.
In a right triangle with legs 5 and 12 and hypotenuse 13, find all six trigonometric ratios for the angle $\theta$ opposite the side of length 5.
Solution:
First, let us verify this is indeed a right triangle: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Yes, it is!
For angle $\theta$:
- Opposite = 5
- Adjacent = 12
- Hypotenuse = 13
The three main ratios:
$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}$$
$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{12}{13}$$
$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{12}$$
The three reciprocal ratios:
$$\csc \theta = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{13}{5}$$
$$\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{13}{12}$$
$$\cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{12}{5}$$
Verification: Notice that $\csc \theta = \frac{1}{\sin \theta}$: we have $\frac{13}{5} = \frac{1}{5/13}$. Check! The same relationship holds for the other pairs.
Find the exact values (no decimals) of $\sin 45°$, $\cos 45°$, and $\tan 60°$.
Solution:
For 45 degrees, we use the 45-45-90 triangle with legs of 1 and hypotenuse of $\sqrt{2}$:
$$\sin 45° = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$$
To rationalize (eliminate the square root from the denominator): $$\sin 45° = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\cos 45° = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
(Since the two legs are equal in a 45-45-90 triangle, sine and cosine are the same.)
For 60 degrees, we use the 30-60-90 triangle. Remember, if the short leg (opposite 30 degrees) is 1, then the long leg (opposite 60 degrees) is $\sqrt{3}$, and the hypotenuse is 2.
For the 60-degree angle:
- Opposite = $\sqrt{3}$
- Adjacent = 1
- Hypotenuse = 2
$$\tan 60° = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$
Summary of exact values:
- $\sin 45° = \frac{\sqrt{2}}{2}$ (approximately 0.707)
- $\cos 45° = \frac{\sqrt{2}}{2}$ (approximately 0.707)
- $\tan 60° = \sqrt{3}$ (approximately 1.732)
In right triangle $ABC$ with the right angle at $C$, angle $A = 37°$ and side $c$ (the hypotenuse) $= 15$. Find all missing parts of the triangle: angle $B$ and sides $a$ and $b$.
Solution:
Step 1: Find angle $B$.
In any triangle, the angles sum to 180 degrees. In a right triangle, the two acute angles sum to 90 degrees.
$$B = 90° - A = 90° - 37° = 53°$$
Step 2: Find side $a$ (opposite angle $A$).
We know angle $A$ and the hypotenuse $c$. The side opposite $A$ is $a$.
$$\sin A = \frac{a}{c}$$ $$\sin 37° = \frac{a}{15}$$
Using a calculator: $\sin 37° \approx 0.6018$
$$0.6018 = \frac{a}{15}$$ $$a = 15 \times 0.6018 \approx 9.03$$
Step 3: Find side $b$ (adjacent to angle $A$).
$$\cos A = \frac{b}{c}$$ $$\cos 37° = \frac{b}{15}$$
Using a calculator: $\cos 37° \approx 0.7986$
$$0.7986 = \frac{b}{15}$$ $$b = 15 \times 0.7986 \approx 11.98$$
Step 4: Verify with the Pythagorean theorem.
$$a^2 + b^2 = c^2$$ $$9.03^2 + 11.98^2 = 81.54 + 143.52 = 225.06$$ $$c^2 = 15^2 = 225$$
The small difference (0.06) is due to rounding. Our answers check out!
Final answer:
- Angle $B = 53°$
- Side $a \approx 9.03$
- Side $b \approx 11.98$
Key Properties and Rules
The Fundamental Trig Ratios (SOH-CAH-TOA)
$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \qquad \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \qquad \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$$
The Reciprocal Ratios
$$\csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Opposite}}$$
$$\sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}}$$
$$\cot \theta = \frac{1}{\tan \theta} = \frac{\text{Adjacent}}{\text{Opposite}}$$
Tangent in Terms of Sine and Cosine
$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
This is because $\frac{\sin \theta}{\cos \theta} = \frac{\text{Opp}/\text{Hyp}}{\text{Adj}/\text{Hyp}} = \frac{\text{Opp}}{\text{Adj}} = \tan \theta$.
Cofunction Identities
For complementary angles (angles that add to 90 degrees):
$$\sin \theta = \cos(90° - \theta)$$ $$\cos \theta = \sin(90° - \theta)$$ $$\tan \theta = \cot(90° - \theta)$$
This is why “cosine” has “co” in the name: it is the “complement’s sine.”
Special Angle Values (Memorize These!)
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |
| $45°$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
The Pythagorean Identity
For any angle $\theta$:
$$\sin^2 \theta + \cos^2 \theta = 1$$
This comes directly from the Pythagorean theorem! If the hypotenuse is 1, then $\text{opp}^2 + \text{adj}^2 = 1$, which is the same as $\sin^2 \theta + \cos^2 \theta = 1$.
Range of Values
For acute angles in a right triangle:
- $0 < \sin \theta < 1$
- $0 < \cos \theta < 1$
- $0 < \tan \theta < \infty$ (tangent can be any positive number)
Real-World Applications
Surveying and Measuring Heights
Surveyors use trigonometry constantly. To measure the height of a building, cliff, or any tall structure:
- Stand a known distance from the base.
- Measure the angle of elevation to the top.
- Use tangent: $\text{height} = \text{distance} \times \tan(\text{angle})$.
This same technique allowed ancient Greeks to estimate the heights of mountains and even the circumference of the Earth.
Navigation and Bearing
Pilots and sailors use trigonometry to calculate distances and bearings. If a ship sails 100 miles on a bearing of 30 degrees east of north, how far north and how far east has it traveled?
- Northward distance = $100 \times \cos 30° = 100 \times \frac{\sqrt{3}}{2} \approx 86.6$ miles
- Eastward distance = $100 \times \sin 30° = 100 \times \frac{1}{2} = 50$ miles
This decomposition of motion into perpendicular components is fundamental to navigation, physics, and engineering.
Ramp and Incline Calculations
When designing wheelchair ramps, loading docks, or driveways, engineers need to calculate angles and lengths. The Americans with Disabilities Act specifies that ramps should have a slope of no more than 1:12 (1 inch of rise for every 12 inches of run).
What angle does this represent? $$\tan \theta = \frac{1}{12}$$ $$\theta = \tan^{-1}\left(\frac{1}{12}\right) \approx 4.8°$$
Roof Pitch in Construction
Roof pitch is often expressed as a ratio (like “6:12” meaning the roof rises 6 inches for every 12 inches of horizontal run). To find the angle:
$$\tan \theta = \frac{6}{12} = 0.5$$ $$\theta = \tan^{-1}(0.5) \approx 26.6°$$
Carpenters and roofers use these calculations to cut rafters at the correct angles.
Shadow Problems
On a sunny day, objects cast shadows. If you know the angle of the sun above the horizon and the length of a shadow, you can find the height of the object:
A flagpole casts a 25-foot shadow when the sun is 55 degrees above the horizon. How tall is the flagpole?
$$\tan 55° = \frac{\text{height}}{25}$$ $$\text{height} = 25 \times \tan 55° \approx 25 \times 1.428 \approx 35.7 \text{ feet}$$
Ancient astronomers used shadow measurements to calculate the heights of pyramids and the tilt of Earth’s axis.
Self-Test Problems
Problem 1: In a right triangle, the side opposite angle $\theta$ is 8 and the hypotenuse is 17. Find $\sin \theta$, $\cos \theta$, and $\tan \theta$.
Show Answer
First, find the adjacent side using the Pythagorean theorem: $$\text{adj}^2 + 8^2 = 17^2$$ $$\text{adj}^2 = 289 - 64 = 225$$ $$\text{adj} = 15$$
So we have an 8-15-17 triangle.
$$\sin \theta = \frac{8}{17}$$ $$\cos \theta = \frac{15}{17}$$ $$\tan \theta = \frac{8}{15}$$
Problem 2: Find the exact value of $\sin 60° + \cos 60°$.
Show Answer
Using the special angle values: $$\sin 60° = \frac{\sqrt{3}}{2}$$ $$\cos 60° = \frac{1}{2}$$
$$\sin 60° + \cos 60° = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2}$$
As a decimal, this is approximately $\frac{1.732 + 1}{2} = \frac{2.732}{2} \approx 1.366$.
Problem 3: A ladder leans against a wall, making a 65-degree angle with the ground. If the foot of the ladder is 6 feet from the wall, how long is the ladder?
Show Answer
The 6 feet is adjacent to the 65-degree angle, and the ladder is the hypotenuse.
$$\cos 65° = \frac{6}{\text{ladder}}$$
$$\text{ladder} = \frac{6}{\cos 65°} = \frac{6}{0.4226} \approx 14.2 \text{ feet}$$
Problem 4: If $\tan \theta = 2$, find $\sin \theta$ and $\cos \theta$. (Hint: draw a triangle where opposite = 2 and adjacent = 1.)
Show Answer
If $\tan \theta = \frac{\text{opp}}{\text{adj}} = 2 = \frac{2}{1}$, we can use opposite = 2 and adjacent = 1.
Find the hypotenuse: $$\text{hyp} = \sqrt{2^2 + 1^2} = \sqrt{5}$$
Therefore: $$\sin \theta = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$ $$\cos \theta = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$
Problem 5: From the top of a 200-foot lighthouse, the angle of depression to a boat is 25 degrees. How far is the boat from the base of the lighthouse?
Show Answer
The angle of depression from the lighthouse equals the angle of elevation from the boat (alternate interior angles). So the boat sees the top of the lighthouse at a 25-degree angle.
The lighthouse height (200 ft) is opposite the 25-degree angle, and the distance to the boat is adjacent.
$$\tan 25° = \frac{200}{\text{distance}}$$
$$\text{distance} = \frac{200}{\tan 25°} = \frac{200}{0.4663} \approx 429 \text{ feet}$$
Problem 6: Find all six trigonometric ratios for a 45-degree angle. Give exact values.
Show Answer
Using the 45-45-90 triangle (legs = 1, hypotenuse = $\sqrt{2}$):
$$\sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\tan 45° = \frac{1}{1} = 1$$
For the reciprocals: $$\csc 45° = \frac{\sqrt{2}}{1} = \sqrt{2}$$ $$\sec 45° = \frac{\sqrt{2}}{1} = \sqrt{2}$$ $$\cot 45° = \frac{1}{1} = 1$$
Problem 7: Solve the right triangle where angle $A = 52°$ and the side opposite angle $A$ is 18. Find angle $B$, the adjacent side, and the hypotenuse.
Show Answer
Angle B: $$B = 90° - 52° = 38°$$
Hypotenuse (using sine): $$\sin 52° = \frac{18}{\text{hyp}}$$ $$\text{hyp} = \frac{18}{\sin 52°} = \frac{18}{0.788} \approx 22.8$$
Adjacent side (using tangent): $$\tan 52° = \frac{18}{\text{adj}}$$ $$\text{adj} = \frac{18}{\tan 52°} = \frac{18}{1.28} \approx 14.1$$
Check with Pythagorean theorem: $18^2 + 14.1^2 = 324 + 198.8 = 522.8$ $22.8^2 = 519.8$ (close enough, considering rounding)
Summary
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Trigonometry is about ratios. The three main trigonometric ratios (sine, cosine, tangent) compare the sides of a right triangle relative to an acute angle.
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SOH-CAH-TOA is your friend: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent.
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“Opposite” and “adjacent” are relative to the angle you are examining. The hypotenuse is always the same (the longest side, opposite the right angle).
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The trig ratios depend only on the angle, not on the size of the triangle. This is what makes them so powerful.
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To find a missing side: identify what you know and what you want, choose the appropriate trig ratio, set up the equation, and solve.
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To find a missing angle: use inverse trig functions ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$).
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The reciprocal ratios (cosecant, secant, cotangent) are simply the flipped versions of sine, cosine, and tangent.
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Special angles (30, 45, 60 degrees) have exact values that come from the 45-45-90 and 30-60-90 triangles. Memorize these!
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Exact values use radicals ($\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$, etc.) and are mathematically precise. Calculator approximations are useful for practical measurements.
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Real-world applications are everywhere: measuring heights, navigation, construction, ramps, and shadow problems.
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The Pythagorean identity ($\sin^2 \theta + \cos^2 \theta = 1$) connects sine and cosine directly to the Pythagorean theorem.
Once you understand right triangle trigonometry, you have the foundation for everything that comes next: the unit circle, graphing trig functions, identities, and applications in calculus and physics. This ancient tool is as practical today as it was thousands of years ago when astronomers first used it to map the stars.