🎯⭐ INTERACTIVE LESSON

Trigonometric Ratios

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Trigonometric Ratios - Complete Interactive Lesson

Part 1: Labeling a Right Triangle (Opposite/Adjacent/Hypotenuse)

📐 Trigonometric Ratios

Part 1 of 5 — Labeling a Right Triangle

Topics in This Part

Section
Opposite, Adjacent, and Hypotenuse
Why Labels Depend on the Angle
Practice Identifying Sides

🔑 Key Concept: Trigonometry connects the angles of a right triangle to the ratios of its sides. Before we can write any ratio, we must label the sides correctly — and the labels change depending on which acute angle we're looking at.

Naming the Three Sides

Every right triangle has one $90^\circ$ angle. Pick one of the acute angles (call it $\theta$ ). Relative to $\theta$ , the three sides get names:

Hypotenuse — the longest side, always across from the right angle. It never changes.
Opposite — the side that does not touch $\theta$ ; it sits directly across from it.
Adjacent — the side (other than the hypotenuse) that does touch $\theta$ .

Side	How to find it
Hypotenuse	Across from the $90^\circ$ angle
Opposite	Across from the angle $\theta$
Adjacent	Next to $\theta$ , but not the hypotenuse

💡 Memory hook: The hypotenuse is the "ramp" of the triangle — the longest, slanted side. The other two legs split into opposite and adjacent based on where $\theta$ is.

Concept Check 🎯

The Labels Switch with the Angle

This is the idea students miss most: opposite and adjacent are not fixed sides — they depend on which acute angle you choose.

Consider a right triangle with the right angle at $C$ , and legs $a$ and $b$ with hypotenuse $c$ .

If your angle is...	Opposite	Adjacent
Angle $A$

Label the Sides 🔽

A right triangle has its right angle at $C$ . The legs are $\overline{BC} = 6$ and $\overline{AC} = 8$ , with hypotenuse .

One More Look at the Same Triangle

Using the same triangle (right angle at $C$ , with $\overline{BC}=6$ , $\overline{AC}=8$ , ), let's now describe the sides from the acute angle, .

Concept Check 🎯

You're Ready for Ratios

You can now look at any right triangle, pick an acute angle, and name all three sides relative to it. That skill is the foundation for everything that follows.

In Part 2 we turn these labels into the three core ratios — sine, cosine, and tangent — and learn the famous mnemonic that locks them in.

Part 2: Sine, Cosine & Tangent (SOH-CAH-TOA)

📐 Trigonometric Ratios

Part 2 of 5 — Sine, Cosine & Tangent (SOH-CAH-TOA)

🔑 The Idea: Each trig ratio is just a fraction of two side lengths. Once the sides are labeled, the ratio is fixed by the angle — not by the size of the triangle.

The Three Ratios

For an acute angle $\theta$ in a right triangle:

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

Part 3: Solving for Missing Sides

📐 Trigonometric Ratios

Part 3 of 5 — Solving for Missing Sides

🔑 Why it matters: If you know one acute angle and one side of a right triangle, a single trig ratio lets you find any other side. This is how surveyors, builders, and navigators measure distances they can't reach.

A Reliable Strategy

To find a missing side:

Label the sides relative to the known angle (opposite / adjacent / hypotenuse).
Pick the ratio that uses the side you have and the side you want.
Set up the equation and solve for the unknown.

Worked Example: Find $x$

A right triangle has a $30^\circ$ angle. The hypotenuse is , and is the side the angle.

Part 4: Finding Angles & Special Triangles (inverse trig, 30-60-90 & 45-45-90)

📐 Trigonometric Ratios

Part 4 of 5 — Finding Angles & Special Triangles

🔑 Big Idea: If a trig ratio turns an angle into a number, then the inverse trig functions ( $\sin^{-1}$ , $\cos^{-1}$ , ) turn a back into an .

Part 5: Applications (elevation/depression) & Mastery Check

📐 Trigonometric Ratios

Part 5 of 5 — Applications & Mastery Check

You can now (1) label a right triangle, (2) write the three ratios, (3) solve for missing sides, and (4) use inverses and special triangles to find angles. Let's apply it to the real world and lock it in.

Angle of Elevation & Depression

A key application is measuring heights and distances you can't reach directly.

Angle of elevation: the angle up from the horizontal to an object (e.g., looking up at a treetop).
Angle of depression: the angle down from the horizontal to an object (e.g., looking down from a cliff).

Worked Example: Height of a Tree

You stand $50$ ft from the base of a tree. The angle of elevation to the top is $32^\circ$ . How tall is the tree?

The height is opposite the angle, and ft is → tangent:

The hypotenuse $c$ stays the hypotenuse for both angles. Only the two legs swap roles.

⚠️ Watch out: A leg that is "opposite" for one angle is "adjacent" for the other. Always re-check the labels when the angle changes.

\overline{AB} = 10

Vertex $B$ touches legs $\overline{BC}$ and the hypotenuse $\overline{AB}$ . So relative to angle $B$ :

the opposite side is $\overline{AC} = 8$ (it does not touch $B$ ),
the adjacent side is $\overline{BC} = 6$ (the leg touching $B$ ).

🔑 Compare with the drill above: for angle $A$ the opposite was $6$ , but for angle $B$ the opposite is $8$ . Same triangle, different labels.

hypotenuseadjacenttanθ=

The mnemonic SOH-CAH-TOA stores all three:

Mnemonic	Meaning
SOH	$\textbf{S}$ ine $=$ $\textbf{O}$ pposite over $\textbf{H}$ ypotenuse
CAH	$\textbf{C}$ osine $=$ $\textbf{A}$ djacent over $\textbf{H}$ ypotenuse
TOA	$\textbf{T}$ angent $=$ $\textbf{O}$ pposite over $\textbf{A}$ djacent

💡 Tip: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ , since $\dfrac{\text{opp}/\text{hyp}}{\text{adj}/\text{hyp}} = \dfrac{\text{opp}}{\text{adj}}$ . The hypotenuses cancel.

Worked Example: the 3-4-5 Triangle

A right triangle has legs $3$ and $4$ and hypotenuse $5$ . Let $\theta$ be the angle opposite the side of length $3$ . Then relative to $\theta$ :

opposite $= 3$ , adjacent $= 4$ , hypotenuse $= 5$ .

$\sin\theta = \frac{3}{5} = 0.6 \qquad \cos\theta = \frac{4}{5} = 0.8 \qquad \tan\theta = \frac{3}{4} = 0.75$

✅ Check with the Pythagorean identity: $\sin^2\theta + \cos^2\theta = 0.6^2 + 0.8^2 = 0.36 + 0.64 = 1$ ✓

Concept Check 🎯

Now Build All Three Yourself

When you write the three ratios for a triangle, keep the hypotenuse on the bottom for sine and cosine, and remember tangent never uses the hypotenuse at all.

The next drill uses the 5-12-13 right triangle — another set of whole-number side lengths that satisfies $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ .

Compute the Ratios 🧮

A right triangle has opposite $= 5$ , adjacent $= 12$ , and hypotenuse $= 13$ (relative to angle $\theta$ ). Enter each ratio as a fraction like 5/13.

1) $\sin\theta = \,?$ 2) $\cos\theta = \,?$ 3) $\tan\theta = \,?$

A Neat Pattern: Cofunctions

Look again at the 3-4-5 triangle. For the angle opposite $3$ we found $\sin\theta = \frac{3}{5}$ . For the other acute angle, the opposite and adjacent sides swap, so its cosine is also $\frac{3}{5}$ .

That's the cofunction relationship: in a right triangle the two acute angles add to $90^\circ$ , and

$\sin\theta = \cos(90^\circ - \theta)$

💡 This is why it's called "co"-sine — the complement's sine. We'll use it again when we work with special angles in Part 4.

Match the Ratio 🔽

For the 3-4-5 triangle, let $\theta$ be the angle opposite the side of length $3$ (so opposite $=3$ , adjacent $=4$ , hypotenuse $=5$ ).

We have the hypotenuse, we want the opposite — that's sine (SOH):

$\sin 30^\circ = \frac{x}{20}$

Since $\sin 30^\circ = \dfrac{1}{2}$ :

$x = 20 \cdot \sin 30^\circ = 20 \cdot \tfrac{1}{2} = 10$

✅ Sanity check: The side opposite the smallest angle should be the shortest. $10 < 20$ ✓

Pick the Right Ratio 🔽

In a right triangle you know angle $\theta$ and one side, and you want another side. Choose the ratio that connects them.

When the Unknown Is in the Denominator

Sometimes the side you want is the hypotenuse (the bottom of the fraction).

Worked Example: Find the hypotenuse $h$

A right triangle has a $40^\circ$ angle. The side adjacent to it is $12$ , and $h$ is the hypotenuse.

Have adjacent, want hypotenuse → cosine (CAH):

$\cos 40^\circ = \frac{12}{h}$

Multiply both sides by $h$ , then divide by $\cos 40^\circ$ :

$h = \frac{12}{\cos 40^\circ} \approx \frac{12}{0.766} \approx 15.7$

⚠️ Common mistake: When the unknown is in the denominator, you must divide the known side by the trig value — not multiply. Multiplying gives the wrong answer.

Solve for the Side 🧮

Use the given exact trig values. Enter a number (round decimals to one decimal place).

1) $\sin 30^\circ = \dfrac{x}{16}$ , and $\sin 30^\circ = 0.5$ . Find $x$ . 2) $\cos 60^\circ = \dfrac{y}{18}$ , and $\cos 60^\circ = 0.5$ . Find $y$ . 3) $\tan 45^\circ = \dfrac{z}{7}$ , and $\tan 45^\circ = 1$ . Find $z$ .

Setting Up the Equation Correctly

The hardest part is usually choosing the ratio and writing the equation — the algebra afterward is easy. Always ask two questions:

Which side do I have? (opposite, adjacent, or hypotenuse)
Which side do I want?

Then pick the single ratio that contains both of those sides. The last check makes you do exactly that.

Concept Check 🎯

Inverse Trig: From Ratio to Angle

If $\sin\theta = 0.5$ , what is $\theta$ ? Undo the sine with inverse sine:

$\theta = \sin^{-1}(0.5) = 30^\circ$

Each ratio has its own inverse:

Known ratio	Find the angle with
$\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$	$\theta = \sin^{-1}\!\left(\dfrac{\text{opp}}{\text{hyp}}\right)$

Worked Example

A right triangle has opposite $= 3$ and adjacent $= 4$ . Find $\theta$ .

$\tan\theta = \frac{3}{4} = 0.75 \;\Rightarrow\; \theta = \tan^{-1}(0.75) \approx 36.9^\circ$

⚠️ $\sin^{-1}$ is not $\dfrac{1}{\sin}$ . The " $-1$ " means , not a reciprocal exponent.

The Special Right Triangles

Two triangles appear so often that their exact ratios are worth memorizing.

The $30^\circ$ - $60^\circ$ - $90^\circ$ triangle has sides in ratio $1 : \sqrt{3} : 2$ (short leg : long leg : hypotenuse).

The $45^\circ$ - $45^\circ$ - $90^\circ$ triangle has sides in ratio $1$ (leg : leg : hypotenuse).

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$^{}$

💡 Notice the cofunction symmetry from Part 2: $\sin 30^\circ = \cos 60^\circ = \frac{1}{2}$ , and .

Special Angle Values 🔽

Recall the exact values from the table above.

Reading the Table Backwards

The drill above gave you an angle and asked for a ratio. The special angles also work the other way: given a familiar ratio, you can name the angle without a calculator.

For example, if $\tan\theta = 1$ , scan the table for a tangent of $1$ — that's $45^\circ$ . The next drill is all reverse lookups like this.

Find the Angle 🧮

Use the special-angle table to read off $\theta$ (in degrees). Enter just the number.

1) $\cos\theta = \dfrac{1}{2} \;\Rightarrow\; \theta = \,?^\circ$ 2) $\tan\theta = 1 \;\Rightarrow\; \theta = \,?^\circ$ 3) $\sin\theta = \dfrac{\sqrt{3}}{2} \;\Rightarrow\; \theta = \,?^\circ$

When the Ratio Isn't a "Nice" Number

Special angles are the exceptions — most ratios don't land on $30^\circ$ , $45^\circ$ , or $60^\circ$ . For those, you use a calculator's $\sin^{-1}$ , $\cos^{-1}$ , or $\tan^{-1}$ button.

The setup is identical: build the ratio from the sides you know, then apply the matching inverse. The final check tests whether you can pick the right inverse.

Concept Check 🎯

$\tan 32^\circ = \frac{h}{50} \;\Rightarrow\; h = 50\tan 32^\circ \approx 50(0.625) \approx 31.2 \text{ ft}$

💡 The angle of elevation from the ground and the angle of depression from the top are equal — they're alternate interior angles between two parallel horizontal lines.

Application Practice 🧮

A $20$ -ft ladder leans against a wall, making a $60^\circ$ angle with the ground.

Use exact special-angle values: $\sin 60^\circ = \dfrac{\sqrt 3}{2}\approx 0.866$ , $\cos 60^\circ = 0.5$ .

1) How high up the wall does the ladder reach? (height $=$ opposite the $60^\circ$ angle; round to one decimal place) 2) How far is the base of the ladder from the wall? (distance $=$ adjacent; exact whole number)

Quick Reference

Goal	Key move
Name a side	Opposite faces $\theta$ ; adjacent touches $\theta$ ; hypotenuse faces the right angle
Find a ratio	SOH-CAH-TOA
Find a side (in numerator)	side $=$ (known side) $\times$ (trig value)
Find a side (in denominator)	side $=$ (known side) $\div$ (trig value)
Find an angle	$\theta = \sin^{-1}, \cos^{-1},$ or $\tan^{-1}$ of the ratio
Special angles	$30$ - $60$ - $90$ is $1{:}\sqrt3{:}2$ ; -- is

⚠️ Remember the two classic traps: don't multiply when the unknown is a denominator, and $\sin^{-1}$ is an inverse function, not a reciprocal.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

θ = sin^{- 1} (\frac{opp}{hyp})

\sin 60^\circ = \cos 30^\circ = \frac{\sqrt 3}{2}

Trigonometric Ratios

Trigonometric Ratios - Complete Interactive Lesson

Part 1: Labeling a Right Triangle (Opposite/Adjacent/Hypotenuse)

📐 Trigonometric Ratios

Topics in This Part

Naming the Three Sides

The Labels Switch with the Angle

One More Look at the Same Triangle

You're Ready for Ratios

Part 2: Sine, Cosine & Tangent (SOH-CAH-TOA)

📐 Trigonometric Ratios

The Three Ratios

Part 3: Solving for Missing Sides

📐 Trigonometric Ratios

A Reliable Strategy

Worked Example: Find xxx

Part 4: Finding Angles & Special Triangles (inverse trig, 30-60-90 & 45-45-90)

📐 Trigonometric Ratios

Part 5: Applications (elevation/depression) & Mastery Check

📐 Trigonometric Ratios

Angle of Elevation & Depression

Worked Example: Height of a Tree

Worked Example: the 3-4-5 Triangle

Now Build All Three Yourself

A Neat Pattern: Cofunctions

When the Unknown Is in the Denominator

Worked Example: Find the hypotenuse hhh

Setting Up the Equation Correctly

Inverse Trig: From Ratio to Angle

Worked Example

The Special Right Triangles

Reading the Table Backwards

When the Ratio Isn't a "Nice" Number

Quick Reference

Worked Example: Find $x$

Worked Example: Find the hypotenuse $h$