Introduction to Trigonometry

📐 Introduction to Trigonometry

Part 1 of 5 — Right Triangles & the Three Ratios

Topics in This Part

🔑 Key Concept: Trigonometry studies the relationships between the angles and side lengths of triangles. In a right triangle, three special ratios — sine, cosine, and tangent — connect one acute angle to two of its sides.

Labeling the Sides

Every right triangle has a $90^\circ$ angle. We name the three sides relative to one chosen acute angle, usually called $\theta$ (theta):

• Hypotenuse — the longest side, always opposite the right angle. Its position never changes.
• Opposite — the side across from $\theta$.
• Adjacent — the side next to $\theta$ that is not the hypotenuse.

$\underbrace{\text{hypotenuse}}_{\text{opposite the }90^\circ} \qquad \underbrace{\text{opposite}}_{\text{across from }\theta} \qquad \underbrace{\text{adjacent}}_{\text{next to }\theta}$

⚠️ Watch out: "Opposite" and "adjacent" depend on which acute angle you pick. The hypotenuse, however, is always the side facing the right angle — it never switches.

Concept Check 🎯

The Three Ratios: SOH-CAH-TOA

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

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

The classic memory phrase is SOH-CAH-TOA:

💡 Tip: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, since $\dfrac{O/H}{A/H} = \dfrac{O}{A}$. The hypotenuses cancel.

Writing Ratios From a Triangle

Worked Example: the 3–4–5 triangle

A right triangle has legs $3$ and $4$ and hypotenuse $5$. Let $\theta$ be the angle whose opposite side is $3$ and whose adjacent side is $4$. Then:

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

If instead we pick the other acute angle (opposite $= 4$, adjacent $= 3$):

$\sin = \frac{4}{5} \qquad \cos = \frac{3}{5} \qquad \tan = \frac{4}{3}$

🔑 Key Idea: Opposite and adjacent swap when you switch acute angles, which is exactly why $\sin$ and $\cos$ trade places. The hypotenuse $5$ stays put.

Write the Ratios 🧮

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

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

Pick the Right Ratio 🔽

For each piece of information, choose which trig ratio connects them.

📐 Introduction to Trigonometry

Part 2 of 5 — Finding a Missing Side

🔑 The Idea: If you know one acute angle and one side, you can use $\sin$, $\cos$, or $\tan$ to find any other side. Pick the ratio that uses the side you have and the side you want.

The Strategy

To find a missing side:

1. Label the sides (opposite, adjacent, hypotenuse) relative to the known angle.
2. Choose the ratio that contains the known side and the unknown side.
3. Set up the equation and solve for the unknown.

Worked Example: side opposite a $30^\circ$ angle

A right triangle has a $30^\circ$ angle and a hypotenuse of $10$. Find the side opposite the $30^\circ$ angle.

We know the hypotenuse and want the opposite side → use $\sin$ (SOH):

$\sin 30^\circ = \frac{\text{opposite}}{10} \;\Rightarrow\; \text{opposite} = 10\sin 30^\circ = 10 \cdot \tfrac{1}{2} = 5$

✅ Check: $\sin 30^\circ = 0.5$, and $10 \cdot 0.5 = 5$. ✓

Worked Example: unknown in the denominator

A right triangle has a $40^\circ$ angle. The side adjacent to it is $8$. Find the hypotenuse $h$.

Adjacent + hypotenuse → use $\cos$ (CAH):

$\cos 40^\circ = \frac{8}{h}$

When the unknown is in the denominator, multiply both sides by $h$, then divide:

$h = \frac{8}{\cos 40^\circ} \approx \frac{8}{0.766} \approx 10.4$

💡 Rule of thumb: If the unknown side is on top of the fraction, you multiply. If it is on the bottom, you divide the known value by the trig value.

Concept Check 🎯

Solve for the Side 🧮

Use a calculator (degree mode) and round to the nearest tenth.

1) Find the side opposite a $30^\circ$ angle when the hypotenuse is $14$: $\;14\sin 30^\circ = \,?$ 2) Find the side adjacent to a $60^\circ$ angle when the hypotenuse is $14$: $\;14\cos 60^\circ = \,?$ 3) Find the side opposite a $45^\circ$ angle when the adjacent side is $9$: $\;9\tan 45^\circ = \,?$

Multiply or Divide? 🔽

For each setup, decide how to isolate the unknown side $x$.

📐 Introduction to Trigonometry

Part 3 of 5 — Finding a Missing Angle

🔑 Why it works: If you know two sides, you can find the angle. The inverse trig functions $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ undo sine, cosine, and tangent — they take a ratio and return the angle.

Inverse Trig Functions

A normal trig function takes an angle and returns a ratio: $\sin 30^\circ = 0.5$

An inverse trig function reverses this — it takes a ratio and returns an angle: $\sin^{-1}(0.5) = 30^\circ$

⚠️ Notation alert: $\sin^{-1}\theta$ means the inverse sine (also written $\arcsin$). It does not mean $\dfrac{1}{\sin\theta}$. On most calculators it is the 2nd → sin key.

Worked Example

A right triangle has a side of length $3$ opposite angle $\theta$ and a hypotenuse of $5$. Find $\theta$.

We have opposite and hypotenuse → use $\sin^{-1}$:

$\sin\theta = \frac{3}{5} = 0.6 \;\Rightarrow\; \theta = \sin^{-1}(0.6) \approx 36.9^\circ$

Second Example: using tangent

Legs $4$ (opposite) and $4$ (adjacent), so:

$\tan\theta = \frac{4}{4} = 1 \;\Rightarrow\; \theta = \tan^{-1}(1) = 45^\circ$

✅ Check: Equal legs make an isosceles right triangle, whose acute angles are both $45^\circ$. ✓

Concept Check 🎯

Choose the Setup 🔽

For each pair of known sides, pick the correct inverse-function setup for $\theta$.

Find the Angle 🧮

Use a calculator in degree mode and round to the nearest whole degree. (Enter just the number, e.g. 37.)

1) $\theta = \sin^{-1}(0.5) = \,?$ degrees 2) $\theta = \cos^{-1}(0.5) = \,?$ degrees 3) $\theta = \tan^{-1}\!\left(\dfrac{3}{4}\right) = \,?$ degrees

📐 Introduction to Trigonometry

Part 4 of 5 — Special Right Triangles & Exact Values

🔑 Big Payoff: Two special triangles — the $45^\circ$–$45^\circ$–$90^\circ$ and the $30^\circ$–$60^\circ$–$90^\circ$ — have trig values you can find exactly, no calculator needed.

The $45^\circ$–$45^\circ$–$90^\circ$ Triangle

This is an isosceles right triangle: both legs are equal, and the hypotenuse is $\sqrt{2}$ times a leg.

$\text{legs } 1, 1 \qquad \text{hypotenuse } \sqrt{2}$

So for $\theta = 45^\circ$:

$\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \qquad \cos 45^\circ = \frac{\sqrt{2}}{2} \qquad \tan 45^\circ = \frac{1}{1} = 1$

💡 Rationalizing: $\dfrac{1}{\sqrt{2}}$ is usually written $\dfrac{\sqrt{2}}{2}$ by multiplying top and bottom by $\sqrt{2}$. As a decimal, $\dfrac{\sqrt{2}}{2} \approx 0.707$.

The $30^\circ$–$60^\circ$–$90^\circ$ Triangle

The sides are in the ratio $1 : \sqrt{3} : 2$, where:

• the side opposite $30^\circ$ is the shortest, $1$,
• the side opposite $60^\circ$ is $\sqrt{3}$,
• the hypotenuse is $2$.

🔑 Notice the symmetry: $\sin 30^\circ = \cos 60^\circ$ and $\cos 30^\circ = \sin 60^\circ$. The sine of an angle equals the cosine of its complement ($30^\circ + 60^\circ = 90^\circ$).

Concept Check 🎯

Recall the Exact Values 🔽

Fill in each exact value from memory.

Exact-Value Practice 🧮

A $30^\circ$–$60^\circ$–$90^\circ$ triangle has a hypotenuse of $8$. Use the side ratios to find each side. (The shortest side is opposite $30^\circ$.)

1) Side opposite $30^\circ$ (the short leg) $= \,?$ 2) Hypotenuse divided by short leg $= \,?$ (the constant ratio) 3) $\tan 45^\circ = \,?$

📐 Introduction to Trigonometry

Part 5 of 5 — Applications & Mastery Check

You can now (1) write the three ratios, (2) find a missing side, (3) find a missing angle, and (4) recall exact values. Let's apply them to real problems and finish with an exit quiz.

Angles of Elevation & Depression

• The angle of elevation is measured upward from the horizontal to a line of sight (looking up at a treetop).
• The angle of depression is measured downward from the horizontal (looking down from a cliff at a boat).

Worked Example: height of a tree

You stand $50$ ft from the base of a tree. The angle of elevation to the top is $35^\circ$. How tall is the tree?

The height is opposite the $35^\circ$ angle, and $50$ ft is adjacent → use $\tan$ (TOA):

$\tan 35^\circ = \frac{\text{height}}{50} \;\Rightarrow\; \text{height} = 50\tan 35^\circ \approx 50(0.700) \approx 35.0 \text{ ft}$

💡 Setup tip: Draw a right triangle. The horizontal distance is the adjacent leg, the height is the opposite leg, and the angle sits at your eye.

Concept Check 🎯

Apply It 🧮

Round to the nearest tenth. Use degree mode.

1) A $20$ ft ladder leans against a wall at a $60^\circ$ angle of elevation. How high up the wall does it reach? $\;20\sin 60^\circ = \,?$ (use $\sin 60^\circ \approx 0.866$) 2) A kite string is $100$ ft long at a $30^\circ$ angle of elevation. How high is the kite? $\;100\sin 30^\circ = \,?$ 3) A guy-wire meets the ground $9$ ft from a pole and reaches $12$ ft up. What angle does it make with the ground? $\;\tan^{-1}\!\left(\frac{12}{9}\right) = \,?$ degrees

Quick Reference

⚠️ Before the quiz: Make sure your calculator is in degree mode, not radians, and remember that $\sin^{-1}$ means inverse sine — not $\frac{1}{\sin}$.

Exit Quiz ✅

Answer all three to finish the lesson.