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Trig Ratios

⚗ Dr. Möbius, from the lab

Every trigonometry course on earth starts with "SOH-CAH-TOA" like it's a goddamn magic spell. It is not a magic spell. It is an index. The actual understanding — the reason those ratios exist at all, the reason they're functions of an angle rather than properties of a triangle — takes one sentence, and almost nobody says it. I am going to say it, loudly, in this lab, with the particle accelerator running in the background, and then we're going to use it to destroy right triangles on demand.

THE BIG IDEA

Because all right triangles with the same acute angle are similar, the ratios of their sides depend only on that angle — that's why sin, cos, and tan are well-defined functions of angle alone.

The one sentence that explains all of trigonometry

Here it is. This is the sentence:

Because all right triangles with the same acute angle are similar, the ratio of any two sides depends only on that angle — not on the size of the triangle.

Let me unpack it. Fix an acute angle $\theta$ . Draw any right triangle containing $\theta$ . Now draw a different right triangle — larger or smaller — also containing $\theta$ . Both triangles have a $90°$ angle and $\theta$ , so by AA similarity (previous lesson), they are similar. Similar triangles have proportional sides, so the ratio $\frac{\text{opposite}}{\text{hypotenuse}}$ is the same in both triangles. It doesn't matter how big the triangle is. It doesn't matter where you drew it. That ratio belongs to $\theta$ alone.

That's why we can speak of "the sine of 30°" without specifying a triangle. Any triangle with a $30°$ angle will give the same ratio. Trigonometric functions are the names we give to these angle-only ratios. That's it. That's the whole mystery, solved in two sentences — something the Federation of Boring Textbook Authors apparently couldn't manage in three hundred pages.

SOH-CAH-TOA: the index

For a right triangle with acute angle $\theta$ , label the sides relative to $\theta$ :

Opposite ( $\text{opp}$ ): the leg across from $\theta$ .
Adjacent ( $\text{adj}$ ): the leg next to $\theta$ (the one that's not the hypotenuse).
Hypotenuse ( $\text{hyp}$ ): the side opposite the right angle (always the longest side).

The three primary trig ratios:

$\sin\theta = \frac{\text{opp}}{\text{hyp}}, \qquad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \qquad \tan\theta = \frac{\text{opp}}{\text{adj}}.$

"SOH-CAH-TOA" is just these three, rearranged as a mnemonic. It's the index. Useful. Not the understanding. Don't mistake the map for the territory, you hear me?

Note also: $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , which follows directly from dividing the first ratio by the second.

The two sacred triangles

Two right triangles give exact, clean trig values and I want you to derive them from geometry, not a damn lookup table. If the power goes out in the lab and you have only chalk and a right angle, you should still be able to get these.

The 45-45-90 triangle (half a square)

Cut a unit square along the diagonal. You get a right isosceles triangle with legs $1, 1$ . The hypotenuse is $\sqrt{2}$ (Pythagorean theorem: $1^2 + 1^2 = 2$ ).

Both acute angles are $45°$ (since the base angles of an isosceles right triangle are equal, and they sum to $90°$ ).

$\sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \tan 45° = \frac{1}{1} = 1.$

The 30-60-90 triangle (half an equilateral)

Start with an equilateral triangle with all sides $2$ . Drop the altitude from the top vertex to the base. By the symmetry of the equilateral triangle (and the AA argument), this altitude bisects the base and creates two congruent right triangles.

Each right triangle has: hypotenuse $= 2$ (half of the equilateral's side...no, the full side), short leg $= 1$ (half the base), and acute angles $30°$ and $60°$ . The remaining leg by the Pythagorean theorem: $h = \sqrt{2^2 - 1^2} = \sqrt{3}$ .

Angles and their ratios:

$\sin 30° = \frac{1}{2}, \qquad \cos 30° = \frac{\sqrt{3}}{2}, \qquad \tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.$

$\sin 60° = \frac{\sqrt{3}}{2}, \qquad \cos 60° = \frac{1}{2}, \qquad \tan 60° = \sqrt{3}.$

The pattern to remember: $\sin 30° = \cos 60°$ and $\sin 60° = \cos 30°$ . This is because $30°$ and $60°$ are complementary — the opposite/adjacent roles swap.

Summary table (memorize or re-derive on demand):

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$30°$	$1/2$	$\sqrt{3}/2$	$1/\sqrt{3}$
$45°$	$\sqrt{2}/2$	$\sqrt{2}/2$	$1$
$60°$	$\sqrt{3}/2$	$1/2$	$\sqrt{3}$

Solving right triangles

"Solving a triangle" means finding all unknown sides and angles. The toolkit:

Finding a side: if you know an angle $\theta$ and one side, set up a trig ratio equation and solve.
Finding an angle: if you know two sides, compute the ratio and use the inverse trig function ( $\arcsin$ , $\arccos$ , $\arctan$ ) to get the angle.

The inverse functions answer the question: "which angle has this ratio?" $\theta = \arcsin(x)$ means $\sin\theta = x$ for $\theta \in [0°, 90°]$ .

Angle of elevation and depression

These are the applied trig problems you'll see everywhere:

Angle of elevation: the angle above horizontal to a line of sight going up to something.
Angle of depression: the angle below horizontal to a line of sight going down to something.

In both cases, the angle equals the alternate interior angle formed with the horizontal — and you get a right triangle to solve.

🔬 SPECIMENS (worked examples)

Worked example 1 — finding a side with a known angle▸

A right triangle has a $30°$ angle and hypotenuse $10$ . Find the side opposite the $30°$ angle.

Step 1. Identify the known quantities relative to the $30°$ angle:

Hypotenuse $= 10$ (known)
Opposite side $= ?$

Step 2. The ratio involving opposite and hypotenuse is sine: $\sin 30° = \frac{\text{opp}}{\text{hyp}} = \frac{\text{opp}}{10}.$

Step 3. Exact value: $\sin 30° = \frac{1}{2}$ .

Step 4. Solve: $\frac{\text{opp}}{10} = \frac{1}{2} \implies \text{opp} = 5.$

The side opposite the $30°$ angle has length $5$ . The other leg is $\sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3}$ — consistent with $\cos 30° = \sqrt{3}/2$ giving adjacent $= 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

Worked example 2 — finding an angle from two sides▸

A right triangle has legs $7$ and $24$ . Find the angle at the vertex between the leg of length $24$ and the hypotenuse. Give an exact expression.

Step 1. Find the hypotenuse. This is a 7-24-25 Pythagorean triple ( $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ ), so hypotenuse $= 25$ .

Step 2. Label the angle we want: $\theta$ at the vertex between the leg $24$ and the hypotenuse. The leg $24$ is adjacent to $\theta$ , and the leg $7$ is opposite to $\theta$ .

Step 3. Set up a ratio: $\tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{7}{24}.$

Step 4. Apply the inverse function: $\theta = \arctan\!\left(\frac{7}{24}\right) \approx 16.26°.$

Alternatively using sine: $\sin\theta = 7/25$ , giving $\theta = \arcsin(7/25)$ . All three inverse functions give the same angle.

Worked example 3 — the trap: neither 30-60-90 nor 45-45-90, deal with it▸

From the top of a lighthouse $40$ m tall, the angle of depression to a boat is $25°$ . How far is the boat from the base of the lighthouse (to the nearest meter)?

Step 1. Draw the setup. The horizontal line from the lighthouse top and the line of sight to the boat form the angle of depression of $25°$ . By alternate interior angles (the horizontal at lighthouse height and the horizontal at sea level are parallel), the angle at the boat looking up to the lighthouse is also $25°$ .

Step 2. We have a right triangle:

Vertical leg (lighthouse) $= 40$ m.
Horizontal leg (distance to boat) $= d$ .
Angle at the boat $= 25°$ .

Relative to the $25°$ angle at the boat: the vertical leg (40 m) is opposite, the horizontal distance is adjacent.

Step 3. Use tangent: $\tan 25° = \frac{40}{d} \implies d = \frac{40}{\tan 25°}.$

Step 4. Compute: $\tan 25° \approx 0.4663$ , so $d \approx 40 / 0.4663 \approx 85.8$ m, which rounds to $86$ m.

The trap: students forget to use the alternate-interior-angle relationship, or mistakenly place the $25°$ inside the triangle incorrectly. Draw the damn picture first, always. The angle of depression at the top equals the angle of elevation at the bottom — no exceptions, no shortcuts, no bullshit.

☠ KNOWN HAZARDS

Forgetting which sides are opposite/adjacent changes with the angle. In a triangle with angles $30°$ , $60°$ , $90°$ , the side opposite $30°$ is the adjacent side for $60°$ and vice versa. Label relative to the angle you're working with, every time.
Using SOH-CAH-TOA without understanding it. If you forget the mnemonic, you can re-derive the ratios by remembering the two sacred triangles. Never memorize without a backup derivation strategy.
Writing $\sin^{-1}$ to mean $1/\sin$ . It doesn't. $\sin^{-1}(x) = \arcsin(x)$ is the inverse function (undoes sine). $1/\sin\theta = \csc\theta$ (cosecant, a different function). The exponent-style notation is genuinely, infuriatingly confusing — complain to whoever invented it, not to me. Know the difference or suffer.
Forgetting the Pythagorean theorem as a third tool. When solving a right triangle, you have: two trig equations, the angle sum ( $90° +$ two acute angles $= 180°$ ), and $a^2 + b^2 = c^2$ . Use all four.

TL;DR

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The trig ratios exist because similar right triangles have proportional sides, so the ratios $\text{opp}/\text{hyp}$ , $\text{adj}/\text{hyp}$ , $\text{opp}/\text{adj}$ depend only on the angle — not the triangle size.
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$\sin\theta = \text{opp}/\text{hyp}$ , $\cos\theta = \text{adj}/\text{hyp}$ , $\tan\theta = \text{opp}/\text{adj} = \sin\theta/\cos\theta$ .
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Exact values come from two triangles: the $45$ - $45$ - $90$ (half a unit square, hypotenuse $\sqrt{2}$ ) and the $30$ - $60$ - $90$ (half an equilateral with side $2$ , longer leg $\sqrt{3}$ ).
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Inverse trig ( $\arcsin$ , $\arccos$ , $\arctan$ ) undoes the ratio: they answer "which angle gives this ratio?" and complete the toolkit for solving any right triangle.

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