Worked examples — Trig ratios of standard angles — 0°, 30°, 45°, 60°, 90° (derive, don't memorize blindly)

You already know the five standard angles and how to derive them. This page is a workout. We hunt down every kind of problem these angles can appear in — every degenerate case, every word-problem twist — and solve one example of each so you never meet a scenario you haven't seen.

Before we begin, the tools we will lean on:

The scenario matrix

Every problem this topic throws at you (within our $[0°,90°]$ domain) falls into one of the cells below. Figure 1 lays the whole grid out visually so you can see the space we are about to cover; each example is also labelled [Cell X].

We hit all nine cells with the examples below. Let's go.

[!example] Example 1 — Pure evaluation [Cell A]

Statement. Compute exactly: $\sin 30°+\cos 60°+\tan 45°.$

Forecast: guess before reading. Two of these terms are equal — spot them. Will the answer be a whole number or have a root?

Steps.

1. Replace each with its exact value: $\sin 30°=\frac{1}{2}$, $\cos 60°=\frac{1}{2}$, $\tan 45°=1$. Why this step? Every entry is a standard angle, so we swap symbols for their derived exact values — no calculator needed.
2. Add: $\frac{1}{2}+\frac{1}{2}+1$. Why this step? Like fractions ($\frac{1}{2}+\frac{1}{2}=1$) combine cleanly.
3. Result: $1+1=2$.

Verify: decimals — $0.5+0.5+1=2.0$. ✓ A whole number, no root, exactly as the symmetry ($\sin 30°=\cos 60°$, from Complementary Angles) hinted.

[!example] Example 2 — Rationalize & simplify [Cell B]

Statement. Simplify exactly: $\frac{1}{\tan 30°}+\tan 60°.$

Forecast: $\tan 30°$ and $\tan 60°$ are reciprocals of a sort — guess whether the two terms are actually the same number.

Steps.

1. $\tan 30°=\frac{1}{\sqrt{3}}$, so $\frac{1}{\tan 30°}=\sqrt{3}$. Why this step? Dividing 1 by $\frac{1}{\sqrt{3}}$ flips the fraction: $\frac{1}{1/\sqrt{3}}=\sqrt{3}$.
2. $\tan 60°=\sqrt{3}$. Why this step? Standard value, from the $30\text{-}60\text{-}90$ triangle where the long leg is $\sqrt{3}$.
3. Add: $\sqrt{3}+\sqrt{3}=2\sqrt{3}$. Why this step? Same radical, so they add like $x+x=2x$.

Verify: $2\sqrt{3}\approx 2(1.732)=3.464$. Check original: $\frac{1}{0.5774}+1.732=1.732+1.732=3.464$. ✓ Note the denominator $\sqrt{3}$ never survives into the final answer — see Rationalization.

[!example] Example 3 — Solve for an unknown side [Cell C]

Statement. A right triangle has a $30°$ angle and its hypotenuse is $10$ cm. Find the side opposite the $30°$ angle exactly.

Forecast: the opposite side of the small angle is the short leg. Will your answer be more or less than half the hypotenuse?

Steps.

1. Look at Figure 2: the orange side is the hypotenuse ($10$ cm), the green vertical side is the opposite of the $30°$ angle marked in red at the lower-right corner. Pick the ratio linking opposite and hypotenuse: that is sine. $\sin 30°=\frac{\text{opposite}}{\text{hypotenuse}}$. Why this step? We know the hypotenuse and want the opposite — sine is the only ratio built from exactly those two sides, and the figure shows which side is which.
2. Substitute: $\frac{1}{2}=\frac{\text{opp}}{10}$ (using $\sin 30°=\frac{1}{2}$). Why this step? Plug the exact standard value in so the equation has one unknown.
3. Solve: $\text{opp}=10\times \frac{1}{2}=5$ cm.

Verify: in Figure 2 the green opposite side is exactly half the orange hypotenuse — that's the defining property of a $30\text{-}60\text{-}90$ triangle (short leg = half the longest side). $5=10/2$. ✓ Units: cm ÷ (unitless ratio) = cm. ✓

[!example] Example 4 — Solve for an unknown angle via $\arctan$ [Cell D]

Statement. In a right triangle the side opposite angle $\theta$ is $\sqrt{3}$ and the adjacent side is $1$. Which standard angle is $\theta$?

Forecast: $\tan \theta$ = opposite/adjacent. Guess: is $\theta$ shallow (small) or steep (large)?

Steps.

1. Form the tangent, the ratio of opposite to adjacent: $\tan \theta =\frac{\sqrt{3}}{1}=\sqrt{3}$. Why this step? We're given opposite AND adjacent, and tangent is exactly opposite ÷ adjacent — the natural fit.
2. Apply $\arctan$: $\theta =\arctan (\sqrt{3})$, i.e. ask "which standard angle has $\tan =\sqrt{3}$?" From our table, $\tan 60°=\sqrt{3}$, so $\theta =60°$. Why this step? $\arctan$ undoes $\tan$ — reading the table backwards from value to angle is exactly what finding an angle means. Within our $[0°,90°]$ domain there is only one such angle, so no ambiguity.

Verify: rebuild the triangle — legs $1$ and $\sqrt{3}$, hypotenuse $\sqrt{1^2+(\sqrt{3}{)}^2}=\sqrt{4}=2$ via Pythagorean Theorem. Then $\cos \theta =\frac{1}{2}=\cos 60°$. ✓ Consistent.

[!example] Example 4b — Same cell, but via $\arcsin$ and $\arccos$ [Cell D]

Statement. In a right triangle the hypotenuse is $2$ and the side opposite angle $\theta$ is $1$. Find $\theta$ two ways: from the sine, and from the cosine.

Forecast: opposite is half the hypotenuse. Which standard angle has that property?

Steps.

1. Via sine. $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1}{2}$. Apply $\arcsin$: $\theta =\arcsin \left(\frac{1}{2}\right)$ = "which angle has sine $\frac{1}{2}$?" From the table, $\sin 30°=\frac{1}{2}$, so $\theta =30°$. Why this step? We were handed opposite and hypotenuse — those are exactly the two sides sine is built from, and $\arcsin$ turns that known ratio back into the angle.
2. Cross-check via cosine. First find the adjacent side: by Pythagorean Theorem, $\text{adjacent}=\sqrt{2^2-1^2}=\sqrt{3}$. Then $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}$, so $\theta =\arccos \left(\frac{\sqrt{3}}{2}\right)$. From the table $\cos 30°=\frac{\sqrt{3}}{2}$, giving $\theta =30°$ again. Why this step? Using a second ratio ($\arccos$) on the same triangle confirms the angle independently — a built-in safety net.

Verify: both routes give $\theta =30°$. Sanity: ${\sin }^{2}30°+{\cos }^{2}30°={\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1$. ✓

[!example] Example 5 — Zero / degenerate input [Cell E]

Statement. A straight ramp of length $6$ m (this is the ramp itself — the hypotenuse) is laid down at an angle of $0°$, i.e. flat on the ground. How high does its far end rise?

Forecast: the ramp is flat. Before computing — what should the height be, physically?

Steps.

1. The height gained is the opposite side, and $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$ rearranges to $\text{opposite}=\text{hypotenuse}\times \sin \theta$. Here the hypotenuse is the $6$ m ramp: height $=6\times \sin 0°$. Why this step? Sine converts a slanted length (the ramp) into its vertical part (the height).
2. Substitute $\sin 0°=0$: height $=6\times 0=0$ m. Why this step? At $0°$ the triangle has collapsed flat — the hypotenuse now lies exactly on top of the ground, so the vertical leg has shrunk to nothing.
3. Notice in Figure 3: the blue ramp and the ground are the same line — the "opposite" side has vanished. That is what degenerate means: two sides merge and the triangle stops being a triangle. Why this step? It shows precisely why the naming "opposite vs adjacent" stops mattering — at $0°$ the hypotenuse and adjacent coincide and the opposite is zero.

Verify: physically a flat ramp gains no height. $0$ m matches intuition and $\sin 0°=0$. ✓ This is the limiting case where "triangle" degenerates into a line segment.

[!example] Example 6 — Undefined / division-by-zero [Cell F]

Statement. Evaluate $\frac{\sin 90°}{\cos 90°}$ and explain the result.

Forecast: $\cos 90°=0$. Guess what happens when it sits in a denominator.

Steps.

1. Recall the definition from the top of this page: $\tan \theta =\frac{\sin \theta }{\cos \theta }$. So this expression is $\tan 90°$. Substitute $\sin 90°=1$, $\cos 90°=0$, giving $\frac{1}{0}$. Why this step? Recognising the expression as $\tan 90°$ tells us what geometric object we're really probing.
2. $\frac{1}{0}$ is undefined — no number times $0$ gives $1$. Why this step? Division asks "how many zeros make 1?" — impossible, so the operation has no answer.
3. Geometry: at $90°$ the line to the point is vertical; a vertical line has infinite steepness (slope), which is exactly why tangent has no finite value here.

Verify: approach it: $\tan 89°\approx 57.3$, $\tan 89.9°\approx 573$, growing without bound as we near $90°$. This confirms "undefined" means "runs off to infinity," not "equals some hidden number." ✓

[!example] Example 7 — Complementary / symmetry shortcut [Cell G]

Statement. Compute exactly: ${\sin }^{2}30°+{\sin }^{2}60°.$ (Here ${\sin }^2\theta$ means $(\sin \theta {)}^2$.)

Forecast: $30°$ and $60°$ add to $90°$ — they're complementary. Guess: does the answer simplify to something famous?

Steps.

1. Use the complement rule: $\sin 60°=\cos 30°$, because sine of an angle equals cosine of its complement (see Symmetry in Trigonometry). Why this step? It converts the second term into a cosine of the same angle $30°$, unlocking an identity.
2. Rewrite: ${\sin }^{2}30°+{\cos }^{2}30°$. Why this step? Now both terms share the angle $30°$.
3. Apply ${\sin }^2\theta +{\cos }^2\theta =1$ (the Pythagorean identity): the sum is $1$.

Verify: brute force — ${\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1$. ✓ The shortcut and the direct computation agree.

[!example] Example 8 — Real-world word problem [Cell H]

Statement. A kite string is $20$ m long and makes a $60°$ angle with the horizontal ground. Assuming the string is straight, how high is the kite? Give the exact height and a decimal.

Forecast: $60°$ is steep. Guess: is the kite higher or lower than half the string length ($10$ m)?

Steps.

1. Read Figure 4: the orange string is the hypotenuse ($20$ m), the green vertical side is the height we want (the opposite of the $60°$ red angle), and the grey dashed line is the ground (adjacent). We know the hypotenuse and want the opposite. Why this step? Translating the story into the labelled right triangle tells us which sides we know and want.
2. Choose sine, since it links opposite (height) to hypotenuse (string): $\sin 60°=\frac{h}{20}$. Why this step? We know the hypotenuse and want the opposite — sine is the connector.
3. Substitute $\sin 60°=\frac{\sqrt{3}}{2}$: $\frac{\sqrt{3}}{2}=\frac{h}{20}$.
4. Solve: $h=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}$ m. Why this step? Multiply both sides by $20$ to isolate $h$.

Verify: $10\sqrt{3}\approx 10(1.732)=17.32$ m. That's the green side, higher than the $10$ m half-string — sensible for a steep $60°$. Units: m × (unitless) = m. ✓

[!example] Example 9 — Exam twist: verify an identity [Cell I]

Statement. Show that $\frac{\tan 45°+\tan 60°}{\tan 30°}=3+\sqrt{3}.$

Forecast: the numerator has a $\sqrt{3}$; the denominator is tiny ($\frac{1}{\sqrt{3}}$). Guess: dividing by a small number makes the answer bigger or smaller?

Steps.

1. Substitute exact values: $\tan 45°=1$, $\tan 60°=\sqrt{3}$, $\tan 30°=\frac{1}{\sqrt{3}}$. Why this step? Standard values turn the whole thing into pure algebra.
2. The expression becomes $\frac{1+\sqrt{3}}{\frac{1}{\sqrt{3}}}$.
3. Dividing by $\frac{1}{\sqrt{3}}$ = multiplying by $\sqrt{3}$: $(1+\sqrt{3})\cdot \sqrt{3}$. Why this step? Reciprocal of $\frac{1}{\sqrt{3}}$ is $\sqrt{3}$; dividing by a fraction means multiply by its flip.
4. Distribute: $\sqrt{3}+\sqrt{3}\cdot \sqrt{3}=\sqrt{3}+3$. Why this step? $\sqrt{3}\cdot \sqrt{3}=3$ removes the root.
5. Rearrange: $3+\sqrt{3}$. ✓ Matches the claim.

Verify: $3+\sqrt{3}\approx 3+1.732=4.732$. Direct: $\frac{1+1.732}{0.5774}=\frac{2.732}{0.5774}=4.732$. ✓

[!recall]- Quick self-test

Connections

• Trig ratios of standard angles — 0°, 30°, 45°, 60°, 90° (derive, don't memorize blindly) — the parent this page drills.
• Pythagorean Theorem — used to rebuild triangles in the verify steps.
• Special Triangles — the $45\text{-}45\text{-}90$ and $30\text{-}60\text{-}90$ blocks behind every value.
• Complementary Angles · Symmetry in Trigonometry — the shortcut in Example 7 and the quadrant-sign story beyond this page.
• Rationalization · Exact vs Approximate Values — cleaning and interpreting answers.
• Unit Circle — where the degenerate $0°/90°$ cases and the sign rules live.