Worked examples — SOH-CAH-TOA mnemonic

This page is a drill. The parent note taught you the three ratios. Here we throw every kind of triangle problem at you — steep angles, shallow angles, angles you must find, sides you must find, degenerate triangles where one side shrinks to zero, and word problems wearing disguises. By the end, no problem type should surprise you.

The scenario matrix

Cell	What you're given	What you want	Tool	Example
A	hypotenuse + angle	opposite side	SOH (sine)	Ex 1
B	adjacent + angle	opposite side	TOA (tangent)	Ex 2
C	opposite + adjacent	the angle	inverse tan	Ex 3
D	opposite + hypotenuse	the angle	inverse sine	Ex 4
E	limiting / degenerate (a side → 0)	behaviour	limits	Ex 5
F	angle → 0° or → 90°	what ratios do	limits	Ex 6
G	word problem in disguise	pick the right side	classify	Ex 7
H	exam twist (two triangles glued)	chain the ratios	combine	Ex 8

We cover A–H. Each example says which cell it hits.

Before we start, a shared picture of the naming — because every mistake below comes from mislabelling a side.

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Cell A — hypotenuse + angle → opposite side

1. Label. The ladder is the hypotenuse ($=10$). The wall-height is across from the ground-angle $\theta =60°$, so height is the opposite side. Why this step? Every problem starts by pinning each number to a side-name; that is what selects the ratio.
2. Pick the ratio. We have hypotenuse, want opposite → SOH: $\sin \theta =\frac{\text{opp}}{\text{hyp}}$. Why this step? Sine is the only ratio that pairs opposite with hypotenuse — exactly the two sides in play.
3. Substitute and solve. $\sin 60°=\frac{h}{10}\Rightarrow h=10\sin 60°=10\cdot \frac{\sqrt{3}}{2}=5\sqrt{3}\approx 8.66\text{ m}.$ Why this step? Multiplying both sides by $10$ isolates the unknown $h$. The exact value $\frac{\sqrt{3}}{2}$ comes from the 30-60-90 triangle (see 2.4.05-special-angles).

Verify: $8.66<10$ ✓ (a side can't exceed the hypotenuse). It's near the top of our forecast band because $60°$ is steep. Units: $\text{m}$. ✓

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Cell B — adjacent + angle → opposite side

1. Label. The $50\text{ m}$ ground distance touches the angle and is not the hypotenuse → adjacent. The height sits across from the angle → opposite. We know nothing about the hypotenuse (the line of sight). Why this step? Noticing the hypotenuse is neither known nor wanted is what kills the temptation to use sine or cosine.
2. Pick the ratio. Have adjacent, want opposite → TOA: $\tan \theta =\frac{\text{opp}}{\text{adj}}$. Why this step? Tangent is the one ratio that never mentions the hypotenuse — perfect when the hypotenuse is absent.
3. Solve. $\tan 35°=\frac{h}{50}\Rightarrow h=50\tan 35°\approx 50(0.7002)=35.0\text{ m}.$

Verify: $35<50$ ✓ matches the "shallow angle → shorter than the run" forecast. If we'd wrongly used $\cos$, we'd have invented a hypotenuse we never measured.

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Cell C — two sides → the angle (inverse tangent)

1. Label. Rise $=3$ is opposite the incline angle; run $=8$ is adjacent. Why this step? Same first move as always — name the sides relative to the angle we want.
2. Pick the ratio. Have opposite and adjacent → TOA: $\tan \theta =\frac{3}{8}$. Why this step? Two legs, no hypotenuse → tangent.
3. Undo the tangent. The unknown $\theta$ is trapped inside $\tan$. To free it we apply the inverse tangent ${\tan }^{-1}$, which asks "which angle has this tangent?" $\theta ={\tan }^{-1}\left(\frac{3}{8}\right)={\tan }^{-1}(0.375)\approx 20.6°.$ Why this step? ${\tan }^{-1}$ is the "undo" button for $\tan$; it is not $1/\tan$. See 2.5.01-inverse-trig-functions.

Verify: feed it back: $\tan 20.6°\approx 0.375$ ✓. And $20.6°<45°$ matches "gentle" ✓.

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Cell D — opposite + hypotenuse → the angle (inverse sine)

1. Label. Height $8.66$ is opposite the ground angle; ladder $10$ is the hypotenuse.
2. Pick the ratio. Have opposite and hypotenuse → SOH: $\sin \theta =\frac{8.66}{10}=0.866$. Why this step? Opposite-over-hypotenuse is exactly sine.
3. Undo the sine. $\theta$ is inside $\sin$; apply the inverse sine ${\sin }^{-1}$ ("which angle has this sine?"). $\theta ={\sin }^{-1}(0.866)\approx 60.0°.$

Verify: matches Ex 1's input angle ✓. Consistency between the forward (Cell A) and backward (Cell D) directions is the strongest possible sanity check.

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Cell E — degenerate triangle: a side shrinks to zero

1. Angle from sine. $\sin \theta =\frac{\text{opp}}{1}=\text{opp}$, so $\theta ={\sin }^{-1}(\text{opp})$.
   • $\text{opp}=0.5\Rightarrow \theta =30°$
   • $\text{opp}=0.1\Rightarrow \theta \approx 5.74°$
   • $\text{opp}\to 0\Rightarrow \theta \to 0°$ Why this step? Sine directly reads the shrinking opposite side, so watching sine is watching the flattening.
2. Adjacent from Pythagoras. With hypotenuse $1$: $\text{adj}=\sqrt{1-{\text{opp}}^2}$ (from 2.4.02-pythagorean-theorem). As $\text{opp}\to 0$, $\text{adj}\to 1$. Why this step? We need the adjacent to see cosine's limit.
3. Cosine's limit. $\cos \theta =\frac{\text{adj}}{1}\to 1$ as $\theta \to 0°$. Why this step? A degenerate flat triangle has its adjacent leg equal to the whole hypotenuse.

Verify: $\cos 0°=1$, $\sin 0°=0$ — the endpoints of every quadrant table. A "triangle" with a zero side isn't really a triangle, so we treat $0°$ as the limit it approaches, not a triangle you can draw. ✓

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Cell F — angle pushed to the extremes $0°$ and $90°$

1. At $0°$: opposite $\to 0$, adjacent $\to$ hypotenuse. So $\sin 0°=0$, $\cos 0°=1$, and $\tan 0°=\frac{0}{\text{adj}}=0$. Why this step? Reading the flat triangle directly gives all three at once.
2. At $90°$: opposite $\to$ hypotenuse, adjacent $\to 0$. So $\sin 90°=1$, $\cos 90°=0$. Why this step? The tall triangle mirrors the flat one with opposite/adjacent swapped.
3. The blow-up. $\tan 90°=\frac{\text{opp}}{\text{adj}}=\frac{\text{(nonzero)}}{0}$ — division by zero. So $\tan 90°$ is undefined; as $\theta \to 90°$, $\tan \theta \to +\infty$. Why this step? Tangent's denominator is the adjacent leg, which vanishes at $90°$ — that's precisely when tangent explodes.

Verify: table $\sin 0=0,\cos 0=1,\tan 0=0,\sin 90°=1,\cos 90°=0$ ✓. The vertical asymptote of $\tan$ at $90°$ is exactly the spike you'll meet in 3.2.01-sine-cosine-graphs.

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Cell G — word problem in disguise

1. Set unknowns. Let $h$ be the cliff height and $x$ be the horizontal distance from the nearer point to the cliff base. The far point is $x+40$ away. Why this step? Two unknowns need two equations — the two angle readings supply them.
2. Two tangents. Height is opposite each ground-angle; the horizontal distances are adjacent → TOA twice: $\tan 47°=\frac{h}{x},\tan 28°=\frac{h}{x+40}.$ Why this step? Both triangles share the vertical height $h$; tangent avoids the (unknown) slanted sightlines.
3. Eliminate $x$. From the first, $x=\frac{h}{\tan 47°}$. Substitute into the second: $\tan 28°=\frac{h}{\frac{h}{\tan 47°}+40}.$ Solving for $h$: $h=\frac{40\tan 28°\tan 47°}{\tan 47°-\tan 28°}\approx \frac{40(0.5317)(1.0724)}{1.0724-0.5317}\approx 42.2\text{ m}.$ Why this step? Clearing the fraction and grouping $h$ isolates it; the $40$ (our measured walk) is what sets the scale.

Verify: back-check with the near point: $x=\frac{42.2}{\tan 47°}\approx 39.4\text{ m}$, then $\tan 28°=\frac{42.2}{39.4+40}=\frac{42.2}{79.4}\approx 0.531$ ✓ ($\tan 28°=0.5317$). Units: $\text{m}$. ✓

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Cell H — exam twist: two triangles sharing a side

1. Left triangle → find the shared side $v$. Have adjacent $6$ and angle $50°$, want opposite $v$ → TOA: $\tan 50°=\frac{v}{6}\Rightarrow v=6\tan 50°\approx 6(1.1918)=\mathrm{7.15.}$ Why this step? Tangent links the known adjacent to the wanted opposite in one move.
2. Carry $v$ across. The same strut is opposite the $30°$ angle in the right triangle. Why this step? Shared-side problems always chain through the one length both triangles own.
3. Right triangle → find $b$. Have opposite $v$ and angle $30°$, want adjacent $b$ → TOA rearranged: $\tan 30°=\frac{v}{b}\Rightarrow b=\frac{v}{\tan 30°}=\frac{7.15}{0.5774}\approx \mathrm{12.4.}$

Verify: $b\approx 12.4>6$ ✓ (shallower $30°$ → longer base, as forecast). Cross-check $v$: $\tan 30°\cdot b=0.5774(12.4)\approx 7.15=v$ ✓.

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Recall Quick self-test (reveal after guessing)

Which cell is each? "Given hyp and angle, find a leg" ::: Cell A (sine) "Given both legs, find the angle" ::: Cell C (inverse tangent) "What is $\tan 90°$?" ::: Undefined — adjacent leg is zero, tangent → $\infty$ (Cell F) "Angle from opposite and hypotenuse" ::: Cell D (inverse sine)

Related: 2.4.04-unit-circle-definition extends these ratios beyond $90°$; 3.2.01-sine-cosine-graphs shows the whole story as curves.