Visual walkthrough — Applications — heights and distances problems

This page rebuilds the single most important formula of heights and distances — $h=d\tan \theta$ — from absolutely nothing. No triangle, no ratio, no tan is used before you have seen it. Follow the pictures in order; every red arrow, every labelled corner is doing a job.

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Step 1 — Draw the situation as points and lines

WHAT: We turn the real scene into three labelled points.

• $A$ — where you stand (your feet / eye on the ground).
• $C$ — the base of the tower, on the ground.
• $B$ — the top of the tower.

WHY: Maths cannot chew on "a tower and a person." It can chew on points joined by lines. Naming points lets us name the lines between them: $AC$ (the ground stretch), $CB$ (the tower going straight up), and $AB$ (your straight line of sight to the top).

PICTURE: In the figure, the blue line $AC$ is the flat ground you paced out. The green line $CB$ is the tower rising vertically. The orange line $AB$ is your gaze skimming from your eye to the very top.

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Step 2 — Find the right angle (this is the secret ingredient)

WHAT: We mark that the corner at $C$ is exactly $90^{\circ }$.

WHY: A tower stands straight up from flat ground. "Straight up" meeting "flat" is the definition of a right angle. This single fact — the corner at $C$ is square — is what makes all of trigonometry apply. Without it, no ratio below is allowed.

PICTURE: The small square drawn in the corner at $C$ is the universal symbol for "$90^{\circ }$ exactly." The tower ($h$, green) and the ground ($d$, blue) meet like the corner of a sheet of paper.

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Step 3 — Name the angle you actually measured

WHAT: We label the corner at $A$ — where your eye is — with the Greek letter $\theta$ (say "theta"), just a name for "the angle I measured."

WHY: This is the one number your instrument gives you. It sits between the flat ground ($AC$) and your gaze ($AB$). It is called the angle of elevation because your line of sight elevates — rises — above the horizontal. (When you instead look down from a height, the twin angle below the horizontal is the angle of depression — see complementary angles and the parent note for why they match.)

PICTURE: The orange wedge at $A$ between the ground and your gaze is $\theta$. Notice: the bigger $\theta$ is, the steeper your gaze — and the taller the tower must be to force that steep a look.

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Step 4 — Why the shape is locked by $\theta$ alone

WHAT: We claim: once $\theta$ is fixed, the triangle's shape is fixed, even if its size changes.

WHY: Suppose you paced twice as far from the tower but kept the same eye-tilt $\theta$ (impossible with a real tower, but imagine it). You would get a bigger triangle that is an exact scaled-up copy of the small one — same angles, all sides multiplied by the same number. This is the idea of similar triangles: same angles ⇒ same shape ⇒ the ratio of any two sides stays constant.

PICTURE: Two nested triangles share the corner $\theta$. The big one is the small one blown up. Look at $\frac{\text{height}}{\text{ground}}$ in each — both dashed measurements give the same fraction. The tower may be near or far, but $\frac{h}{d}$ depends only on $\theta$.

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Step 5 — Define $\tan \theta$ as "opposite over adjacent"

WHAT: We give the constant fraction from Step 4 its name.

$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{d}$

Reading it term by term:

• $\tan \theta$ — the steepness number of the angle $\theta$. Big when the gaze is steep, small when nearly flat.
• $h$ (numerator) — the opposite side, the vertical rise.
• $d$ (denominator) — the adjacent side, the horizontal run.

WHY tan and not sin or cos? Because tan is the only one of the three built from the two sides we care about — the vertical ($h$, unknown, wanted) and the horizontal ($d$, known). Sin and cos each involve the line of sight $AB$, which we never measured and never needed. Choosing tan avoids dragging in an unknown. (Definitions of all three live in trigonometric ratios.)

PICTURE: The figure isolates the two legs — the green rise $h$ over the blue run $d$ — and shows $\tan \theta$ literally as rise ÷ run, the same "slope" idea you'd feel walking up a ramp.

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Step 6 — Solve for the height

WHAT: We rearrange to get $h$ alone.

Start from Step 5 and multiply both sides by $d$: $\tan \theta =\frac{h}{d}\stackrel{\times d}{\to }h=d\tan \theta$

Term by term in the final formula:

• $h$ — the height we could not reach, now handed to us.
• $d$ — the distance you paced (a plain length, in metres).
• $\tan \theta$ — the pure number belonging to your measured angle; multiply the run by it to get the rise.

WHY multiply by $d$? $h$ was trapped in a fraction being divided by $d$. Multiplying both sides by $d$ frees it — the $d$'s on the right cancel, leaving $h$ alone. Nothing else changes because we did the identical thing to both sides.

PICTURE: The run $d$ (blue) is stretched by the factor $\tan \theta$ to become the rise $h$ (green). Tan is the stretch factor that converts horizontal into vertical.

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Step 7 — The edge cases: what tan does at the extremes

WHAT: We check the formula at the boundary angles so no scenario surprises you.

Case $\theta \to 0^{\circ }$ (you stand infinitely far / tower is a speck): Your gaze is almost flat. Rise is tiny compared to run, so $\tan 0^{\circ }=0$, giving $h=d\cdot 0=0$. A flat gaze means no measurable height — correct.

Case $\theta =45^{\circ }$ (gaze at the halfway tilt): Rise equals run exactly, so $\tan 45^{\circ }=1$ and $h=d$. At $45^{\circ }$, height equals distance — a handy mental checkpoint.

Case $\theta \to 90^{\circ }$ (you stand almost at the base, craning straight up): The run $d$ shrinks toward zero while the rise stays large, so the fraction $h/d$ blows up. $\tan \theta$ grows without bound toward a vertical asymptote at $90^{\circ }$. The formula says: to force a straight-up gaze, the tower would need infinite height — you can never actually reach $90^{\circ }$.

WHY this matters: This growth is not linear (doubling $\theta$ does not double $h$) and not exponential — it is tan's own peculiar rush to infinity near $90^{\circ }$. Knowing it stops you trusting silly outputs when angles get extreme.

PICTURE: The graph of $\tan \theta$ from $0^{\circ }$ to $90^{\circ }$: hugging zero at the left, passing through the height $1$ at $45^{\circ }$, then curving up to shoot past the ceiling as it nears $90^{\circ }$ (dashed red asymptote).

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The one-picture summary

Everything above, compressed: pace the ground $d$, measure the tilt $\theta$, and multiply — the run stretches by $\tan \theta$ into the height $h$.

Recall Feynman retelling — say it back in plain words

I stand on flat ground and mark three dots: my feet ($A$), the tower's base ($C$), its top ($B$). Because the tower rises straight up from flat ground, the corner at $C$ is a perfect right angle — that's what lets me use trig at all. My eyes tilt up by an angle I call $\theta$; the ground beside that angle is adjacent, the tower across from it is opposite. Since the same $\theta$ always makes the same triangle shape, the fraction opposite-over-adjacent is fixed by $\theta$ alone — I name it $\tan \theta$, "the steepness number." Because tan uses only the two sides I care about (and skips the slanted gaze I never measured), it's the right tool. Writing $\tan \theta =h/d$ and freeing $h$ gives $h=d\tan \theta$: tan is just the stretch factor that turns my paced-out horizontal distance into the tower's vertical height. At $0^{\circ }$ the height is nothing, at $45^{\circ }$ height equals distance, and as I crane toward $90^{\circ }$ tan races off to infinity — so a real tower never lets my gaze go fully vertical.

Recall Quick self-test

Why do we use tan and not sin here? ::: Because tan is built only from the two sides we deal with (opposite $h$ and adjacent $d$); sin and cos both use the unmeasured line of sight. In $h=d\tan \theta$, what is $\tan \theta$ physically? ::: A pure stretch factor — multiply the horizontal run $d$ by it to get the vertical rise $h$. What is $h$ when $\theta =45^{\circ }$? ::: Exactly $h=d$, because $\tan 45^{\circ }=1$. Why can $\theta$ never truly reach $90^{\circ }$? ::: $\tan \theta \to \infty$ there; a finite tower can't force a perfectly vertical gaze.

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Connected ideas: 2.4.01-Trigonometric-ratios · 2.4.03-Complementary-angles · 1.3.07-Pythagoras-theorem · 3.2.04-Similar-triangles · 2.5.02-Surveying-and-navigation