Worked examples — Applications — heights and distances problems

This page is the drill hall. The parent note taught the single move — turn an angle into a length with a right triangle. Here we hunt down every kind of problem this topic can hand you, so no exam question can surprise you.

Before we start, one promise: every symbol is earned. We treat three tools as known — read the trig ratios note if any are new:

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The scenario matrix

Every heights-and-distances problem is one (or a mix) of these cells. We will hit all of them.

Cell	What makes it different	Example that hits it
A. Find height	angle + ground distance known, find vertical	Ex 1
B. Find distance	height known, find horizontal	Ex 2
C. Angle of depression	looking down from a height	Ex 2
D. Two positions (observer moves)	two angles, unknown base	Ex 3
E. Stacked objects (two angles, one point)	pole on building	Ex 4
F. Special-angle limits (45°, and the 0°/90° edges)	degenerate / limiting behaviour	Ex 5
G. Non-special angle (real survey)	must use a calculator value	Ex 6
H. Word problem / real world	translate English → triangle	Ex 7
I. Exam twist (ill-posed check)	data may not be consistent	Ex 8

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[!example] Example 1 — Cell A: find the height

Statement: You stand $40\text{ m}$ from a mast. The angle of elevation to its top is $60°$. How tall is the mast?

Forecast: $60°$ is a steep look-up. Guess: is the mast taller or shorter than $40\text{ m}$? Write your guess down.

Figure s02 — Example 1. The navy base is labelled $40\text{ m}$; the orange arc at the eye is $60°$; the violet vertical leg is the unknown $h$ we chase; the magenta hypotenuse is the (unused) line of sight. Notice the violet leg is clearly taller than the navy base — the picture predicts $h>40$ before any arithmetic.

1. Identify the sides. The ground $40\text{ m}$ is next to the $60°$ angle → adjacent. The height $h$ is across from the angle → opposite. Why this step? Naming sides relative to the angle is the whole game; get it wrong and you pick the wrong ratio.
2. Pick tan. We have adjacent, want opposite — that is exactly $\tan$. $\tan 60°=\frac{h}{40}$ Why this step? $\sin$ and $\cos$ drag in the hypotenuse (the line of sight), which we neither know nor need.
3. Solve. $\tan 60°=\sqrt{3}$, so $h=40\sqrt{3}\approx 69.28\text{ m}.$ Why this step? Multiply both sides by $40$ to free $h$.

Verify: A steep $60°$ should make the height bigger than the base. $69.28>40$ ✓. Units: metres × (pure ratio) = metres ✓.

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[!example] Example 2 — Cells B + C: angle of depression, find distance

Statement: From the top of a $90\text{ m}$ lighthouse, the angle of depression to a boat is $45°$. How far is the boat from the foot of the lighthouse?

Forecast: At exactly $45°$, something special happens. Guess the distance before reading on.

Figure s03 — Example 2. The violet vertical is the $90\text{ m}$ lighthouse. A dashed navy horizontal juts out from the top; the orange arc between that dashed line and the magenta line of sight is the $45°$ depression. Follow the magenta sight down to the boat (magenta dot) sitting distance $d$ along the navy ground. The dashed top-horizontal and the solid ground are parallel, so the depression at the top equals the elevation at the boat — that is the trick the figure makes visible.

1. Bring the angle down to the boat. The angle of depression at the top equals the angle of elevation at the boat — they are alternate interior angles between two parallel horizontals (see parent note and angle facts). Why this step? It relocates the $45°$ into a triangle whose right angle sits at the lighthouse foot.
2. Name sides from the boat's angle. Height $90\text{ m}$ is opposite; distance $d$ is adjacent. $\tan 45°=\frac{90}{d}$ Why this step? Opposite-over-adjacent again → $\tan$.
3. Solve. $\tan 45°=1$, so $d=90\text{ m}$. Why this step? At $45°$ the two legs are equal — the triangle is isosceles.

Verify: $\tan 45°=1$ means height = distance, so $d=90$ ✓. Sanity: a moderate $45°$ gives distance comparable to height ✓.

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[!example] Example 3 — Cell D: observer walks closer

Statement: A man sees a tower top at $30°$. He walks $30\text{ m}$ towards it; the angle rises to $60°$. Find the tower height $h$.

Forecast: Walking closer makes the tower look taller (steeper). Will $h$ be more or less than the $30\text{ m}$ he walked?

Figure s04 — Example 3. One violet tower of height $h$ on the right. Two magenta/orange lines of sight fan out to its top from two eye positions on the navy ground: the far one makes the shallow $30°$, the near one (after walking $30\text{ m}$, shown by the navy arrow labelled "walk 30 m") makes the steep $60°$. Seeing both triangles share the same violet height is exactly why we can link them into two equations.

1. First position, distance $x$. $\tan 30°=\frac{h}{x}⟹x=h\sqrt{3}.(1)$ Why this step? We know only the angle, so we tie the unknown distance to the unknown height.
2. Second position, distance $x-30$. $\tan 60°=\frac{h}{x-30}⟹x-30=\frac{h}{\sqrt{3}}.(2)$ Why this step? Same tower height $h$, shorter base — one shared unknown links the two.
3. Subtract (2) from (1) to kill $x$. $30=h\sqrt{3}-\frac{h}{\sqrt{3}}=h\cdot \frac{3-1}{\sqrt{3}}=\frac{2h}{\sqrt{3}}.$ Why this step? Eliminating $x$ leaves a single equation in $h$.
4. Solve. $h=30\cdot \frac{\sqrt{3}}{2}=15\sqrt{3}\approx 25.98\text{ m}.$

Verify: Height ($25.98$) is comparable to the walk ($30$) ✓. Back-check the base: $x=h\sqrt{3}=45$, and $x-30=15$; then $\tan 60°=25.98/15=1.732=\sqrt{3}$ ✓.

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[!example] Example 4 — Cell E: pole stacked on a building

Statement: From a ground point, the elevation to the top of a building is $30°$ and to the top of a flagpole on it is $45°$. The building is $30\text{ m}$ tall. Find the flagpole height $f$.

Forecast: At $45°$, total height equals distance. Keep that in your pocket.

Figure s05 — Example 4. On the right stands a violet building with a magenta flagpole stacked on top of it. From one eye point on the navy ground two sight lines rise: the orange one to the building top ($30°$) and the magenta one to the pole top ($45°$). The ground leg $d$ is shared by both triangles — that shared base is what lets one measurement ($d$) unlock the other height ($f$).

1. Angle to building top ($30°$). $\tan 30°=\frac{30}{d}⟹d=30\sqrt{3}.(1)$ Why this step? This lone equation fixes the ground distance $d$.
2. Angle to flagpole top ($45°$). $\tan 45°=\frac{30+f}{d}⟹30+f=d.(2)$ Why this step? $\tan 45°=1$, so the whole height equals the distance.
3. Substitute (1) into (2). $30+f=30\sqrt{3}⟹f=30(\sqrt{3}-1)\approx 21.96\text{ m}.$

Verify: Total height $30+f=30\sqrt{3}\approx 51.96$, which must equal $d$ ✓. Since the look-angle grew ($30°\to 45°$) as we scanned upward, total height beats the building's $30\text{ m}$ ✓.

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[!example] Example 5 — Cell F: the limiting / degenerate edges

Statement: Explore what $h=d\tan \theta$ does at the edges: $\theta \to 0°$, $\theta =45°$, and $\theta \to 90°$, for a fixed base $d=50\text{ m}$.

Forecast: One of these gives $0$, one gives $50$, one blows up. Predict which is which.

Figure s06 — Example 5. This one is a graph, not a triangle: the horizontal axis is the angle $\theta$ in degrees, the vertical axis is the computed height $h=50\tan \theta$. The magenta curve starts flat at $0$, passes through the orange dot at $(45°,50)$ where the violet dashed line marks $h=d=50$, then rockets upward towards the navy dotted asymptote at $90°$. The picture is the argument: gentle near $0$, balanced at $45$, unbounded at $90$.

1. $\theta \to 0°$ (object on the horizon). $\tan 0°=0⟹h=0$. Why this step? A flat line of sight sees nothing above ground — degenerate zero height.
2. $\theta =45°$ (balance point). $\tan 45°=1⟹h=50\text{ m}=d$. Why this step? Equal legs — the "height equals distance" landmark you keep meeting.
3. $\theta \to 90°$ (looking straight up). $\tan \theta \to \infty$, so $h\to \infty$. Why this step? The line of sight becomes vertical; the top is directly overhead and the model's height runs off to infinity along the tangent's asymptote — not exponential growth, just the tangent's own blow-up.

Verify: $\tan 0°=0$, $\tan 45°=1$, and $\tan 89°\approx 57.29$ giving $h\approx 2864.6\text{ m}$ — already enormous, confirming the runaway ✓.

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[!example] Example 6 — Cell G: a real survey (non-special angle)

Statement: A surveyor measures the elevation to a hill top as $52°$ from a point $200\text{ m}$ from its base (see surveying). Find the height to the nearest metre.

Forecast: $52°$ is past $45°$, so expect height greater than $200\text{ m}$.

1. Set up. Adjacent $=200$, opposite $=h$. $h=200\tan 52°.$ Why this step? Nothing special about $52°$ — we just read $\tan 52°$ from a calculator ($\approx 1.2799$).
2. Compute. $h=200\times 1.2799\approx 255.98\approx 256\text{ m}.$ Why this step? Rounding to the nearest metre matches survey precision.

Verify: $52°>45°$ so $h>200$ ✓ ($256>200$). Reverse-check with the undo button: $\arctan (256/200)=\arctan (1.28)\approx 52.0°$ ✓ (this uses $\arctan$ exactly as defined at the top — the angle whose tangent is $1.28$).

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[!example] Example 7 — Cell H: pure word problem

Statement: A $1.6\text{ m}$-tall girl looks up at a kite. Her eyes make a $30°$ elevation with the horizontal, and the string (line of sight) is taut at $120\text{ m}$. How high is the kite above the ground?

Forecast: The line of sight is the hypotenuse here (not a leg!), and don't forget to add her eye height.

Figure s07 — Example 7. The magenta hypotenuse is the $120\text{ m}$ string running from the girl's eye up to the kite (magenta dot). The orange arc at the eye is the $30°$ elevation, the violet vertical is the rise from eye-level to the kite, and the navy horizontal is the ground run. The note "eye 1.6 m up" reminds you the whole triangle floats $1.6\text{ m}$ above the ground, so you must add that back at the end.

1. Choose the ratio. We know the hypotenuse ($120$) and want the vertical opposite side. Opposite-over-hypotenuse is $\sin$ (defined at the top of this page). $\sin 30°=\frac{\text{vertical rise}}{120}.$ Why this step? This is the one time $\tan$ is wrong — we were handed the hypotenuse, so $\sin$ fits.
2. Rise above eye level. $\sin 30°=\frac{1}{2}$, so rise $=120\times \frac{1}{2}=60\text{ m}.$ Why this step? That is height above her eyes, not the ground.
3. Add eye height. Ground height $=60+1.6=61.6\text{ m}.$ Why this step? Her eyes start $1.6\text{ m}$ up; the triangle only measured from there.

Verify: Rise ($60$) must be less than the string ($120$) since $\sin <1$ ✓. Add-on gives $61.6\text{ m}$ ✓. (By Pythagoras the horizontal run is $\sqrt{120^2-60^2}=103.9\text{ m}$, all consistent.)

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[!example] Example 8 — Cell I: the exam twist (spot the missing information)

Statement: "A bird flying at a constant but unknown height is seen at an angle of depression $45°$ from the top of a $60\text{ m}$ tower. After the bird flies $100\text{ m}$ horizontally away, the depression is $30°$. Find the bird's height above the ground." Solve — or prove the numbers describe no real situation.

Forecast: Two angles look like enough, but the bird has its own unknown altitude. Guess: how many unknowns are really in play?

Figure s08 — Example 8. A violet tower of height $60\text{ m}$ on the left. From its top a dashed navy horizontal runs right; two orange depression arcs ($45°$ then $30°$) open below it to two magenta sight-lines reaching the bird at its first and second positions (magenta dots), which sit at the same violet dashed altitude $H_b$. The navy arrow between the two bird positions is the $100\text{ m}$ flight. Follow the figure while reading the steps: the vertical side of each triangle is the orange gap $60-H_b$ from the tower top down to the bird's altitude — not the full tower — and the horizontals $a$ and $a+100$ are the adjacent sides.

1. Count the unknowns honestly. Let the bird's constant height above the ground be $H_b$, and let the horizontal distance from the tower to the bird's first position be $a$. The vertical drop from the tower top to the bird is $60-H_b$ (the bird is below the top, since we look down). Why this step? Depression is measured against the tower-top horizontal (the dashed navy line in Figure s08), so the opposite side of each angle is this drop $60-H_b$, not the full tower height. Naming the drop correctly is what the flawed shortcut skips. We now have two unknowns, $H_b$ and $a$, so we will need two equations.
2. First position ($45°$). In the near triangle of Figure s08, opposite $=60-H_b$, adjacent $=a$: $\tan 45°=\frac{60-H_b}{a}⟹60-H_b=a.(1)$ Why this step? Opposite over adjacent is $\tan$; $\tan 45°=1$ makes the drop equal the near horizontal.
3. Second position ($30°$), horizontal $a+100$. In the far triangle, the vertical is the same orange gap $60-H_b$ (constant altitude), adjacent is now $a+100$: $\tan 30°=\frac{60-H_b}{a+100}⟹a+100=(60-H_b)\sqrt{3}.(2)$ Why this step? The bird's altitude did not change, so the opposite side is identical — only the base grew by the $100\text{ m}$ flight shown by the navy arrow.
4. Substitute (1) into (2) to eliminate $a$. Replace $a$ by $60-H_b$ from (1): $(60-H_b)+100=(60-H_b)\sqrt{3}.$ Why this step? Now the only unknown is $H_b$ — this is exactly where the tower's given $60\text{ m}$ pins the third quantity down. Without that number the equation could not close.
5. Isolate the drop $60-H_b$. Move the $(60-H_b)$ terms together: $100=(60-H_b)\sqrt{3}-(60-H_b)=(60-H_b)(\sqrt{3}-1).$ Why this step? Factoring out $(60-H_b)$ leaves a single linear equation.
6. Solve for the drop, then the height. $60-H_b=\frac{100}{\sqrt{3}-1}=50(\sqrt{3}+1)\approx 136.6\text{ m},$ $H_b=60-136.6=-76.6\text{ m}.$ Why this step? A negative height is geometrically impossible — the bird would be underground.

Verify: The algebra is sound but the answer is negative, so these numbers describe no real situation. Reading Figure s08 explains why: a $45°$-then-$30°$ depression over a $100\text{ m}$ flight forces the orange drop to be $136.6\text{ m}$, yet the violet tower supplies only $60\text{ m}$ — the sight lines can never fall that far. Lesson: count unknowns first (here $H_b$ and $a$), use the given length ($60\text{ m}$) to close the system, and if the solved height comes out negative (or the required drop exceeds the tower), the problem is ill-posed — reject it rather than report nonsense.

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[!recall]- Rapid self-check

Related: similar triangles explain why a fixed angle fixes the shape and lets one measured side scale all the others.