Exercises — Applications — heights and distances problems

This page is your self-test. Read each problem, try it on paper, THEN open the collapsible solution. Problems climb five rungs: L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery. Every solution is complete — no "left as an exercise".

Before you start, make sure you're comfortable with the parent note Applications — heights and distances problems and the raw ratios in 2.4.01-Trigonometric-ratios. A few problems lean on 2.4.03-Complementary-angles, 1.3.07-Pythagoras-theorem and 3.2.04-Similar-triangles; one echoes real fieldwork from 2.5.02-Surveying-and-navigation.

Look at the triangle above. The angle $\theta$ lives at the observer's eye. The side straight across from $\theta$ (never touching it) is the opposite side — here the height. The side that runs along the angle to the right angle is the adjacent side — here the ground distance. The slanted side facing the right angle is the hypotenuse — the line of sight. Keep this labelling in front of you.

Exact values you'll reuse (memorise these three angles):

$\theta$	$\sin \theta$	$\cos \theta$	$\tan \theta$
$30°$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}\approx 0.577$
$45°$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$
$60°$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}\approx 1.732$

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Level 1 — Recognition

Can you spot the triangle and pick the right ratio?

Exercise L1.1

A pole is 12 m tall. From a point on the ground the angle of elevation to its top is $45°$. How far is the point from the base of the pole?

Recall Solution

What we know: height (opposite the angle) $=12$, angle $=45°$, want the ground distance $d$ (adjacent). Why $\tan$: we have opposite, we want adjacent — the two legs. That's exactly $\tan$. $\tan 45°=\frac{\text{opposite}}{\text{adjacent}}=\frac{12}{d}$ Since $\tan 45°=1$: $1=\frac{12}{d}⟹d=12\text{ m}$ Sanity check: at $45°$ the triangle is isosceles — height equals distance. $12=12$. ✓

Exercise L1.2

From the top of a lighthouse the angle of depression to a boat is $60°$. In the right triangle formed by the lighthouse height, the sea-level distance, and the line of sight to the boat, which side is opposite the $60°$ angle measured at the boat, and which is adjacent?

Recall Solution

Key idea (from the parent note): the angle of depression at the top equals the angle of elevation at the boat — they are alternate interior angles across the two horizontal (parallel) lines cut by the line of sight.

Placing the angle at the boat (elevation $=60°$):

• Opposite the $60°$ angle = the lighthouse height (straight up, not touching the boat's angle).
• Adjacent to the $60°$ angle = the sea-level distance (runs along the water from the boat to the base). No arithmetic needed — this is pure recognition. Getting this right is what makes every later problem trivial.

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Level 2 — Application

Plug into $\tan$ and solve cleanly.

Exercise L2.1

A student stands 40 m from a tower. The angle of elevation to the top is $60°$. Find the height of the tower.

Recall Solution

Setup: adjacent (ground) $=40$, want opposite (height) $h$, angle $=60°$. Why $\tan$: we KNOW the ground leg (40 m, adjacent) and WANT the vertical leg (height, opposite). Two legs, no line of sight involved — that is precisely what $\tan$ relates. Reaching for $\sin$ or $\cos$ would drag in the unknown hypotenuse for nothing. $\tan 60°=\frac{h}{40}⟹h=40\tan 60°=40\sqrt{3}$ $h=40\times 1.732\approx 69.28\text{ m}$ Sanity check: the angle is steep ($>45°$), so the height should exceed the distance. $69.28>40$. ✓

Exercise L2.2

From the top of a 90 m cliff the angle of depression to a fishing boat is $30°$. How far is the boat from the foot of the cliff?

Recall Solution

Setup: depression $30°$ = elevation $30°$ at the boat. Height (opposite) $=90$, want distance (adjacent) $d$. Why $\tan$: we KNOW the vertical leg (90 m, opposite) and WANT the horizontal leg (distance, adjacent). Again two legs meet at the angle — the ratio that ties them together is $\tan$, so no hypotenuse is needed. $\tan 30°=\frac{90}{d}⟹d=\frac{90}{\tan 30°}=90\sqrt{3}$ $d=90\times 1.732\approx 155.88\text{ m}$ Sanity check: shallow angle ($<45°$) → the boat is far, distance bigger than height. $155.88>90$. ✓

Exercise L2.3

A ladder 10 m long leans against a wall, making an angle of $60°$ with the ground. How high up the wall does it reach?

Recall Solution

Careful: here we DO know the slanted side — the ladder is the hypotenuse ($=10$). We want the vertical reach, which is opposite the $60°$ angle. Why $\sin$ not $\tan$: we have hypotenuse and want opposite → that's $\sin$. $\sin 60°=\frac{\text{height}}{10}⟹\text{height}=10\sin 60°=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}$ $\text{height}=5\times 1.732\approx 8.66\text{ m}$ Sanity check: the reach must be less than the ladder length. $8.66<10$. ✓

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Level 3 — Analysis

Two conditions, one unknown to eliminate.

Exercise L3.1

A man observes the top of a tower at an elevation of $30°$. He walks 30 m directly toward the tower; the elevation becomes $60°$. Find the height of the tower.

Recall Solution

Let height $=h$, first distance $=x$, second distance $=x-30$. First position ($30°$): $\tan 30°=\frac{h}{x}⟹x=h\sqrt{3}(1)$ Second position ($60°$): $\tan 60°=\frac{h}{x-30}⟹x-30=\frac{h}{\sqrt{3}}(2)$ Why subtract (2) from (1): it kills $x$, leaving only $h$. $x-(x-30)=h\sqrt{3}-\frac{h}{\sqrt{3}}$ $30=h\left(\frac{3-1}{\sqrt{3}}\right)=h\cdot \frac{2}{\sqrt{3}}$ $h=30\times \frac{\sqrt{3}}{2}=15\sqrt{3}\approx 25.98\text{ m}$ Sanity check: he walked 30 m and the height ($\approx 26$ m) is comparable — moving closer made the angle jump $30°\to 60°$, exactly as expected. ✓

Exercise L3.2

The angle of elevation of a cloud from a point 60 m above a lake is $30°$, and the angle of depression of its reflection in the lake is $60°$. Find the height of the cloud above the lake surface.

Recall Solution

The trick — the reflection is as far below the water as the cloud is above. Let the cloud be $H$ above the lake; the observer sits 60 m above the lake. Let horizontal distance from observer to the cloud's vertical line $=d$.

• Cloud is $H-60$ above the observer’s eye level ⇒ $\tan 30°=\frac{H-60}{d}⟹d=(H-60)\sqrt{3}(1)$
• Reflection is $H$ below the lake, so it is $H+60$ below the eye ⇒ $\tan 60°=\frac{H+60}{d}⟹d=\frac{H+60}{\sqrt{3}}(2)$ Set (1) = (2) to eliminate $d$: $(H-60)\sqrt{3}=\frac{H+60}{\sqrt{3}}$ Multiply both sides by $\sqrt{3}$: $3(H-60)=H+60⟹3H-180=H+60⟹2H=240$ $H=120\text{ m}$ Sanity check: the depression to the reflection ($60°$) is steeper than the elevation to the cloud ($30°$) because the reflection is genuinely lower (further below) than the cloud is high. Consistent. ✓

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Level 4 — Synthesis

Combine two heights on one vertical line, or two triangles sharing a base.

Exercise L4.1

From a point on the ground, the elevation to the bottom of a flagpole standing on a building is $30°$, and to the top of the flagpole is $45°$. The building is 30 m tall. Find the length of the flagpole.

Recall Solution

Let flagpole length $=f$, ground distance $=d$. To the top of the building (30 m), $30°$: $\tan 30°=\frac{30}{d}⟹d=30\sqrt{3}(1)$ To the top of the flagpole ($30+f$), $45°$: $\tan 45°=\frac{30+f}{d}⟹30+f=d(2)$ Substitute (1) into (2): $30+f=30\sqrt{3}⟹f=30(\sqrt{3}-1)=30\times 0.732\approx 21.96\text{ m}$ Sanity check: at $45°$, top height $=$ distance $=30\sqrt{3}\approx 51.96$ m, which indeed exceeds the 30 m building. ✓

Exercise L4.2

A tower and a building stand on opposite sides of a straight 80 m road. From the base of the tower the angle of elevation of the building's top is $45°$; from the base of the building the angle of elevation of the tower's top is $60°$. Find the height of the tower, and use the $45°$ reading to state the building's height as a cross-check.

Recall Solution

Two triangles share the same 80 m horizontal base (the road width). Elevation of tower top from building base ($60°$): the whole road (80 m) is adjacent, tower height $t$ is opposite. $\tan 60°=\frac{t}{80}⟹t=80\sqrt{3}\approx 138.56\text{ m}$ Now use the $45°$ reading (the second, opposite-facing triangle): looking from the tower base across the same 80 m road to the building top, the road is adjacent and the building height $b$ is opposite. $\tan 45°=\frac{b}{80}⟹b=80\times 1=80\text{ m}$ Why this cross-check matters: the two elevations are read in opposite directions across one fixed gap. Confirming $b=80$ m (an $45°$ isosceles triangle: height = road width) proves we assigned the shared 80 m base correctly to both triangles — if we had mistakenly used two different distances, this clean $b=80$ would not appear. Sanity check: the tower is seen at a steep $60°$ across the same 80 m, so it should be taller than the road is wide. $138.56>80$. ✓

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Level 5 — Mastery

Full modelling, exact + decimal, and a consistency check like a real surveyor.

Exercise L5.1

The angle of elevation of the top of a tower from a point $A$ on the ground is $30°$. On walking 40 m toward the tower to a point $B$, the elevation becomes $45°$. Find (a) the height of the tower and (b) the distance from $B$ to the tower base — all in exact surd form, then to 2 decimals.

Recall Solution

Let height $=h$, distance from $B$ to base $=y$, so distance from $A$ $=y+40$. At $B$ ($45°$): $\tan 45°=\frac{h}{y}\Rightarrow y=h(1)$ At $A$ ($30°$): $\tan 30°=\frac{h}{y+40}\Rightarrow y+40=h\sqrt{3}(2)$ Why substitute (1) into (2): equation (1) tells us $y$ is just $h$ in disguise, so replacing it collapses two unknowns into one — leaving a single equation in $h$ alone. $h+40=h\sqrt{3}⟹40=h\sqrt{3}-h=h(\sqrt{3}-1)⟹h=\frac{40}{\sqrt{3}-1}$ Why rationalise: a surd in the denominator is hard to read and to approximate, so multiply top and bottom by the conjugate $\sqrt{3}+1$ to clear it. $h=\frac{40(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{40(\sqrt{3}+1)}{3-1}=\frac{40(\sqrt{3}+1)}{2}=20(\sqrt{3}+1)$ $h=20(\sqrt{3}+1)\approx 20\times (1.732+1)=20\times 2.732\approx 54.64\text{ m}$ (b) From (1), $y=h$, so the distance from $B$ to the base is $y=20(\sqrt{3}+1)\approx 54.64\text{ m}.$ Sanity check: at $45°$ from $B$ the triangle is isosceles, so the height must equal the distance from $B$ — indeed $h=y$. ✓ And $A$ is 40 m further out with a smaller angle ($30°$), consistent with a longer, shallower line of sight. ✓

Exercise L5.2 (surveyor's consistency check)

A drone is spotted at elevation $60°$ from a ground station. It flies horizontally away at constant height; the elevation drops to $30°$ after the drone covers a horizontal distance reported as 100 m by its GPS. Assuming the height stays fixed, is the GPS figure geometrically consistent? If so, what is the drone's height?

Recall Solution

Let height $=h$, first horizontal distance $=d_1$, second $=d_2$. $\tan 60°=\frac{h}{d_1}\Rightarrow d_1=\frac{h}{\sqrt{3}},\tan 30°=\frac{h}{d_2}\Rightarrow d_2=h\sqrt{3}$ The geometry forces the horizontal travel to be $d_2-d_1=h\sqrt{3}-\frac{h}{\sqrt{3}}=h\cdot \frac{3-1}{\sqrt{3}}=\frac{2h}{\sqrt{3}}.$ Answering the consistency question FIRST: the two angles do NOT over-determine anything on their own — for any height $h$ there exists a matching travel $\frac{2h}{\sqrt{3}}$. So the pair "$60°\to 30°$ plus travel $=100$ m" is consistent provided a single positive $h$ satisfies $\frac{2h}{\sqrt{3}}=100$. Solve it: $\frac{2h}{\sqrt{3}}=100⟹h=\frac{100\sqrt{3}}{2}=50\sqrt{3}\approx 86.60\text{ m}.$ Since this gives one clean positive height, YES — the 100 m GPS reading is geometrically consistent, and the drone's height is $50\sqrt{3}\approx 86.60$ m. (Back-check: $d_1=50$, $d_2=150$, difference $=100$. ✓) Contrast (from the parent's steel-man): if instead the height were pinned independently and the angles were $45°\to 30°$, the travel would be $h(\sqrt{3}-1)\approx 0.73h$, which would generally clash with a raw "100 m" claim. Always recompute the implied displacement and compare — that is exactly the field discipline of 2.5.02-Surveying-and-navigation.

Recall Quick recall — pick your ratio

Two legs (vertical + horizontal), no slant known ::: use $\tan$ Hypotenuse (line of sight / ladder) and the opposite side ::: use $\sin$ Hypotenuse and the adjacent side ::: use $\cos$ Angle of depression from the top equals... ::: the angle of elevation from the bottom (alternate interior angles)

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Answer key (exact / decimal)

• L1.1 $=12$ m · L1.2 opposite = height, adjacent = sea distance
• L2.1 $=40\sqrt{3}\approx 69.28$ m · L2.2 $=90\sqrt{3}\approx 155.88$ m · L2.3 $=5\sqrt{3}\approx 8.66$ m
• L3.1 $=15\sqrt{3}\approx 25.98$ m · L3.2 $=120$ m
• L4.1 $=30(\sqrt{3}-1)\approx 21.96$ m · L4.2 tower $=80\sqrt{3}\approx 138.56$ m, building $=80$ m
• L5.1 $h=20(\sqrt{3}+1)\approx 54.64$ m, $y=20(\sqrt{3}+1)\approx 54.64$ m · L5.2 $=50\sqrt{3}\approx 86.60$ m (consistent)