How Surveyors Measure a Mountain They Cannot Climb
The summit of Mount Everest was fixed at $8{,}840$ metres in 1856 — by men standing on the plains of India more than $150$ kilometres away, who never set foot on it. They used theodolites to measure angles, a known baseline distance, and trigonometry. That is the whole promise of heights and distances: if you can see the top of something and measure the angle your eye makes with the horizontal, you can compute its height from solid ground. No ladder, no drone, no climb.
What Are Heights and Distances in Trigonometry?
Heights and distances is the application of trigonometric ratios to find the height of a distant object or the horizontal distance to it, using a measured angle and one known length. It is the practical face of trigonometry — often taught under the heading Some Applications of Trigonometry — and it rests entirely on the right triangle formed by the object, the observer, and the ground.
Three terms carry the whole topic, and each must be clear before any problem makes sense.
Line of sight — the straight line from the observer's eye to the object being viewed.
Angle of elevation — the angle the line of sight makes with the horizontal when the object is above the observer (you look up).
Angle of depression — the angle the line of sight makes with the horizontal when the object is below the observer (you look down).
In every case the object, the observer's position, and the foot of the object form a right triangle, and the six trigonometric ratios connect its angle to its sides. That is the engine; the rest is choosing the right ratio.
Angle of elevation versus angle of depression
The single most useful fact about these two angles: the angle of depression from a high point to a low point equals the angle of elevation from the low point back up to the high point. They are alternate angles between two parallel horizontals cut by the same line of sight. So a depression problem can always be redrawn as an elevation problem — handy when the right angle is easier to spot from the ground. The full definitions live in angle of elevation and angle of depression.
How Do You Find Heights and Distances? The Method
Every heights-and-distances problem yields to the same three steps.
Draw the right triangle. Mark the object as the vertical side, the ground as the horizontal side, and the line of sight as the hypotenuse. Label the given angle and the given length.
Pick the ratio that links what you know to what you want. If you know the angle and the base and want the height, use $\tan$ (opposite over adjacent). If you know the angle and the hypotenuse, use $\sin$ or $\cos$.
Substitute the standard value and solve. The angle is almost always $30°$, $45°$, or $60°$, so its ratio is an exact surd you can read off the table.
The standard-angle values you will reach for constantly:
$\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
$30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |
$45°$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | $1$ |
$60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
The full set is in the trigonometric table. The most common formula, by far, is the elevation case:
$$\tan(\text{angle of elevation}) = \frac{\text{height of object}}{\text{horizontal distance}}$$
where the height is the side opposite the angle and the distance is the side adjacent to it.
Examples of Heights and Distances
Example 1
The angle of elevation of the top of a tower from a point $30$ m away on level ground is $30°$. Find the height of the tower.
Draw the right triangle: tower is the opposite side (height $h$), ground distance $30$ m is the adjacent side, angle is $30°$.
$$\tan 30° = \frac{h}{30}$$
$$\frac{1}{\sqrt{3}} = \frac{h}{30} \implies h = \frac{30}{\sqrt{3}} = 10\sqrt{3}$$
Final answer: $h = 10\sqrt{3} \approx 17.32$ m.
Example 2
A $12$ m ladder leans against a wall and makes a $60°$ angle with the ground. How high up the wall does it reach?
A tempting first move is to multiply the ladder length by the angle's tangent — "$12 \times \tan 60° = 12\sqrt{3} \approx 20.8$ m." Pause and sanity-test that against the picture: the ladder is only $12$ m long, so the height it reaches cannot exceed $12$ m. An answer of $20.8$ m is longer than the ladder itself — impossible.
The error is choosing the wrong ratio. The ladder is the hypotenuse, not the adjacent side, so tangent does not apply here. The height is opposite the $60°$ angle, and the side we know is the hypotenuse — that pairing is sine.
$$\sin 60° = \frac{\text{height}}{12}$$
$$\frac{\sqrt{3}}{2} = \frac{\text{height}}{12} \implies \text{height} = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$$
Final answer: $6\sqrt{3} \approx 10.39$ m — comfortably less than the $12$ m ladder, as it must be.
Example 3
A bird sits on top of a tree. From a point $40$ m from the base of the tree, the angle of elevation of the bird is $45°$. How high is the bird?
At $45°$ the triangle is isosceles — opposite and adjacent are equal.
$$\tan 45° = \frac{h}{40}$$
$$1 = \frac{h}{40} \implies h = 40$$
Final answer: the bird is $40$ m up. Whenever the angle of elevation is exactly $45°$, the height equals the ground distance — a quick check worth remembering.
Example 4
From the top of a $50$ m cliff, the angle of depression of a boat at sea is $30°$. How far is the boat from the foot of the cliff?
The angle of depression from the cliff top equals the angle of elevation from the boat, so the $30°$ sits at the boat, between the horizontal and the line of sight up to the observer. The cliff height ($50$ m) is opposite that angle; the distance $d$ is adjacent.
$$\tan 30° = \frac{50}{d}$$
$$\frac{1}{\sqrt{3}} = \frac{50}{d} \implies d = 50\sqrt{3}$$
Final answer: the boat is $50\sqrt{3} \approx 86.6$ m from the foot of the cliff.
Example 5
Two poles of equal height stand on opposite sides of an $80$ m wide road. From a point between them on the road, the angles of elevation of the tops are $60°$ and $30°$. Find the height of the poles and the position of the point.
Let the point be $x$ m from the foot of the first pole, so it is $(80 - x)$ m from the second. Both poles have height $h$. Set up one equation per pole, each on its own line.
From the first pole ($60°$):
$$\tan 60° = \frac{h}{x} \implies h = x\sqrt{3}$$
From the second pole ($30°$):
$$\tan 30° = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}}$$
Set the two expressions for $h$ equal:
$$x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$$
$$3x = 80 - x \implies 4x = 80 \implies x = 20$$
Then $h = 20\sqrt{3} \approx 34.64$ m.
Final answer: each pole is $20\sqrt{3} \approx 34.64$ m tall, and the point is $20$ m from the first pole (and $60$ m from the second).
Example 6
A person standing on the ground finds the angle of elevation of the top of a building to be $45°$. On walking $20$ m toward the building, the angle becomes $60°$. Find the height of the building.
Let the height be $h$ and the closer distance be $x$. After walking $20$ m, the person is $x$ m from the base; before, they were $(x + 20)$ m away. Two equations, one per position.
At $60°$ (closer):
$$\tan 60° = \frac{h}{x} \implies h = x\sqrt{3}$$
At $45°$ (farther):
$$\tan 45° = \frac{h}{x + 20} \implies h = x + 20$$
Set equal and solve:
$$x\sqrt{3} = x + 20$$
$$x(\sqrt{3} - 1) = 20 \implies x = \frac{20}{\sqrt{3} - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1)$$
Then:
$$h = x\sqrt{3} = 10\sqrt{3}(\sqrt{3} + 1) = 10(3 + \sqrt{3}) = 30 + 10\sqrt{3}$$
Final answer: $h = 30 + 10\sqrt{3} \approx 47.32$ m.
Why Heights and Distances Matter Beyond the Exam
Heights and distances exists because there are countless objects we need to measure but cannot reach — and an angle plus a known length is almost always cheaper to obtain than a direct measurement.
Surveying and mapping. Every contour line on a topographic map, every road gradient, every plot boundary traces back to angle-and-baseline triangulation — the same method that fixed Everest's height from the plains.
Construction and safety. Crane operators, scaffolders, and ladder-safety codes all work from elevation angles; a ladder set at too shallow an angle slips, too steep tips back.
Navigation and aviation. A pilot's glide path, a ship's distance to a lighthouse, an air-traffic controller's read on an aircraft's altitude — all are depression-angle computations.
The destination this points toward is the wider field of applications of trigonometry: once you can turn one angle and one length into a height, the same triangle solves distance-across-a-river, satellite positioning, and the parallax that measures the distance to nearby stars.
Where Students Slip on Heights and Distances
Mistake 1: Confusing the angle of depression with the wrong horizontal
Where it slips in: Depression problems, where the observer is at the top and the angle is measured at the top.
Don't do this: Placing the depression angle at the foot of the object, or between the line of sight and the vertical instead of the horizontal.
The correct way: The angle of depression is at the observer's eye, measured down from the horizontal. The cleanest fix is to use its equality with the angle of elevation from the object back up — that moves the known angle to the bottom of the triangle, where the right angle is easy to see. The step that looks skippable but isn't: redrawing the depression as an elevation before reaching for a ratio.
Mistake 2: Picking tan when the hypotenuse is the known side
Where it slips in: Ladder, wire, and slant-distance problems, where the given length runs along the line of sight.
Don't do this: Reaching for $\tan = \frac{\text{opposite}}{\text{adjacent}}$ out of habit when the length you actually know is the hypotenuse.
The correct way: Match the ratio to the two sides involved — opposite-and-hypotenuse means $\sin$, adjacent-and-hypotenuse means $\cos$, opposite-and-adjacent means $\tan$. The first-instinct error is defaulting to tangent on every problem; always read off which two sides the question connects before choosing.
Mistake 3: Leaving the surd unsimplified or mishandling the units
Where it slips in: Final answers like $\frac{30}{\sqrt{3}}$, and problems mixing metres with the height of the observer.
Don't do this: Reporting $\frac{30}{\sqrt{3}}$ as a final height, or forgetting to add the observer's eye-height when the problem gives it.
The correct way: Rationalise the surd ($\frac{30}{\sqrt{3}} = 10\sqrt{3}$), and re-read the problem for any height of the observer that must be added to the triangle's result.
Key Takeaways
Heights and distances uses one measured angle and one known length to find an unreachable height or distance through a right triangle.
The angle of elevation is measured up from the horizontal; the angle of depression is measured down — and for the same two points they are equal.
The method is always: draw the triangle, choose the ratio that links the known and unknown sides, substitute the standard value, solve.
$\tan\theta = \frac{\text{height}}{\text{distance}}$ is the workhorse formula; use $\sin$ or $\cos$ whenever the hypotenuse is the known side.
The same triangle that fixed Everest's height from the plains powers surveying, navigation, and construction today.
Practice Before Moving On
Draw the triangle for each before you compute.
The angle of elevation of the top of a $15$ m pole from a point on the ground is $30°$. How far is the point from the foot of the pole?
From the top of a $20$ m building, the angle of depression of a car is $45°$. How far is the car from the building?
A kite is flying at a height of $60$ m. The string makes a $60°$ angle with the ground. Find the length of the string (assume it is straight).
Answer to Question 1: $\tan 30° = \frac{15}{d}$, so $d = 15\sqrt{3} \approx 25.98$ m. Answer to Question 2: at $45°$ the distance equals the height, so $20$ m. Answer to Question 3: $\sin 60° = \frac{60}{\text{string}}$, so string $= \frac{60}{\sqrt{3}/2} = 40\sqrt{3} \approx 69.28$ m — re-check the sine step if you didn't land here.
To work through tower, pole, and observer-walks problems with a teacher who builds the diagram-first habit, explore Bhanzu's trigonometry tutor, the high school math tutor track, or math tutoring online.
Want a Bhanzu trainer to coach your child through the elevation-and-depression method until the triangle draws itself? Book a free demo class.
Read More
Trigonometric ratios of specific angles — the exact $30°$, $45°$, $60°$ values these problems use.
Sin cos tan — the three ratios that solve every elevation and depression triangle.
Trigonometric functions — the functions behind the ratios, extended to any angle.
Trigonometry formulas — the full formula reference for trigonometry.
Pythagoras theorem — the right-triangle result that underpins the side lengths.
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