Heights and distances

Stand at the foot of a tall tower and try to measure its height with a tape and you will fail — the top is simply out of reach. Trigonometry solves this without ever touching the object: look up at the top, measure the angle your line of sight makes with the horizontal, measure (or already know) one distance along the ground, and a single ratio hands you the height. This is heights and distances, the applied half of the Class 10 trigonometry unit — the same $\sin$ , $\cos$ , $\tan$ you already know, now aimed at towers, cliffs, boats and buildings. Every problem reduces to drawing one clean right-angled triangle (or, at most, two sharing a common side) and picking the ratio that connects what you know to what you want. This article extends the tower-and-lighthouse examples from your syllabus notes into a fuller toolkit: the two-point elevation method for when you cannot walk right up to a tower's foot, a fully worked derivation, and the traps examiners love to set.

The core insight

Every heights-and-distances problem is a right-angled triangle in disguise, solved by matching the known angle of elevation or depression to the correct ratio among opposite, adjacent and hypotenuse.

The core idea

Elevation and depression are mirror angles. The angle of elevation is measured upward from the horizontal at the observer's eye to an object above; the angle of depression is measured downward from the horizontal to an object below. Because the horizontal at the observer and the horizontal at the object are parallel lines cut by the same line of sight, alternate angles make them equal: the angle of depression from a cliff top to a boat exactly equals the angle of elevation from the boat back up to the cliff top. You are never given both angles separately — geometry hands you the second one free.

One triangle, one ratio. Once the diagram is drawn — horizontal ground, vertical height, line of sight, marked angle — the whole problem collapses to choosing which trigonometric ratio links the side you know to the side you want:

\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height}}{\text{horizontal distance}}, \qquad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

where: $\theta$ is the angle of elevation or depression, "opposite" and "adjacent" are the triangle's legs relative to $\theta$ , and "hypotenuse" is the slanted line of sight. Most heights-and-distances problems give a horizontal distance and want a vertical height (or vice versa), so $\tan$ is used far more often than $\sin$ or $\cos$ .

Two triangles when one point is not enough. Sometimes you cannot stand exactly where the diagram needs you to, or the problem deliberately gives two observation points to test whether you can combine equations. The ICSE syllabus allows at most two right-angled triangles sharing a common vertical side (the unknown height) — this is exactly the case built into the derivation below.

Derivation AHeight of a tower from two points on the same line

The goal is to find the height of a tower when you cannot reach its foot but can stand at two points on the same straight line through the foot, a known distance $d$ apart, and measure the angle of elevation from each — a genuinely two-triangle ICSE problem that cannot be solved from a single right-angled triangle alone. Let the foot of the tower be $O$ , the top be $T$ , the height $OT = h$ , the nearer point $B$ at distance $OB = x$ from the foot with elevation angle $\beta$ , and the farther point $A$ at distance $OA = x + d$ with elevation angle $\alpha$ , where $\alpha < \beta$ because $A$ is farther away.

Step 1. Write the tangent ratio in triangle $OBT$ , right-angled at $O$ . The height $OT$ is opposite $\beta$ and $OB = x$ is adjacent, so $\tan\beta = h/x$ , which gives $x = h/\tan\beta$ .

Step 2. Write the tangent ratio in triangle $OAT$ , right-angled at $O$ . Here $OA = x + d$ is adjacent to $\alpha$ , so $\tan\alpha = h/(x+d)$ , which gives $x + d = h/\tan\alpha$ .

Step 3. Eliminate $x$ by subtracting Step 1 from Step 2. Since $(x+d) - x = d$ , subtracting the two expressions for $x$ and $x+d$ gives $d = \dfrac{h}{\tan\alpha} - \dfrac{h}{\tan\beta} = h\left(\dfrac{1}{\tan\alpha} - \dfrac{1}{\tan\beta}\right)$ . The unknown $x$ has vanished — this is exactly why the "common side" trick works: both triangles share the same height $h$ , so eliminating the other unknown $x$ leaves one equation purely in $h$ .

Step 4. Solve for $h$ . Rearranging, $h = \dfrac{d}{\dfrac{1}{\tan\alpha} - \dfrac{1}{\tan\beta}} = \dfrac{d\,\tan\alpha\,\tan\beta}{\tan\beta - \tan\alpha}$ . This is the general two-point elevation formula: given the walked distance $d$ and the two angles, the height follows directly, with no need to ever find $x$ itself.

Worked example

Standing at a point $A$ on level ground, the angle of elevation of the top of a tower is $30^\circ$ . Walking $40$ m closer, in a straight line towards the tower, to a point $B$ , the angle of elevation becomes $60^\circ$ . Find the height of the tower.

Step 1. Match the problem to the derivation. Here $\alpha = 30^\circ$ (farther point $A$ ), $\beta = 60^\circ$ (nearer point $B$ ), and $d = 40$ m, the distance walked between them.

Step 2. Substitute into $h = \dfrac{d\,\tan\alpha\,\tan\beta}{\tan\beta - \tan\alpha}$ , using $\tan 30^\circ = \dfrac{1}{\sqrt3}$ and $\tan 60^\circ = \sqrt3$ : the product $\tan\alpha\tan\beta = \dfrac{1}{\sqrt3}\times\sqrt3 = 1$ , and the difference $\tan\beta - \tan\alpha = \sqrt3 - \dfrac{1}{\sqrt3} = \dfrac{3-1}{\sqrt3} = \dfrac{2}{\sqrt3}$ .

Step 3. Combine: $h = \dfrac{40 \times 1}{2/\sqrt3} = \dfrac{40\sqrt3}{2} = 20\sqrt3 \approx 34.64$ m.

Step 4. Sanity-check. At the nearer point $B$ , the distance to the foot should be $x = h/\tan 60^\circ = 20\sqrt3/\sqrt3 = 20$ m; at the farther point $A$ , the distance is $x + d = 20 + 40 = 60$ m, and indeed $h/\tan 30^\circ = 20\sqrt3 \times \sqrt3 = 60$ m — the two triangles agree. The tower is approximately 34.6 m tall.

Common traps

Mixing up which angle belongs to which point. The smaller angle always goes with the farther point and the larger angle with the nearer point — check this before substituting, since swapping $\alpha$ and $\beta$ flips the sign of the answer.
Reaching for $\sin$ or $\cos$ out of habit. If the two known/unknown quantities are a horizontal distance and a vertical height, the ratio is $\tan$ , not $\sin$ or $\cos$ — always name the opposite and adjacent sides first.
Forgetting depression equals elevation. The alternate-angles property means an angle of depression given at the top is numerically the same angle of elevation at the bottom — do not treat it as a separate unknown.
Rounding surds too early. Keep $\sqrt3$ , $\dfrac{1}{\sqrt3}$ and similar exact until the very last step; rounding $\tan 30^\circ$ to $0.577$ partway through a two-triangle problem compounds errors.

Ask the tutor: Want to work through a flagpole-on-a-building or a river-width problem step by step, or check your own two-triangle setup before you submit it? Ask the AI tutor to walk through it with you.

Why does walking closer to a tower increase the angle of elevation?

Picture the right-angled triangle formed by the horizontal distance to the foot and the vertical height. As you walk closer, the horizontal distance (the adjacent side) shrinks while the height (the opposite side) stays fixed, so the ratio opposite over adjacent grows, and since that ratio is $\tan\theta$ , the angle $\theta$ itself must grow. At the extreme, if you stood right at the foot the angle would approach $90^\circ$ ; far away it would approach $0^\circ$ .

Do I always need two right-angled triangles, or is one usually enough?

One triangle is enough whenever a single distance and a single angle are given directly, as in the classic tower-and-30-degree example. You need two triangles only when the problem hands you a second angle from a second point (or a second height, like a flagpole on a building) without directly giving you the distance from either point to the foot — then the two equations are needed together to eliminate that unknown distance, exactly as in the derivation above.

Is it fine to just measure the diagram with a protractor and ruler instead of calculating?

No — ICSE heights-and-distances problems are meant to be solved exactly using trigonometric ratios and standard angle values (or a trigonometric table for non-standard angles), not read off a scale drawing. A diagram is essential for setting up the triangle correctly, but the final height or distance must come from the ratio equation, kept in exact surd form until the last step, then rounded only in the final answer.

Heights and distances

The core idea

Worked example

Common traps

Before this

Where this leads