Q.Find the height of a given tower using mathematical modelling.

Let the observer's eye be at height $h$ and horizontal distance $l$ from the tower, and $\alpha$ the angle of elevation of the top. In the right triangle at eye level,

If the base is unreachable, measure the angle of depression $\beta$ of the foot: $\tan\beta = h/l$, so $l = h\cot\beta$ and $H = h(1 + \tan\alpha\,\cot\beta)$.

Step 1 — Identify the situation. We want the height $H$ of a tower we cannot climb, so it must be found indirectly by measuring angles.

Step 2 — Set up variables and assumptions. Let $AB$ be the tower with foot $B$ and top $A$. An observer stands with his eye at $P$, at height $PQ = h$ above the ground, at horizontal distance $l$ from the tower. Let $C$ be the point on the tower at the observer's eye level, so $PC = QB = l$ and $AC = H - h$. Assume the ground is horizontal, the tower is vertical, and the angles are read with a sextant.

Step 3 — Mathematical formulation. Let $\alpha$ be the angle of elevation of the top $A$ from the eye $P$. In right triangle $PCA$,

Step 4 — Solve. If the base is accessible, $l$ is measured directly and $(1)$ gives $H$ at once. If the base is not accessible, let $\beta$ be the angle of depression from $P$ to the foot $B$. In right triangle $PQB$,

Substituting into $(1)$,

Step 5 — Interpret and validate. The quantities $h$ (eye height), $\alpha$ and $\beta$ are all measurable from the ground, so $H$ is fully determined without ever touching the tower. Since $h$, $l$, $\alpha$, $\beta$ are known exactly, no further refinement of assumptions is needed; the model can be checked against a tower of known height.

Method: Modelling an Inaccessible Height with Angles of Elevation/Depression

This method applies whenever a real-world quantity (like a tower's height) cannot be measured directly and must be found from angles measured at ground level.

Steps

Step 1: Identify what can and cannot be measured directly

Recognise that the tower's height $H$ cannot be measured directly, but an observer's eye height $h$ and certain angles (of elevation/depression) can be measured with an instrument. This decides what the model's inputs will be.

Step 2: Set up the geometry — a right triangle from eye level to the top

Draw the vertical tower, the observer's eye at height $h$, and the horizontal distance $l$ to the tower. The segment from eye level to the top of the tower, together with the horizontal and vertical legs, forms a right triangle.

Step 3: Translate the angle of elevation into an equation

If $\alpha$ is the angle of elevation of the top from the eye,

Step 4: Handle the case where the base is inaccessible

If $l$ cannot be measured directly (the foot of the tower is unreachable), introduce the angle of depression $\beta$ to the foot and use the second right triangle:

Substitute this expression for $l$ back into the Step 3 formula.

Step 5: Interpret the final model

Express $H$ purely in terms of measurable quantities ($h$, $\alpha$, and either $l$ or $\beta$), and check that every symbol in the final formula is something you could actually read off an instrument in the field.

Common Mistakes

Mistake 1: Forgetting to add back the observer's eye height $h$

Why it's wrong: The right triangle formed by the line of sight only gives $H - h$ (the height of the tower above eye level), not the tower's full height $H$. Stopping at $l\tan\alpha$ and forgetting the "$+h$" undercounts the height by exactly the observer's eye level. Correct approach: always write $H = h + l\tan\alpha$, not $H = l\tan\alpha$.

Mistake 2: Confusing angle of elevation with angle of depression

Why it's wrong: The angle of elevation $\alpha$ (looking up to the top) and the angle of depression $\beta$ (looking down to the foot) are measured from the same eye position but relate to different right triangles and different tangent ratios ($\tan\alpha = (H-h)/l$ versus $\tan\beta = h/l$). Mixing them up gives a formula that doesn't match the geometry. Correct approach: keep each angle tied to its own triangle and ratio before combining them.

Mistake 3: Using $\tan\beta$ instead of $\cot\beta$ when eliminating $l$

Why it's wrong: From $\tan\beta = h/l$, solving for $l$ gives $l = h/\tan\beta = h\cot\beta$ — a common slip is to instead write $l = h\tan\beta$, inverting the ratio. Correct approach: solve $\tan\beta = h/l$ for $l$ algebraically (cross-multiply) rather than guessing which trig ratio to use.