Heights and Distances — chapter notes

The angle of elevation is the angle between the horizontal through the observer's eye and the line of sight to an object above eye level. The angle of depression is the angle between that horizontal and the line of sight to an object below eye level. Both are measured from the horizontal, never from the vertical.

Because the horizontal through the observer is parallel to the ground, the angle of depression of an object equals the angle of elevation of the observer (alternate angles).

Example: from a $45$ m deck a jogger is seen at a depression of $25^\circ$, so the angle at the jogger is $25^\circ$ inside the right triangle.

At $30^\circ$, $45^\circ$ and $60^\circ$ use exact values: $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, $\tan 45^\circ = 1$, $\tan 60^\circ = \sqrt{3}$, and the sines and cosines from the table. For other angles, read the value from the tables, which are rounded to four decimal places, for example $\tan 35^\circ = 0.7002$.

An answer found with table values is approximate. Say how you rounded: "correct to the nearest $0.1$ m".

Example: with $\tan 40^\circ = 0.8391$, $18\tan 40^\circ = 15.1$ m.

Draw the figure, mark the angle from the horizontal, and name opposite, adjacent and hypotenuse for that angle.

• height and ground distance: $\tan\theta = \dfrac{\text{height}}{\text{distance}}$;
• a rope, string or line of sight (hypotenuse): $\text{height} = \text{slant}\times\sin\theta$ and $\text{distance} = \text{slant}\times\cos\theta$;
• an unknown distance with a known height: $\text{distance} = \dfrac{\text{height}}{\tan\theta}$.

Example: a taut kite string of $80$ m at $30^\circ$: height $80\sin 30^\circ = 40$ m.

The angle is measured at the observer's eye. If the eye is $e$ m above the ground, the right triangle gives the height above eye level, and

Keep surds or extra decimals in the working, and round once, at the end, to the precision asked. A table-based answer and an exact answer agree after rounding.

Example: $e = 1.6$ m, distance $30$ m, $\theta = 40^\circ$: $1.6 + 30\times 0.8391 = 26.8$ m.

Two points $P$ and $Q$ in line with the foot $F$ of a tower, on the same side, give two right triangles with the same height $h$. With $QF$ nearer, elevations $\alpha$ at $P$ and $\beta$ at $Q$, and $PQ = d$:

Example: $d = 24$, $\alpha = 30^\circ$, $\beta = 45^\circ$ gives $h = 12(\sqrt{3} + 1) = 32.8$ m. A walker who moves $24$ m towards the tower is the same problem: $24$ m is a gap, not the distance to the foot.

Opposite sides. From a height $h$ the angles of depression to two points on opposite sides of the foot are $\alpha$ and $\beta$. By alternate angles the distances from the foot are $h\cot\alpha$ and $h\cot\beta$, and they add:

An object on top. From a point $x$ from the foot, the bottom and the top of the object are seen at $\alpha$ and $\beta$. Its length is $x\tan\beta - x\tan\alpha$.

Example: $48$ m, depressions $60^\circ$ and $45^\circ$: $AB = 16\sqrt{3} + 48 = 75.7$ m.

1. Draw a labelled figure; mark the horizontal and the angle.
2. Find the shared side (height or distance).
3. Same side: subtract the distances. Opposite sides: add them.
4. Include the eye height if the observer is not at ground level.
5. Check the answer is sensible; give units and the stated precision.

Doubling the distance does not halve the angle: use the ratio.

Not in the ICSE 2027 syllabus: problems involving more than two right-angled triangles, and three-dimensional problems. Every problem in this chapter uses at most two right-angled triangles in one vertical plane.