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AFCAT Numerical Ability · Heights and Distances

Two Heights from One Point: Flagstaffs, Two Buildings and Reflections

Two triangles share one horizontal distance: find that shared side first from the triangle you know, then use it in the other.

Why this matters

This is the chapter's hardest page: more than half its questions are HARD. A flagstaff stands on a tower, one building is seen from the top or a window of another, or a cloud is seen together with its reflection in a lake. Marks go when the shared distance is not found first, or when the observer's own height is left out at the end.

Concept 1 of 3: A flagstaff on a tower, seen from one point

From one point on the ground you see the foot of a flagstaff (the top of the tower) at one angle and the top of the flagstaff at a larger angle. Both triangles have the same horizontal side, the distance to the tower. Write both heights with that distance and subtract: the difference is the flagstaff.

Definition

  • Tower HH, flagstaff hh, distance dd: H=dtan⁡αH = d\tan\alpha and H+h=dtan⁡βH + h = d\tan\beta.
  • Subtract to get dd, then HH. In letters, H=htan⁡αtan⁡β−tan⁡αH = \dfrac{h\tan\alpha}{\tan\beta - \tan\alpha}.
  • A statue on a pedestal, an antenna on a building or a pillar to be raised are the same picture.
  • If the flagstaff and the tower are said to subtend equal angles at a point, the whole height subtends twice the angle: use tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}.

Two elevations, one distance

H=dtan⁡α,H+h=dtan⁡βH = d\tan\alpha, \qquad H + h = d\tan\beta

Worked example

A statue 8 m tall stands on a pedestal. From a point on the ground, the elevation of the top of the statue is 60° and of its feet is 30°. How tall is the pedestal?
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The same idea in a practice question:

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Example 1 · Heights and Distances · Two Buildings, Flagstaffs and Reflections

A radio transmitter antenna of height 100 m stands at the top of a tall building. At a point on the ground, the angle of elevation of bottom of the antenna is 45∘45^\circ and that of top of antenna is 60∘60^\circ. What is the height of the building?

The flagstaff is a difference

The top angle gives the tower plus the flagstaff, not the flagstaff alone. Subtract the tower's height before answering.

One distance for both angles

Both elevations are taken from the same point, so they share dd. Giving each angle its own unknown distance leaves too many unknowns.

Concept 2 of 3: Two buildings across a gap

An observer in one building looks across at another. The gap between them is the horizontal side of both triangles. If the observer sees the foot of the other building (depression) and its top (elevation), the depression fixes the gap from the observer's height, and the elevation then gives the part above the observer. If two observers stand at the foot and the top of one building, both look at one top across the same gap.

Definition

  • From a window ww high: the depression ϕ\phi of the foot gives the gap g=wcot⁡ϕg = w\cot\phi; the elevation θ\theta of the top gives gtan⁡θg\tan\theta above the window.
  • Height of the other building =w+gtan⁡θ= w + g\tan\theta.
  • From the foot (elevation β\beta) and the top (elevation α\alpha) of a building bb tall: H=gtan⁡βH = g\tan\beta and H−b=gtan⁡αH - b = g\tan\alpha.
  • Find the gap first from whichever triangle has two known parts.

Height seen from a window

H=w+wcot⁡ϕtan⁡θH = w + w\cot\phi\tan\theta

Worked example

From a window 15 m above the street, the top of the house opposite is at an elevation of 60° and its foot at a depression of 45°. How tall is the house opposite?
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The same idea in a practice question:

Practice questionModerate

Example 2 · Heights and Distances · Two Buildings, Flagstaffs and Reflections

The angle of elevation of the top of a tower from the bottom of a building is twice that from its top. What is the height of the building, if the height of the tower is 75 m and the angle of elevation of the top of the tower from the bottom of the building is 60∘60^\circ?

Add the observer's height back

The elevation gives only the part of the other building above the observer's eye. The building's height is that part plus the height of the window or roof.

The depression finds the gap

Start with the angle of depression of the foot, which has two known parts (the observer's height and the angle). Starting with the elevation leaves two unknowns.

Concept 3 of 3: A cloud and its reflection, a balloon and its size

Two set-ups need one extra fact each. A still lake works as a mirror: the reflection of a cloud HH above the water appears HH below it. A round balloon has width: the two lines from the eye that just touch it make an angle, and the line to its centre cuts that angle in half.

Definition

  • Reflection: from a point hh above the water, the cloud is H−hH - h above the eye and its reflection H+hH + h below it, both at the same horizontal distance xx.
  • So H−h=xtan⁡(elevation)H - h = x\tan(\text{elevation}) and H+h=xtan⁡(depression)H + h = x\tan(\text{depression}); divide to remove xx.
  • Balloon: if a balloon of radius rr subtends an angle α\alpha at the eye, the radius to a touching point is perpendicular to the line of sight, so the distance from the eye to the centre is rsin⁡(α/2)\dfrac{r}{\sin(\alpha/2)}.
  • The height of the centre is that distance times the sine of the centre's elevation.

Cloud and reflection

H+hH−h=tan⁡(depression)tan⁡(elevation)\dfrac{H + h}{H - h} = \dfrac{\tan(\text{depression})}{\tan(\text{elevation})}

Worked example

A balloon of radius 5 m subtends an angle of 60° at an observer's eye, and the elevation of its centre is 60°. How high is the centre above the eye?
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The same idea in a practice question:

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Example 3 · Heights and Distances · Two Buildings, Flagstaffs and Reflections

A balloon of radius rr makes an angle α\alpha at the eye of an observer and the angle of elevation of its centre is β\beta. The height of its centre from the ground level is given by

The reflection is below the water, not the eye

The reflection is HH below the surface, so it is H+hH + h below an eye hh above the water. Using H−hH - h for both makes the two angles equal.

Halve the subtended angle

A balloon subtending 60° puts the centre at rsin⁡30∘\tfrac{r}{\sin 30^\circ}, not rsin⁡60∘\tfrac{r}{\sin 60^\circ}. The line to the centre splits the angle in two.

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