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AFCAT Numerical Ability · Heights and Distances

Two Sightings of One Height: Same Side, Opposite Sides and Speed

One height seen from two points: write each distance as the height times the cotangent of its angle, then subtract the distances for points on the same side or add them for opposite sides.

All Heights and Distances notes

Why this matters

Most questions here are MODERATE and a quarter are HARD: an observer walks towards a tower, two ships sit on either side of a lighthouse, or a plane moves between two sightings. Marks are lost on the sign (same side subtracts, opposite sides add) and on the last step, where the answer has to be rationalised to match the options.

Concept 1 of 4: Two points on the same side

Walking towards a tower raises the angle of elevation. Each point gives its own triangle, and the two triangles share the tower. Write both distances in terms of the height hh: the far one is hcot⁡αh\cot\alpha, the near one hcot⁡βh\cot\beta. The distance walked is their difference, which gives one equation in hh.

Definition

  • Far point: angle α\alpha, distance hcot⁡αh\cot\alpha. Near point: angle β\beta (larger), distance hcot⁡βh\cot\beta.
  • Distance walked d=h(cot⁡α−cot⁡β)d = h(\cot\alpha - \cot\beta), so h=dcot⁡α−cot⁡βh = \dfrac{d}{\cot\alpha - \cot\beta}.
  • Two shadows of one pole at two sun elevations work the same way: the difference of the shadows is the dd.
  • Rationalise the end: a3−1=a(3+1)2\dfrac{a}{\sqrt3 - 1} = \dfrac{a(\sqrt3 + 1)}{2}.

Height from the distance walked

h=dcot⁡α−cot⁡βh = \dfrac{d}{\cot\alpha - \cot\beta}
h = 10√3 m30°60°ABwalked 20 m10 mh cot 30° minus h cot 60° = 20 m

Worked example

From a point on level ground the top of a tower is at an elevation of 30°. After walking 20 m towards it, the elevation is 60°. Find the height of the tower.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Heights and Distances · Two Angles from Two Points

A man observes a church spire at 30°30° elevation. After moving 5050 m closer, elevation becomes 45°45°. Find spire height.

The far point has the smaller angle

Walking towards a tower raises the angle. If the angle falls, the observer moved away, and the distance walked is still the difference of the two distances.

Do not stop at the first distance

Solving for the distance from the near point and giving it as the height is right only at 45°. At any other angle multiply by the tangent.
Drill 4 more on two points on the same side

Concept 2 of 4: The same set-up with letters

When the angles are letters, the method does not change. The distances are hcot⁡αh\cot\alpha and hcot⁡βh\cot\beta, and their difference is the distance walked. So any question that gives the walk in terms of hh is asking for cot⁡α−cot⁡β\cot\alpha - \cot\beta.

Definition

  • Distance from the foot at elevation θ\theta: hcot⁡θ=htan⁡θh\cot\theta = \dfrac{h}{\tan\theta}.
  • Walk of dd: cot⁡α−cot⁡β=dh\cot\alpha - \cot\beta = \dfrac{d}{h}. If dd is a fraction of hh, the answer is that fraction.
  • Shadows pp and qq of a pole ss tall: p−q=s(cot⁡α−cot⁡β)p - q = s(\cot\alpha - \cot\beta).
  • To match options in tangents: cot⁡α−cot⁡β=tan⁡β−tan⁡αtan⁡αtan⁡β\cot\alpha - \cot\beta = \dfrac{\tan\beta - \tan\alpha}{\tan\alpha\tan\beta}.

The cotangent difference

cot⁡α−cot⁡β=dh\cot\alpha - \cot\beta = \dfrac{d}{h}

Worked example

A tower is hh m tall. From a point A its top is at elevation α\alpha. After walking h3\tfrac{h}{3} towards it, the elevation is β\beta. Find cot⁡α−cot⁡β\cot\alpha - \cot\beta, and cot⁡β\cot\beta when α=45∘\alpha = 45^\circ.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Heights and Distances · Two Angles from Two Points

The height of a tower is hh and the angle of elevation of the top of the tower is α\alpha. When observer on moving a distance h/2h/2 towards the tower, the angle of elevation becomes β\beta. What is the value of cot⁡α−cot⁡β\cot\alpha - \cot\beta?

Distances use cotangent, not tangent

The distance is hcot⁡θh\cot\theta. Subtracting tan⁡α−tan⁡β\tan\alpha - \tan\beta gives a number with no meaning as a length.

Convert to the options' form

1tan⁡α−1tan⁡2α=tan⁡2α−tan⁡αtan⁡αtan⁡2α\tfrac{1}{\tan\alpha} - \tfrac{1}{\tan 2\alpha} = \tfrac{\tan 2\alpha - \tan\alpha}{\tan\alpha\tan 2\alpha}. The right option often prints this form, with the subtraction in the right order on top.
Drill 3 more on the same set-up with letters

Concept 3 of 4: Two points on opposite sides

When the two points are on either side of the tower, east and west, or two ships on either side of a lighthouse, the tower stands between them. The distance between the points is the sum of the two distances, not the difference.

Definition

  • Distance between the points =hcot⁡α+hcot⁡β=h(cot⁡α+cot⁡β)= h\cot\alpha + h\cot\beta = h(\cot\alpha + \cot\beta).
  • Given the distance between them, h=Dcot⁡α+cot⁡βh = \dfrac{D}{\cot\alpha + \cot\beta}.
  • Angles of depression from the top work the same way, since each equals the elevation from the ship.
  • Complementary angles: if the two elevations add to 90°, their tangents multiply to 1, because tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ - \theta) = \cot\theta.

Opposite sides add

D=h(cot⁡α+cot⁡β)D = h(\cot\alpha + \cot\beta)

Worked example

A tower 90 m tall is seen from two points on opposite sides of it, at elevations of 45° and 60°. How far apart are the points?
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Heights and Distances · Two Angles from Two Points

Suppose the angle of elevation of the top of a tree at a point EE due East of the tree is 60∘60^\circ and that at a point FF due West of the tree is 30∘30^\circ. If the distance between the points EE and FF is 160 ft, then what is the height of the tree?

Opposite sides add, same side subtracts

Read for 'opposite sides', 'east and west' or 'on either side' before writing the equation. Subtracting here gives the option for points on the same side.

Both depressions are measured from one top

Two angles of depression from one lighthouse both use the lighthouse's height. Do not use a different height for each ship.
Drill 5 more on two points on opposite sides

Concept 4 of 4: Speed from two sightings

A plane flying level, or a boat or car moving on the ground, is seen at two angles a known time apart. The height is fixed, so each angle gives a horizontal distance, and the change in that distance is how far the object moved. Distance over time is the speed.

Definition

  • Distance moved =h ∣cot⁡α−cot⁡β∣= h\,|\cot\alpha - \cot\beta| when the object moves along a line through the foot.
  • Speed =distance movedtime= \dfrac{\text{distance moved}}{\text{time}}. To change m/s to km/h multiply by 185=3.6\tfrac{18}{5} = 3.6.
  • A car approaching at a steady speed: time is in proportion to distance, so the time still needed is the time taken ×distance leftdistance covered\times \dfrac{\text{distance left}}{\text{distance covered}}.
  • 'Directly overhead' means an elevation of 90°, a horizontal distance of 0.

Speed

v=h(cot⁡α−cot⁡β)tv = \dfrac{h(\cot\alpha - \cot\beta)}{t}

Worked example

A plane flies level at a height of 1500 m. Its angle of elevation changes from 45° to 30° in 15 seconds. Find its speed.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Heights and Distances · Two Angles from Two Points

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60∘60^\circ. After 10 s, its elevation is observed to be 30∘30^\circ, the uniform speed of the aeroplane in km/h is

Use the change in horizontal distance

The object moves along the ground or along a level flight path. The lines of sight are slant distances; their difference is not the distance moved.

Check the units of the options

A distance in km covered in tt seconds is a speed of 3600t\tfrac{3600}{t} times that many km/h. Options in m/s and km/h differ by a factor of 3.6.
Drill 3 more on speed from two sightings

Summary — formulas & gotchas at a glance

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