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SSC CGL Quantitative Aptitude · Formula sheet

Heights and Distances formulas

5 formulas and 14 common traps for SSC CGL Quantitative Aptitude Heights and Distances, grouped by subtopic.

Full notes with worked examples

One Line of Sight

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Height and ground distance from one angle

Height and distance from the angle of elevation

h=dtan⁡θd=htan⁡θtan⁡θ=hdh = d \tan\theta \qquad d = \frac{h}{\tan\theta} \qquad \tan\theta = \frac{h}{d}
  • hhthe height of the object
  • ddthe distance along the ground to its foot, or the length of its shadow
  • θθthe angle of elevation (or of depression, seen from the top)

A ladder, rope or wire as the slant side

Slant side, height and ground distance

h=Lsin⁡θd=Lcos⁡θL=hsin⁡θ=dcos⁡θh = L\sin\theta \qquad d = L\cos\theta \qquad L = \frac{h}{\sin\theta} = \frac{d}{\cos\theta}
  • LLthe slanting length: ladder, rope, wire or ramp
  • θθthe angle the slant side makes with the ground

Common traps

Measuring the angle of depression from the vertical

The angle of depression is between the horizontal and the line of sight. A depression of 30° means the line of sight makes 60° with the vertical, and using 60° gives the wrong triangle.

Swapping the 30° and 60° ratios

tan⁡60∘=3\tan 60^\circ = \sqrt{3} and tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}. A higher sun means a larger angle and a shorter shadow, so at 60° the shadow is shorter than the height.

Expecting an exact answer when the options are rounded

SSC sometimes asks for an approximate distance and prints 5003500\sqrt{3} as 866. Work exactly, then multiply by 1.732 at the end.

Using tan when the slant length is given

When the length of the ladder or rope is known, it is the hypotenuse, so the ratio is sin (for the height) or cos (for the ground distance). Tan links height and ground distance only.

Reading the angle with the wall as the angle with the ground

A ladder at 30° to the wall is at 60° to the ground. Its height up the wall is then Lsin⁡60∘L\sin 60^\circ, not Lsin⁡30∘L\sin 30^\circ.

Rounding too early when the answer is to the nearest metre

525\sqrt{2} is about 7.07, so it rounds to 7. Rounding 2\sqrt{2} to 1.5 first gives 7.5, which rounds to 8 and picks the wrong option.

Two Angles of Elevation or Depression

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Two observers on opposite sides of a tower

Two observers on opposite sides

gap=htan⁡α+htan⁡β\text{gap} = \frac{h}{\tan\alpha} + \frac{h}{\tan\beta}
  • α,β\alpha, \betathe two angles of elevation, one from each side
  • hhthe height of the tower

Two observers on the same side of a tower

Two observers on the same side

gap=htan⁡α−htan⁡β(α<β)\text{gap} = \frac{h}{\tan\alpha} - \frac{h}{\tan\beta} \qquad (\alpha < \beta)
  • α\alphathe smaller angle, seen from the farther point
  • β\betathe larger angle, seen from the nearer point

Complementary angles from two points

Complementary angles of elevation

θ1+θ2=90∘⇒h2=ab\theta_1 + \theta_2 = 90^\circ \Rightarrow h^2 = ab
  • a,ba, bthe distances of the two points from the foot

Common traps

Subtracting the distances for observers on opposite sides

On opposite sides the foot of the tower lies between the observers, so their distances add. Subtracting is only for two observers on the same side.

Leaving a surd in the bottom when matching options

3003+1\frac{300}{\sqrt{3} + 1} is printed as 150(3−1)150(\sqrt{3} - 1). Multiply top and bottom by 3−1\sqrt{3} - 1, since (3+1)(3−1)=2(\sqrt{3} + 1)(\sqrt{3} - 1) = 2.

Giving the larger angle to the farther point

As you walk towards a tower its top rises in your view, so the nearer point has the larger angle. Swap them and the distance walked comes out negative.

Using the 30° and 60° shortcut for other angles

'The far point is 3 times as far' holds only for 30° and 60°. For 30° and 45° the ratio is 3:1\sqrt{3} : 1, so write both distances in terms of hh and subtract.

Treating two printed forms as different answers

1033−1\frac{10\sqrt{3}}{\sqrt{3} - 1} and 5(3+3)5(3 + \sqrt{3}) are the same number. Rationalise before you compare options, or compare their decimal values.

Taking the average of the two distances

For complementary angles the height is ab\sqrt{ab}, the geometric mean, not a+b2\frac{a + b}{2}. From 4 m and 16 m the height is 8 m, not 10 m.

Giving the building the height of the drop

From a tower top, the depression to a building's top gives how far that top is below you. The building's height is the tower's height minus that drop.

Using the wrong angle for the shared distance

Find the ground distance from the triangle whose height you already know: the pedestal, the building or the tower. The other angle only gives a height once that distance is known.

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