Heights and Distances — SSC CGL Quantitative Aptitude

This is trigonometry with one tool: tan θ = opposite ÷ adjacent = height ÷ base. Every question is a right triangle with a known angle (30°, 45° or 60°) and one known side. Draw it, label the angle of elevation or depression, and solve for the missing side. The only "hard" version stacks two such triangles — same height, two angles — and even that is two lines of algebra.

1. What SSC actually asks

Tier 1: ~1 Q · Tier 2: 1 Q. Types: height from an angle of elevation and a distance, distance from an angle of depression, the two-position problem (angle changes as you walk toward/away), sun-and-shadow angles, and equal poles on opposite sides of a road. Always right triangles, always the standard angles.

2. The core setup

• Angle of elevation: you look up at a top; the angle is at your eye, between the horizontal and the line of sight.
• Angle of depression: you look down from a height; it equals the elevation from the object (alternate angles).

$\tan \theta =\frac{\text{height}}{\text{base}}\Rightarrow \text{height}=\text{base}\times \tan \theta ,\text{base}=\frac{\text{height}}{\tan \theta }$

The standard angles — the only ones SSC uses — are:

$\tan 30^{\circ }=\frac{1}{\sqrt{3}},\tan 45^{\circ }=1,\tan 60^{\circ }=\sqrt{3}$

So a 45° elevation means height = base — the single most common shortcut.

3. The two-position problem (drill this)

As you walk a distance $d$ toward a tower, the elevation rises from 30° to 60°. Same height $h$, two bases:

• Far base (30°): $h/\tan 30^{\circ }=h\sqrt{3}$.
• Near base (60°): $h/\tan 60^{\circ }=h/\sqrt{3}$.
• Difference $=h\sqrt{3}-\frac{h}{\sqrt{3}}=\frac{2h}{\sqrt{3}}=d\Rightarrow h=\frac{d\sqrt{3}}{2}$.

Walking 40 m gives $h=20\sqrt{3}\approx 34.6$ m. Memorise the pattern: 30°→60° over distance $d$ ⇒ $h=\frac{d\sqrt{3}}{2}$.

4. Depression and shadows

• From a height: the angle of depression to a point equals the elevation from that point, so distance $=h/\tan \theta$. A 60 m tower with a car at 30° depression: distance $=60/\tan 30^{\circ }=60\sqrt{3}$ m.
• Sun and shadow: if a tower's shadow is $\sqrt{3}$ times its height, then $\tan \theta =\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ }$. Longer shadow ⇒ lower sun ⇒ smaller angle.

5. Equal poles on opposite sides

Two equal poles flank a road of width $w$; from a point between them the elevations are 60° and 30°. Let the point be $x$ from the 60° pole:

$x\sqrt{3}=(w-x)\cdot \frac{1}{\sqrt{3}}\Rightarrow 3x=w-x\Rightarrow x=\frac{w}{4},h=\frac{w\sqrt{3}}{4}.$

For an 80 m road, $h=20\sqrt{3}$ m and the point is 20 m from the taller-angle pole.

6. Solved PYQ-style examples

Q1. The angle of elevation of the top of a tower from a point 30 m away is 45°. The height is… Solution. $h=30\tan 45^{\circ }=$ 30 m (height = base at 45°).

Q2. The elevation changes from 30° to 60° as an observer walks 40 m toward a tower. Its height is… Solution. $h=\frac{d\sqrt{3}}{2}=\frac{40\sqrt{3}}{2}=$ $20\sqrt{3}$ m ($\approx 34.6$).

Q3. From the top of a 60 m tower, the angle of depression of a car is 30°. The car's distance from the base is… Solution. $60/\tan 30^{\circ }=60\sqrt{3}=$ $60\sqrt{3}$ m ($\approx 103.9$).

Q4. The shadow of a tower is $\sqrt{3}$ times its height. The sun's angle of elevation is… Solution. $\tan \theta =\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}\Rightarrow \theta =$ 30°.

Q5. Two equal poles stand on either side of an 80 m road; from a point between them the elevations are 60° and 30°. The height of each pole is… Solution. $x=20$ m, $h=20\tan 60^{\circ }=$ $20\sqrt{3}$ m.

7. Exam protocol

1. Draw the right triangle first; mark the angle at the eye/observer.
2. Use $\tan \theta$ = height ÷ base; 45° means height = base immediately.
3. Depression from a height = elevation from the ground — use $h/\tan \theta$.
4. Two-position 30°→60° over $d$? Height is $\frac{d\sqrt{3}}{2}$ — reach for it directly.
5. Keep answers in surd form ($20\sqrt{3}$, $60\sqrt{3}$); rationalise only if the options demand a decimal.