SSC CGL Maths · Heights and Distances

Two Angles of Elevation or Depression

Two right triangles that share a side: two observers looking at one tower, or one observer looking at two points, with the shared height or distance tying them together.

All Heights and Distances notes

Why this matters

Ten questions in the bank, all from the 2025 shifts, and every MODERATE row of the chapter is on this page (seven of the ten). Each is two one-angle triangles joined by a side they share, so the method is the same every time: write each ground distance in terms of the height, then add or subtract.

Concept 1 of 4: Two observers on opposite sides of a tower

When the two observers stand on opposite sides of the tower, the tower's foot lies between them. So the gap between them is the sum of their two distances from the foot, and the tower's height is shared by both triangles.

Definition

Write each distance from the height: at angle $\theta$ , distance $= \frac{h}{\tan\theta}$ . So 30° gives $h\sqrt{3}$ , 45° gives $h$ and 60° gives $\frac{h}{\sqrt{3}}$ .
Opposite sides: gap $=$ sum of the two distances. Solve for $h$ if the gap is given, or add if the height is given.
The same works from above: two boats or cars seen from a tower top, a cliff or a plane on opposite sides. Each angle of depression is the angle of elevation from that point.
Fast route: for 30° and 45° the gap is $h(\sqrt{3} + 1)$ ; for 30° and 60° it is $h\sqrt{3} + \frac{h}{\sqrt{3}} = \frac{4h}{\sqrt{3}}$ .

Two observers on opposite sides

\text{gap} = \frac{h}{\tan\alpha} + \frac{h}{\tan\beta}

$\alpha, \beta$ the two angles of elevation, one from each side
hthe height of the tower

Worked example

Two points on the ground lie on opposite sides of a 60 m tower, and the angles of elevation of its top from them are 30° and 60°. How far apart are the points?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 26 Sep 2025, Shift 2 · Q54Moderate

Example 1 · Heights and Distances · Two Angles of Elevation or Depression

The angles of elevation of the top of a cliff from two points on the ground, 200 m apart, and on opposite sides of the cliff, are

30^\circ

and

45^\circ

. What is the height of the cliff?

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

\frac{300}{\sqrt{3} + 1}

is printed as

150(\sqrt{3} - 1)

. Multiply top and bottom by

\sqrt{3} - 1

, since

(\sqrt{3} + 1)(\sqrt{3} - 1) = 2

Drill 3 more on two observers on opposite sides of a tower

Concept 2 of 4: Two observers on the same side of a tower

Walking towards a tower makes the angle of elevation grow. Both points lie on one side of the foot, so the distance walked is the far distance minus the near one, and the height is the same in both triangles.

Definition

The nearer point always sees the larger angle.
Same side: gap $=$ far distance $-$ near distance $= \frac{h}{\tan\alpha} - \frac{h}{\tan\beta}$ , with $\alpha$ the smaller angle.
Or set the near distance as $x$ , write the height twice (once from each angle), and equate.
Fast route for 30° and 60° only: the far point is 3 times as far as the near point, so the gap is twice the near distance and $h = \frac{\sqrt{3}}{2} \times$ gap. For other angle pairs, use the full method.

Two observers on the same side

\text{gap} = \frac{h}{\tan\alpha} - \frac{h}{\tan\beta} \qquad (\alpha < \beta)

$\alpha$ the smaller angle, seen from the farther point
$\beta$ the larger angle, seen from the nearer point

Worked example

A man sees the top of a tower at an elevation of 30°. He walks 50 m straight towards it, and the elevation becomes 45°. Find the height of the tower.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 2 · Q65Moderate

Example 2 · Heights and Distances · Two Angles of Elevation or Depression

A vertical pole of height H stands on the ground. From a point P on the ground, the angle of elevation of the top of the pole is

60^\circ

. From a different point Q, located 10 meters away from point P (directly along the line extending from the base of the pole), the angle of elevation to the top of the pole is

30^\circ

.What is the height of the pole?

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

\sqrt{3} : 1

, so write both distances in terms of

h

and subtract.

Treating two printed forms as different answers

\frac{10\sqrt{3}}{\sqrt{3} - 1}

and

5(3 + \sqrt{3})

are the same number. Rationalise before you compare options, or compare their decimal values.

Drill 2 more on two observers on the same side of a tower

Concept 3 of 4: Complementary angles from two points

If the two angles of elevation add up to 90°, the tangent of one is the cotangent of the other, so the two tangents multiply to 1. That turns the two distances into the height directly.

Definition

From points at distances $a$ and $b$ from the foot, the angles are complementary exactly when $h^2 = ab$ .
So $h = \sqrt{ab}$ : the height is the geometric mean of the two distances. Only the distances from the foot matter, not which side each point is on.
Why: $\tan\theta = \frac{h}{a}$ and $\tan(90^\circ - \theta) = \cot\theta = \frac{h}{b}$ . Since $\cot\theta = \frac{1}{\tan\theta}$ , the two multiply to 1, so $\frac{h}{a} \times \frac{h}{b} = 1$ .

Complementary angles of elevation

\theta_1 + \theta_2 = 90^\circ \Rightarrow h^2 = ab

a, bthe distances of the two points from the foot

Worked example

From two points 4 m and 9 m from the foot of a tower, on a line through its foot, the angles of elevation of its top are complementary. Find the height of the tower.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 1 · Q61Easy

Example 3 · Heights and Distances · Two Angles of Elevation or Depression

A vertical pole TF has its base F on the ground. From a point A, 25 meters from F, the angle of elevation of T is

x^\circ

. From another point B, 75 meters from F (and in the same line as A), the angle of elevation of T is

y^\circ

. If the height of the pole is

25\sqrt{3}

meters, what is the value of

x^\circ + y^\circ

Taking the average of the two distances

For complementary angles the height is

\sqrt{ab}

, the geometric mean, not

\frac{a + b}{2}

. From 4 m and 16 m the height is 8 m, not 10 m.

Concept 4 of 4: One observer reading two angles on the same vertical line

One observer reads two angles to two points on the same vertical line: the bottom and top of a statue, or the top and foot of a building. Both triangles share the same ground distance, so find that distance from one angle and use it in the other.

Definition

Step 1: use the angle whose triangle has a known side to find the shared ground distance.
Step 2: use the other angle with that distance to find the second height.
Step 3: the answer is the difference of the two heights (the statue alone, the flagstaff alone) or what is left (the building is the tower minus the drop to its top).
Looking down from a tower top, the depression to the building's foot gives the distance; the depression to its top gives how far the building's top is below you.

Worked example

From the top of a 60 m tower, the angles of depression of the top and the foot of a building are 30° and 60°. Find the height of the building.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 3 · Q61Moderate

Example 4 · Heights and Distances · Two Angles of Elevation or Depression

A statue that is 1.6 meters in height is positioned on top of a pedestal. From a location on the ground, the angle of elevation to the top of the statue measures 60 degrees, while the angle of elevation to the top of the pedestal from the same point is 45 degrees. Determine the height of the pedestal.

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.

Drill 1 more on one observer reading two angles on the same vertical line

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Two observers on opposite sides of a tower
Two observers on opposite sides
$\text{gap} = \frac{h}{\tan\alpha} + \frac{h}{\tan\beta}$
Two observers on the same side of a tower
Two observers on the same side
$\text{gap} = \frac{h}{\tan\alpha} - \frac{h}{\tan\beta} \qquad (\alpha < \beta)$
Complementary angles from two points
Complementary angles of elevation
$\theta_1 + \theta_2 = 90^\circ \Rightarrow h^2 = ab$

Watch out for (8)

Subtracting the distances for observers on opposite sides
→ Two observers on opposite sides of a tower
Leaving a surd in the bottom when matching options
→ Two observers on opposite sides of a tower
Giving the larger angle to the farther point
→ Two observers on the same side of a tower
Using the 30° and 60° shortcut for other angles
→ Two observers on the same side of a tower
Treating two printed forms as different answers
→ Two observers on the same side of a tower
Taking the average of the two distances
→ Complementary angles from two points
Giving the building the height of the drop
→ One observer reading two angles on the same vertical line
Using the wrong angle for the shared distance
→ One observer reading two angles on the same vertical line

Test yourself on Heights and Distances

10 past SSC CGL questions from this chapter, timed at 6 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Heights and Distances chapter test Drill all 10 questions