Heights and distances for SSC CGL

Heights and distances is trigonometry's most practical corner. Nearly every SSC CGL question here is solved in three steps: draw a right triangle, identify the known side and angle, and use tan (or occasionally sin or cos) to find the unknown. Because the angles are almost always 30°, 45° or 60°, remembering the side ratios of these special triangles lets you skip the trigonometry altogether.

Elevation and depression

• Angle of elevation: the angle between the horizontal and your line of sight when you look up at an object.
• Angle of depression: the angle between the horizontal and your line of sight when you look down.
• The angle of depression from A to B equals the angle of elevation from B to A (alternate angles).

The special triangles

Single observation

Worked example: From a point 30 m from the foot of a tower, the angle of elevation of the top is 60°. Find the height.
height = 30 × tan 60° = 30√3 m (about 51.96 m).

Worked example: A 10 m ladder leans against a wall, making a 60° angle with the ground. How high up the wall does it reach?
height = 10 × sin 60° = 5√3 m.

Two observation points

Worked example: The angles of elevation of the top of a tower from two points on the same side, in line with its foot, are 30° and 60°. The points are 40 m apart. Find the height.
Let the height be h. From the nearer point, the distance is h/√3; from the farther point, h√3.
h√3 − h/√3 = 40 → h(3 − 1)/√3 = 40 → h = 40√3/2 = 20√3 m.

A useful formula for angles α and β (α > β) on the same side with gap d:

h = d × tan α × tan β / (tan α − tan β)

Check: d = 40, tan 60° = √3, tan 30° = 1/√3. The numerator is 40 × 1 = 40 and the denominator is √3 − 1/√3 = 2/√3. So h = 40 × √3/2 = 20√3. ✓

Opposite sides

Worked example: Two poles of equal height stand on either side of a road 80 m wide. From a point on the road between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles.
Distances from the point: h/√3 and h√3. So h/√3 + h√3 = 80 → h × 4/√3 = 80 → h = 20√3 m.

Shadows

Worked example: A tower's shadow is 30 m longer when the sun's elevation is 30° than when it is 60°. Find the height of the tower.
The shadows are h√3 and h/√3. The difference is 2h/√3 = 30, so h = 15√3 m.

Depression problems

Worked example: From the top of a 75 m cliff, the angle of depression of a boat is 30°. How far is the boat from the foot of the cliff?
distance = 75 × √3 = 75√3 m.

Worked example: From the top of a building, the angles of depression of two cars in a straight line on the same side are 45° and 30°. The cars are 100 m apart. Find the height of the building.
Distances: h and h√3. h√3 − h = 100 → h = 100/(√3 − 1) = 50(√3 + 1) m (about 136.6 m).

Method for any question

1. Draw the figure; mark the vertical height, the horizontal ground and the angles.
2. Label unknowns with h and x.
3. Write tan for each right triangle.
4. Eliminate x and solve for h.
5. Rationalise if needed and compare with the options.

Common traps

Practice

1. The angle of elevation of the top of a 50 m tower from a point on the ground is 45°. How far is the point from the tower?
2. A kite on a 100 m string makes a 30° angle with the ground. How high is the kite?
3. From a point, the angle of elevation of a tower's top is 30°. After walking 20 m towards the tower, it becomes 60°. Find the height.
4. From the top of a 60 m tower, the angle of depression of a car is 60°. How far is the car from the tower?
5. A 1.5 m tall person stands 28.5 m from a chimney and sees its top at 45°. Find the chimney's height.

Answers: 1. 50 m. 2. 50 m. 3. 10√3 m. 4. 20√3 m. 5. 30 m.