The angle of elevation is simply how far up you tilt your eyes from the flat ground to look at the top of something tall — a tower, a kite, the sun. Every height & distance question, no matter how it is dressed up, is really just one right-angled triangle: the height stands up straight, the ground stretches out flat as the base, and your line of sight is the slanted third side. Spot those three sides and the problem is half solved.
The one ratio that runs the whole topic
Height and base are the two legs of the triangle, joined by the tangent of the elevation angle:
tanθ=baseheight.
Learn the three standard values cold: tan30∘=31, tan45∘=1, tan60∘=3.
• At 45∘ the legs are equal — height : base =1:1. The triangle is a perfect right isosceles, so height equals distance.
• At 30∘ and 60∘ the legs are in the ratio 1:3. At 60∘ the height is the long side (height=3×base); at 30∘ the base is the long side.
Use sin or cos only when the slant line itself (a ladder, a kite string) is involved: sinθ=slantheight and cosθ=slantbase.
• Broken tree: when a tree snaps and the top tips over to touch the ground, the bent-over piece is the slanted hypotenuse and the standing stump is the height. So total height =stump+broken part — never report just one piece.
🗼Heights & Distances LabTower · observer · angle of elevation
Solve for
Known distance d
m
Solution steps
tan 60° = h / d
h = d × tan 60° = 30 × √3
h = 30√3 ≈ 51.96 m
Quick check: at 45∘ the height equals the base, so a tower whose top is seen at45∘ from a point 50 m away is m tall.
📝Practice Questions
Q1A pole 10 m high casts a shadow 10 m long. The angle of elevation of the sun is:
Q2The angle of elevation of the top of a tower from a point 20 m away is 45°. The height of the tower is:
Q3A ladder leans against a wall making 60° with the ground. Its foot is 5 m from the wall. The length of the ladder is:
📚
Real exam questions — Height & Distance Basics
5 question types · 20 solved examples from real SSC papers
Every height and distance question is one right triangle, so label the height, the base and the slanted line of sight, then pick tan, sin or cos to connect what you know to what you want.
How to solve this type
Draw the right triangle: the tower is the vertical (height), the ground is the horizontal (base), the line of sight is the slope. The angle of elevation sits at the observer, between the ground and the line of sight. Height and base are the two legs, so the ratio that links them is the tangent: tanθ=baseheight. Plug the standard value (tan30∘=31, tan45∘=1, tan60∘=3) and solve. Use sin or cos only when the slant line of sight itself is involved.
From a point 30 m away on the ground, the angle of elevation of the top of a tower is 60∘. Find the height of the tower.
A30 mB103 mC303 mD60 m
The angle of elevation of the top of a tower from a point 40 m away is 45∘. Find the height of the tower.
A402 mB40 mC80 mD20 m
A tower is 50 m high. The angle of elevation of its top from a point on the ground is 45∘. Find the distance of the point from the foot of the tower.
A25 mB100 mC503 mD50 m
A building is 1000 m high and its angle of elevation from a point on the ground is 30∘. The distance of the point from the building is approximately (take 3=1.732):