Heights and distances is applied trigonometry — you are never given the actual height of a tower or depth of a valley directly. Instead, you are given angles and horizontal distances, and you have to reconstruct the vertical measurement using the three core trig ratios.
The physical setup is always a right triangle. One side is vertical (the object — tower, pole, cliff), one side is horizontal (the ground distance), and the hypotenuse is either the line of sight or a physical object like a ladder or rope. The angle is always measured from the horizontal.
Two key terms the paper will test:
Angle of Elevation: You are looking up at something. The angle is measured between the horizontal ground level at your feet and your line of sight going upward. Imagine standing at the base of a stadium and looking at a scoreboard mounted on the roof — the angle your eyes make with the ground is the angle of elevation.
Angle of Depression: You are looking down at something. The angle is measured between the horizontal at your eye level and your line of sight going downward. Imagine standing on a rooftop and looking at a car parked below — the angle your eyes drop below horizontal is the angle of depression.
Here is the critical geometric fact most people miss: the angle of elevation from point A to point B equals the angle of depression from point B to point A. This is because the line of sight is the same line, and the two horizontal lines (at A and at B) are parallel. Alternate interior angles are equal. This symmetry saves you from drawing separate diagrams when two-object problems arise.
The ratios you will actually use:
tan θ = opposite/adjacent = height/horizontal distance — this is the workhorse of 80% of problemssin θ = opposite/hypotenuse — used when a ladder length or rope length (the hypotenuse) is givencos θ = adjacent/hypotenuse — used for the horizontal projection of a ladder or ropeKnow these values cold: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, sin 30° = 1/2, sin 60° = √3/2, cos 30° = √3/2, cos 60° = 1/2.
A tower of height h stands vertically. An observer stands on the ground at horizontal distance d from the base. The angle of elevation to the top is θ.
That is the complete model. Every single-tower question is a rearrangement of this one equation. If you are given θ and d, find h = d \cdot \tan\theta. If you are given θ and h, find d = h/\tan\theta = h\cot\theta.
These are the harder variants. A tower of unknown height h casts a shadow. The angle of elevation changes as the sun moves, so the shadow length changes. The classic setup:
Shadow at angle
αhas lengthL₁, shadow at angleβhas lengthL₂. DifferenceL₁ - L₂ = k.
Since tan α = h/L₁ → L₁ = h/\tan\alpha = h\cot\alpha, and similarly L₂ = h\cot\beta:
For the specific case α = 45°, β = 60°: cot 45° = 1, cot 60° = 1/√3.
Rationalise by multiplying numerator and denominator by (√3 + 1): h = k√3(√3+1)/2 = k(3+√3)/2.
Here a tower of known height and a pole of unknown height stand apart. The observer is on top of the tower, so angles of depression are used.
Draw it this way: tower height H, pole height h, horizontal distance between bases d.
tan θ₁ = H/d → d = H/\tan\theta_1tan θ₂ = (H-h)/dSubstitute d from the first equation into the second. Solve for h.
When the problem says "an observer 1.5 m tall", do not ignore that. The angle of elevation is measured from the eye of the observer, not from the ground. So the effective vertical gap between the observer's eye and the top of the tower is (H - 1.5), not H. The horizontal distance stays the same.
This is a frequent trap. SSC CGL puts observer height in problems precisely because many candidates set up tan θ = H/d and get the wrong answer.
When two angles from two points are complementary (sum to 90°), a beautiful shortcut emerges. Let the angles be θ and (90° - θ) from distances a and b:
Multiply the two equations: \tan\theta \cdot \cot\theta = \frac{h}{a} \cdot \frac{h}{b}
The height is the geometric mean of the two distances. This is a formula worth memorising directly — no need to work through angles at all when complementary angles appear.
Ladder length = hypotenuse. If the ladder makes angle θ with the ground and has length L:
L \sin\thetaL \cos\thetaIf no angle is given but the foot distance and ladder length are given, use Pythagoras: h = \sqrt{L^2 - d^2}.
Before writing a single equation, label your triangle: mark the angle, the side Opposite to it (vertical height), the side Adjacent to it (horizontal distance), and the Hypotenuse (ladder/rope/line of sight). Then pick the ratio that connects the two sides you know and the one you want. Most errors in this topic happen because students pick sin when they should pick tan. This labelling step takes 5 seconds and eliminates the wrong-ratio mistake entirely. Standard approach without labelling: 40% error rate on mixed problems. With labelling: near-zero wrong setups.
When the problem states that two angles of elevation are complementary (they add to 90°), immediately write h = √(ab) where a and b are the two distances. Do not solve for the actual angle. Standard method (find θ from tan, use complementary, solve system): 6 steps, ~75 seconds. Direct formula: 1 step, ~8 seconds.
For problems where shadow lengths at two sun angles differ by a known amount k, memorise: h = k / (cot α - cot β) where α is the smaller elevation angle (longer shadow) and β is the larger elevation angle (shorter shadow). For the most common pair 45° and 60°: cot 45° - cot 60° = 1 - 1/√3 = (√3-1)/√3, so h = k√3/(√3-1). Rationalise once and store: h ≈ k × 2.732 / 1 after rationalisation gives h = k(3+√3)/2. Standard derivation each time: 5 steps. Using stored formula: 2 steps.
Whenever an observer height e is mentioned, completely ignore it while setting up tan θ = x/d. Solve for x. Then add e at the very end: actual tower height H = x + e. This prevents the confusion of whether to subtract or add. The line of sight covers (H - e) vertically, so tan θ = (H-e)/d, which means H = d·tan θ + e. Adding e last is the same thing, just sequenced to avoid algebra errors. Saves 1-2 re-works per exam.
When the angle of elevation is 45°, tan 45° = 1, so height = horizontal distance, always. Shadow length = tower height. Ladder's vertical reach = its horizontal foot distance. Any 45° question with one measurement given is answered by reading off the other measurement directly. Standard: write and solve equation. Shortcut: write the answer. Time: 3 seconds vs 20 seconds.
Read the problem and immediately classify it:
Is a physical length given as the hypotenuse (ladder, rope)? → Use sin (for height) or cos (for base). Go to Pythagoras if no angle is given.
Are height and horizontal distance involved with an angle? → Use tan θ = h/d. Identify which two of the three (θ, h, d) are known. Solve for the third.
Are two angles mentioned from two different points? Check: do they add to 90°? If yes — use h = √(ab) immediately. If no — write two tan equations and solve the system.
Is the observer's height mentioned? Add it to your answer at the end.
Are shadow lengths at two different sun elevations given, with a difference? Use h = k / (cot α - cot β).
Is angle of depression used? Redraw: flip the triangle so the observer is at the top. The angle of depression from the top equals the angle of elevation from the bottom. Use tan θ = (vertical drop) / (horizontal distance).
If the numbers are messy, check whether an option matches h = d × (value of tan θ from your table) directly — back-substitution from options is faster than algebra in a timed exam.
Why this question: Tests the foundational Pythagorean right-triangle setup — the bedrock of all ladder problems.
Solving path: Ladder = hypotenuse = 10 m. Foot from wall = base = 6 m. Height on wall = √(10² - 6²) = √(100 - 36) = √64 = 8 m. The 6-8-10 triple is a scaled Pythagorean triple (3-4-5 scaled by 2). If you have memorised 3-4-5 and 5-12-13, you read this answer in 4 seconds.
Why this question: Canonical single-tower, angle of elevation, find height. The most common question template in the chapter.
Solving path: tan 30° = h/30. So h = 30 × (1/√3) = 30/√3. Rationalise: 30/√3 × √3/√3 = 30√3/3 = 10√3. Answer: 10√3 m. Scan options first — only one contains 10√3, so no full calculation needed.
Why this question: Tests sun-elevation / shadow setup. Height and shadow given; find the angle. Reverse of the usual direction.
Solving path: tan θ = height/shadow = 12/12 = 1. tan⁻¹(1) = 45°. When height equals shadow length, the angle is always 45° — zero computation needed once you spot the equality.
Why this question: Two-object (tower + pole) problem with angles of depression from the tower top. The hardest standard variant.
Solving path: Tower height H = 100 m. Let pole height = h, horizontal gap = d.
tan 45° = 100/d → d = 100 m.tan 30° = (100 - h)/d → 1/√3 = (100 - h)/100 → 100 - h = 100/√3 = 100/1.732 ≈ 57.73.h = 100 - 57.73 = 42.27 m.The trap: many students use tan 30° = h/d instead of (H - h)/d. Draw the diagram — the vertical gap from the tower top to the pole top is H - h, not h.
Why this question: Two-shadow, angle-difference problem. Requires setting up simultaneous equations using cotangent.
Solving path: Let height = h.
L₁ = h/tan 45° = h.L₂ = h/tan 60° = h/√3.h - h/√3 = 10 → h(1 - 1/√3) = 10 → h = 10/(1 - 1/√3) = 10√3/(√3 - 1).(√3 + 1)/(√3 + 1): h = 10√3(√3+1)/(3-1) = 10√3(√3+1)/2 = 5√3(√3+1) = 5(3 + √3) = 15 + 5×1.732 = 15 + 8.66 = 23.66 m.Why this question: Observer height adjustment — tests whether you adjust for the observer's eye level correctly.
Solving path: Observer's eye is at 1.5 m height. The line of sight covers (h - 1.5) vertically over 20.5 m horizontally.
tan 45° = (h - 1.5)/20.5 → 1 = (h - 1.5)/20.5 → h - 1.5 = 20.5 → h = 22 m.
Why this question: Ladder with angle given. Tests sin vs tan selection — the angle is with the ground (not the wall), so sin gives the height.
Solving path: Ladder = hypotenuse = 10 m. Angle with ground = 60°. Height = 10 × sin 60° = 10 × √3/2 = 5√3 m. Note: if you use tan 60° here you get 10 × √3 = 10√3, which is option (a) — a well-set trap.
Why this question: Complementary angles — the product formula shortcut. A favourite in SSC because it looks hard but solves in one step.
Solving path: Complementary angles → h = √(ab). Done. If you want the verification: tan θ = h/a and cot θ = h/b. Multiply: 1 = h²/(ab), so h = √(ab). The options list √(ab) directly — scan, identify, mark.
Confusing angle of elevation and angle of depression in two-level problems. When the observer is on top of a tower, the angle to the bottom of a pole is measured from the horizontal going down — angle of depression. Many students draw it as an angle of elevation and get the complement wrong.
Forgetting to add observer height. When an observer of height e is given, tan θ = (H - e)/d, not H/d. The final tower height is d·tan θ + e. This is tested precisely because it is easy to skip.
Using tan instead of sin for ladder/rope hypotenuse problems. When the inclined length (ladder, rope) is given — not the horizontal base — you need sin to find the vertical component, not tan. sin θ = height/hypotenuse. Using tan here gives a structurally wrong equation.
Incorrect vertical gap in tower-and-pole problems. The angle of depression from tower top to pole top corresponds to vertical gap (H - h), not H or h alone. Sketching the figure and labelling all vertical segments before writing equations prevents this.
Not rationalising 30/√3 quickly. 30/√3 = 30√3/3 = 10√3. Students sometimes leave it as 30/√3 and fail to match the option. Always rationalise surds in the denominator.
Misapplying the complementary angles formula when angles are not actually complementary. Check that the two angles genuinely add to 90° before using h = √(ab). If the problem says "angles are 30° and 45°" — they are not complementary, and you need two separate tan equations.