Why this topic matters · 8 min read
Heights and Distances is an applied trigonometry topic that appears in SSC CGL Tier 1 and Tier 2 almost every year, typically contributing 1-2 questions per paper. Questions involve towers, poles, shadows, cliffs, and angles of elevation or depression. The core skill is drawing the right triangle from a word problem and applying the correct trig ratio. Speed depends on memorizing tan values and recognizing standard diagram patterns.
Core Concepts: Angle of Elevation and Depression
Angle of Elevation is the angle your line of sight makes with the horizontal when you look UP at an object. Angle of Depression is the angle your line of sight makes with the horizontal when you look DOWN at an object. Both angles are always measured from the horizontal, never from the vertical. A key fact: the angle of elevation from point A to point B equals the angle of depression from point B to point A. This alternate-angle relationship saves time in many problems.
- Angle of elevation: observer looks upward from horizontal
- Angle of depression: observer looks downward from horizontal
- Both measured from the horizontal line, not from vertical
- Elevation from A to B = Depression from B to A (alternate interior angles)
- Always draw a rough diagram before setting up the equation
Key formulas
tan of angle
tan(theta) = Opposite / Adjacent = Height / Horizontal Distance
When: Use when height and horizontal distance are involved — covers 80% of problems
sin of angle
sin(theta) = Opposite / Hypotenuse = Height / Line of Sight
When: Use when the slant distance (line of sight) is given or asked
cos of angle
cos(theta) = Adjacent / Hypotenuse = Horizontal Distance / Line of Sight
When: Use when horizontal distance and slant line are both involved
Worked examples
A tower is 50 m tall. From a point on the ground, the angle of elevation of the top is 30 degrees. Find the distance from the point to the base. Solution: tan(30) = 50 / d => 1/root(3) = 50/d => d = 50 root(3) meters, approximately 86.6 m.
From the top of a 100 m cliff, the angle of depression to a boat is 45 degrees. Find the horizontal distance to the boat. Solution: tan(45) = 100 / d => 1 = 100/d => d = 100 m.
Must-Know Trigonometric Values
SSC CGL almost exclusively uses angles of 30, 45, and 60 degrees. Rarely, 0 and 90 degrees appear. You must recall sin, cos, and tan for these five angles instantly without thinking. The tan values are the most used because most problems involve height and horizontal distance directly.
- tan 30 = 1/root(3) approx 0.577
- tan 45 = 1
- tan 60 = root(3) approx 1.732
- sin 30 = 0.5, sin 45 = 1/root(2), sin 60 = root(3)/2
- cos 30 = root(3)/2, cos 45 = 1/root(2), cos 60 = 0.5
- Memory trick for sin: 0, 1/2, 1/root(2), root(3)/2, 1 for angles 0, 30, 45, 60, 90
Key formulas
Sin value pattern
sin(theta) = root(n)/2 where n = 0,1,2,3,4 for theta = 0,30,45,60,90
When: Quick recall shortcut for sin values of standard angles
Standard Problem Types and Diagram Patterns
SSC CGL repeats the same 5-6 diagram setups year after year. Recognizing the pattern instantly tells you which formula to pick. The two-position observer problem (observer moves toward or away from a tower and angle changes) is the most common advanced type. The shadow problem (sun angle gives shadow length) is another frequent type.
- Type 1 — Single tower, single observer: one right triangle, use tan directly
- Type 2 — Observer moves, two angles given: set up two tan equations and subtract/divide
- Type 3 — Two towers facing each other: draw two right triangles sharing a base
- Type 4 — Shadow problems: shadow is the base, sun angle is the angle at the tip of shadow
- Type 5 — Pole on top of building: two angles from same point, subtract heights
- In two-observer problems, let the height = h and distances = x and y, then solve simultaneous equations
Key formulas
Two-angle tower formula
h = d * tan(A) * tan(B) / (tan(A) - tan(B)), where d is the distance between two observer positions and A > B
When: When observer walks toward tower and two angles of elevation are given
Shadow length
Shadow length = Height / tan(sun angle)
When: When sun's angle of elevation and height of object are given
Worked example
A man observes a tower at 30 degrees, walks 20 m closer, and sees it at 60 degrees. Find height. Let height = h, closer distance = x. From second position: tan 60 = h/x => x = h/root(3). From first: tan 30 = h/(x+20) => x+20 = h*root(3). Subtract: 20 = h*root(3) - h/root(3) = h*(3-1)/root(3) = 2h/root(3). So h = 10*root(3) meters.
Key Geometry Rules for These Problems
Most mistakes happen not in the trig but in the geometry setup. Always remember that the height of a tower is the perpendicular distance from the base to the top — it is the vertical leg of the right triangle. The horizontal distance is always the flat ground from the observer to the base. Never mix hypotenuse with height or base.
- The right angle is always at the base of the tower or pole
- Angles of depression use a horizontal line at the observer's eye level — draw this line explicitly
- If observer is on a hill, the horizontal line is at the hill's top level, not ground
- Two towers problem: the connecting horizontal line at the top creates two separate triangles
⚠ Common mistakes to avoid
- Confusing angle of elevation with angle of depression — elevation is looking up, depression is looking down. Many students set up the wrong triangle when depression is given.
- Using sin or cos when tan is needed — if slant distance is not given or asked, always default to tan(angle) = height/base.
- In two-observer problems, setting the distance between observers as x instead of the individual distances from the tower, leading to wrong simultaneous equations.
- Forgetting to simplify root(3) expressions — SSC answers are always clean numbers like 10*root(3) or 50, so if your answer looks ugly, you have made an error.
- Shadow problems: treating the shadow as the hypotenuse instead of the base (adjacent side). The sun's ray is the hypotenuse, the object height is opposite, shadow is adjacent.
🧠 Memory aids
- SOHCAHTOA: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent. Chant this as 'Silly Old Harry Caught A Herring Trawling Off America'.
- For standard angles, remember tan values as: 30 is One-Root (1/root3), 45 is One-One (1/1), 60 is Root-One (root3/1). Increasing order: small angle, small tan.
- Elevation = Eyes going UP like an Elevator. Depression = looking DOWN like you are Depressed.
- The two-observer trick: always label the tower height as h and write TWO tan equations, then eliminate the unknown distance by subtraction or division. Think of it as two equations, two unknowns.
🎯 SSC CGL exam tips
- SSC CGL Tier 1 typically has 1 question from this topic; Tier 2 may have 2. Difficulty is easy to medium. A correct diagram gets you 70% of the way to the answer.
- Almost all questions use only 30, 45, and 60 degree angles. If you see any other angle in an option or question, re-read carefully — it may be a distractor or a misprint.
- The two-observer walking problem (person walks x metres toward tower) is the most repeated PYQ pattern — practice it until the equation setup is automatic.
- Time target: 1 to 1.5 minutes per question. If the diagram is not clear in 20 seconds, skip and return — these questions can eat time if you freeze.
- Check your answer by substituting back. If height comes out negative or the trig ratio exceeds 1 in a sin/cos equation, you have set up the triangle incorrectly.
Q1 · hard · AI-verified
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving 20 m towards the base of the tower, the angle of elevation becomes 60°. What is the height of the tower?
- 10√3 m
- 20 m
- 10 m
- 20√3 m
Q2 · medium · AI-verified
The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 100 m towards the tower, the angle becomes 60°. What is the height of the tower?
- 100√3 m
- 100/√3 m
- 50 m
- 50√3 m
Q3 · medium · AI-verified
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string.
- 120 m
- 80 m
- 40√3 m
- 60√3 m
Q4 · medium · AI-verified
A ladder 10 m long makes an angle of 60° with the ground while leaning against a wall. How high on the wall does the ladder reach?
- 10√3 m
- 10/√3 m
- 5 m
- 5√3 m
Q5 · medium · AI-verified
The angles of elevation of the top of a tower from two points at distances 'a' and 'b' from the base (a > b) and in the same horizontal line are complementary. What is the height of the tower?
- √(a/b)
- ab/(a+b)
- (a+b)/2
- √(ab)