Study Material — 9 PYQs (2021–2021) · Concept Notes · Shortcuts
SSC MTS Heights & Distances is a frequently tested subtopic — 9 previous year questions from 2021–2021 papers are included below with concept notes, key rules and shortcut tricks.
9 questions from actual SSC MTS papers · all shown free · click option to reveal solution
Exam Q 12021Previous Year Pattern
From a point on the ground 20 metres away from the base of a vertical pole, the angle of elevation to the top is 30°. What is the height of the pole? (Use √3 = 1.732)
Exam Q 22021Previous Year Pattern
A man standing 30 metres away from the base of a tower observes the angle of elevation to the top of the tower to be 60°. What is the height of the tower? (Use √3 = 1.732)
Exam Q 32021Previous Year Pattern
From the top of a 45-metre-high building, the angle of depression to a point on the ground is 30°. How far is the point from the base of the building?
Exam Q 42021Previous Year Pattern
A person standing on the ground observes the angle of elevation to the top of a tree to be 45°. If the person walks 10 metres closer to the tree, the angle of elevation becomes 60°. What is the height of the tree? (Use √3 = 1.732)
Exam Q 52021Previous Year Pattern
A boy standing 15 metres from a building observes the angle of elevation to the roof to be 60°. What is the height of the building? (Use √3 = 1.732)
Exam Q 62021Previous Year Pattern
A man standing 40 metres away from the base of a tower observes the angle of elevation to the top of the tower as 60°. Find the height of the tower (in metres). [Use √3 = 1.732]
Exam Q 72021Previous Year Pattern
From a point on the ground, the angle of elevation to the top of a 30-metre tall building is 45°. How far is the point from the base of the building?
Exam Q 82021Previous Year Pattern
A ladder leans against a wall making an angle of 60° with the ground. If the ladder is 8 metres long, how high does it reach on the wall? [Use √3 = 1.732]
Exam Q 92021Previous Year Pattern
A person standing on the ground observes the angle of elevation to the top of a tree as 45°. After walking 10 metres closer to the tree, the angle of elevation becomes 60°. Find the height of the tree. [Use √3 = 1.732]
Concept Notes
Heights & Distances— Rules & Concept
Core ConceptRead this first — the foundation of the topic
CORE CONCEPT
When you look up at a tall building, the angle your line of sight makes with the horizontal ground is called the angle of elevation. When you look down from a height, it's called the angle of depression. These angles help us calculate heights and distances we cannot measure directly
KEY RULES
The angle of elevation from point A to point B equals the angle of depression from point B to point A. Always draw a right triangle and identify the opposite side, adjacent side, and hypotenuse clearly. The horizontal distance remains constant in most problems.
Formula BlockMemorise — at least one formula appears in every paper
• tan θ = Height/Base (most used)
• sin θ = Height/Hypotenuse
• cos θ = Base/Hypotenuse
• When angle changes from α to β: New height = Base × (tan β - tan α) + Original height
Exam PatternsWhat examiners ask — read before attempting PYQs
SSC CGL consistently asks 1-2 questions on this topic. Common scenarios include: tower/building height from given distance, finding distance when height is known, problems involving two angles of elevation, lighthouse/ship problems, and ladder-wall problems.
ShortcutsUse these to save 30–60 seconds per question
- ANGLE CHANGE METHOD: When moving closer or farther from an object, use the formula: h = d₁ × tan α = d₂ × tan β, where h is height, d is distance, and α, β are angles. This eliminates the need to calculate height separately.
Worked ExampleSolve this step-by-step before moving on
Effective height to calculate = h - 1.8m (since man has height)
3
Step 3
tan 30° = (h - 1.8)/150
4
Step 4
1/√3 = (h - 1.8)/150
5
Step 5
h - 1.8 = 150/√3 = 150/1.732 = 86.6m
6
Step 6
h = 86.6 + 1.8 = 88.4m
WORKED EXAMPLE 2: From a point on ground, a tree top's angle of elevation is 45°. Moving 20m closer, the angle becomes 60°. Find tree height.
1
Step 1
Let tree height = h, original distance = d
2
Step 2
From original position: tan 45° = h/d, so h = d
3
Step 3
From new position: tan 60° = h/(d-20)
4
Step 4
√3 = h/(d-20) = d/(d-20) [since h = d]
5
Step 5
√3(d-20) = d
6
Step 6
1.732d - 34.64 = d
7
Step 7
0.732d = 34.64, so d = 47.32m
8
Step 8
Tree height h = d = 47.32m
Exam TrapsCommon mistakes students make — avoid these
#1: Students forget to account for the observer's height. When a person observes something, always subtract the person's height from the total height calculated. Many students calculate the total vertical distance but forget the observer is not on the ground level.
ADDITIONAL SHORTCUTS: For 30-60-90 triangles, use ratio 1:√3:2.
For 45-45-90 triangles, use ratio 1:1:√2. When angle of elevation doubles, use the identity tan(2θ) = 2tan(θ)/(1-tan²θ). Remember that complementary angles have reciprocal trigonometric ratios.
Key Points to Remember
Angle of elevation = angle looking up; angle of depression = angle looking down
tan θ = Height/Base is the most frequently used formula in height-distance problems
Always subtract observer's height from total calculated height
Angle of elevation from A to B = Angle of depression from B to A
For 30° angle: tan 30° = 1/√3 = 0.577
For 45° angle: tan 45° = 1
For 60° angle: tan 60° = √3 = 1.732
In two-angle problems, use h = d₁ × tan α = d₂ × tan β shortcut
Draw clear diagrams marking height, base, and angles before solving
Horizontal distance remains same; only vertical measurements change with angle
Exam-Specific Tips
tan 30° = 1/√3 = 0.5774 (exact value)
tan 45° = 1 (exact value)
tan 60° = √3 = 1.732 (exact value)
sin 30° = 1/2 = 0.5
cos 30° = √3/2 = 0.866
sin 45° = cos 45° = 1/√2 = 0.707
sin 60° = √3/2 = 0.866
cos 60° = 1/2 = 0.5
Practice MCQs
Heights & Distances — Practice Questions
5graded MCQs · easy to hard · full solution & trap analysis
A vertical pole of height 12 metres casts a shadow of 12√3 metres on the ground. What is the angle of elevation of the sun at that time?
Practice 2hard
A man standing at a point P observes the angle of elevation to the top of a vertical tower to be 30°. He walks 60 m towards the base of the tower along level ground, and the angle of elevation becomes 60°. Find the height of the tower (in metres).
Practice 3hard
From the top of a cliff 80 m high, the angles of depression to two boats on the water are 30° and 45° respectively. If both boats are on the same side of the cliff and in line with the base, find the distance between the boats (in metres).
Practice 4hard
A ladder leans against a vertical wall. The angle between the ladder and the ground is 60°. If the ladder is 10 m long, and a man climbs to a point 2 m below the top of the ladder, at what height above the ground is the man standing?
Practice 5hard
A vertical pole of height 15 m casts a shadow of 5√3 m on the ground. At the same time, a tree casts a shadow of 20 m. Assuming the sun's rays make the same angle with the ground for both objects, find the height of the tree (in metres).
60-Second Revision — Heights & Distances
Remember: Always draw diagram first and mark given angles and distances clearly
Formula: tan θ = Height/Base is the primary formula for 90% of problems
Trap: Don't forget to subtract observer's height from calculated total height
Shortcut: Use tan 30° = 0.577, tan 45° = 1, tan 60° = 1.732 for quick calculations
Pattern: Two-angle problems use h = d₁ × tan α = d₂ × tan β relationship
Check: Angle of elevation and depression are always measured from horizontal line
Quick tip: 30-60-90 triangle sides are in ratio 1:√3:2