Heights and Distances
Practice MCQsHeights and Distances is an important quantitative aptitude topic based on trigonometric ratios, angle of elevation, angle of depression, shadows, towers, buildings, poles, and distance measurement.
Heights and Distances is an important quantitative aptitude topic based on trigonometric ratios, angle of elevation, angle of depression, shadows, towers, buildings, poles, and distance measurement. These questions are commonly asked in competitive exams and aptitude tests.
What are Heights and Distances?
Heights and Distances problems use basic trigonometry to find unknown heights or distances when an angle and one side of a right triangle are known.
In most exam questions, a tower, pole, building, tree, hill, or object is represented as a vertical line, and the distance from the observer is represented as a horizontal line. This forms a right-angled triangle.
| Term | Meaning | Example |
|---|---|---|
| Height | Vertical measurement of an object | Height of tower or building |
| Distance | Horizontal measurement from observer to object | Distance from observer to tower |
| Angle of Elevation | Angle made when looking upward | Looking at top of a tower |
| Angle of Depression | Angle made when looking downward | Looking down from a building |
“In heights and distances, the diagram is half the solution.”
Key points
- Most problems form a right-angled triangle.
- Height is usually the opposite side.
- Distance is usually the adjacent side.
- Line of sight is usually the hypotenuse.
- Use angle of elevation for upward viewing.
- Use angle of depression for downward viewing.
Visual Understanding
The following diagrams show the most common visual patterns used in heights and distances questions.
Angle of Elevation
When the observer looks upward, the angle is called angle of elevation.
Angle of Depression
The horizontal line at the top is parallel to the ground, so alternate interior angles are equal.
Shadow-Based Problem
Shadow problems also form right triangles.
Important Trigonometric Ratios
Sine Ratio
Use when opposite side and hypotenuse are involved.
Cosine Ratio
Use when adjacent side and hypotenuse are involved.
Tangent Ratio
Most height and distance questions use tangent.
Height Formula
If distance \(d\) and angle \(\theta\) are known:
Distance Formula
If height \(h\) and angle \(\theta\) are known:
Standard Trigonometric Values
These standard values are commonly used in heights and distances problems.
| Angle | \(\sin\theta\) | \(\cos\theta\) | \(\tan\theta\) |
|---|---|---|---|
| \(30^\circ\) | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{3}}\) |
| \(45^\circ\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{\sqrt{2}}\) | \(1\) |
| \(60^\circ\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{2}\) | \(\sqrt{3}\) |
Common Types of Questions
Tower Problems
Find height or distance using angle of elevation.
- Tower height
- Distance from tower
- Observer on ground
- Usually tangent based
Shadow Problems
Use height and shadow length.
- Tree shadow
- Pole shadow
- Sun elevation angle
- Right triangle
Angle of Depression
Observer looks downward from height.
- Building top
- Looking down
- Horizontal reference line
- Same as elevation angle
Two-Point Observation
Object is viewed from two different points.
- Two angles given
- Find height
- Find distance between points
- Higher-order type
Method Bank
When \(d\) and \(\theta\) are known:
When \(h\) and \(\theta\) are known:
Use Pythagoras if height and distance are known:
Convert to angle of elevation:
Tip: If the object is vertical and ground is horizontal, the triangle is right-angled.
Two-Point Observation Diagram
Step-by-Step Solving Method
| Step | Action | Example |
|---|---|---|
| Step 1 | Draw a simple right triangle. | Tower is vertical, ground is horizontal. |
| Step 2 | Mark the given angle. | Angle of elevation \(=45^\circ\). |
| Step 3 | Identify opposite and adjacent sides. | Height = opposite, distance = adjacent. |
| Step 4 | Select trigonometric ratio. | \(\tan\theta = \frac{h}{d}\) |
| Step 5 | Substitute values and solve. | \(\tan45^\circ = \frac{h}{20}\) |
Solved Examples
| Question | Method | Answer |
|---|---|---|
| The angle of elevation of the top of a tower from a point 20 m away is \(45^\circ\). Find the height. |
\[
\tan45^\circ = \frac{h}{20}
\]
\[
1 = \frac{h}{20}
\]
\[
h = 20
\]
|
20 m |
| A pole casts a shadow of 10 m when the angle of elevation of the sun is \(30^\circ\). Find the height of the pole. |
\[
\tan30^\circ = \frac{h}{10}
\]
\[
\frac{1}{\sqrt{3}} = \frac{h}{10}
\]
\[
h = \frac{10}{\sqrt{3}}
\]
|
\(\frac{10}{\sqrt{3}}\) m |
| The height of a tower is 30 m. The angle of elevation from a point on the ground is \(60^\circ\). Find the distance from the tower. |
\[
\tan60^\circ = \frac{30}{d}
\]
\[
\sqrt{3} = \frac{30}{d}
\]
\[
d = \frac{30}{\sqrt{3}} = 10\sqrt{3}
\]
|
\(10\sqrt{3}\) m |
| A man observes the top of a building at an angle of elevation \(45^\circ\). If he is 50 m away, find the height of the building. |
\[
\tan45^\circ = \frac{h}{50}
\]
\[
h = 50
\]
|
50 m |
| From the top of a 40 m high building, the angle of depression of a car is \(45^\circ\). Find the distance of the car from the building. |
Angle of depression equals angle of elevation.
\[
\tan45^\circ = \frac{40}{d}
\]
\[
d = 40
\]
|
40 m |
| A ladder of length 10 m makes an angle of \(60^\circ\) with the ground. Find the height reached by the ladder. |
Here ladder is hypotenuse, height is opposite side.
\[
\sin60^\circ = \frac{h}{10}
\]
\[
\frac{\sqrt{3}}{2} = \frac{h}{10}
\]
\[
h = 5\sqrt{3}
\]
|
\(5\sqrt{3}\) m |
| The angle of elevation of the top of a tower changes from \(30^\circ\) to \(60^\circ\) as a man walks 40 m towards it. Find the height of the tower. |
Let the distance from nearer point be \(x\), height be \(h\).
\[
\tan60^\circ = \frac{h}{x}
\]
\[
h = x\sqrt{3}
\]
From farther point, distance is \(x+40\):
\[
\tan30^\circ = \frac{h}{x+40}
\]
\[
\frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x+40}
\]
\[
x+40 = 3x \Rightarrow x=20
\]
\[
h = 20\sqrt{3}
\]
|
\(20\sqrt{3}\) m |
Note: In two-point observation problems, form two equations using the two angles.
Common Traps and Shortcuts
Common Traps
- Confusing angle of elevation with angle of depression.
- Using sine instead of tangent in height-distance questions.
- Not drawing the right triangle.
- Taking line of sight as distance on ground.
- Forgetting observer height in some real-life problems.
- Using wrong value of \(\tan30^\circ\), \(\tan45^\circ\), or \(\tan60^\circ\).
Useful Shortcuts
- Use tangent when height and ground distance are involved.
- \(\tan45^\circ = 1\), so height equals distance.
- \(\tan60^\circ = \sqrt{3}\).
- \(\tan30^\circ = \frac{1}{\sqrt{3}}\).
- Angle of depression equals corresponding angle of elevation.
- Draw two triangles for two observation points.
Practice
A) Multiple Choice Questions
-
A tower is 30 m high. The angle of elevation from a point on the ground is \(45^\circ\).
Find the distance from the tower.
15 m 30 m 45 m 60 m
-
If \(\tan\theta = \frac{h}{d}\), then \(h\) equals:
\(\frac{d}{\tan\theta}\) \(d\tan\theta\) \(\frac{\tan\theta}{d}\) \(d+\tan\theta\)
-
The value of \(\tan45^\circ\) is:
0 1 \(\sqrt{3}\) \(\frac{1}{\sqrt{3}}\)
-
A pole casts a shadow of 20 m when the angle of elevation of the sun is \(45^\circ\).
Find the height of the pole.
10 m 20 m 30 m 40 m
-
A ladder of length 10 m makes \(30^\circ\) with the ground. Height reached is:
5 m \(5\sqrt{3}\) m 10 m 20 m
B) Solve the Higher-Order Problems
- From a point 50 m away from a tower, the angle of elevation is \(30^\circ\). Find the height of the tower. (Hint: Use \(\tan30^\circ = \frac{1}{\sqrt{3}}\).)
- The height of a building is 60 m. The angle of depression of a car from the top is \(60^\circ\). Find the distance of the car from the building. (Hint: Convert depression to elevation.)
- A ladder of length 20 m makes an angle of \(60^\circ\) with the ground. Find the height reached by the ladder. (Hint: Use \(\sin60^\circ\).)
- A man walks 30 m towards a tower. The angle of elevation changes from \(30^\circ\) to \(60^\circ\). Find the height of the tower. (Hint: Use two tangent equations.)
- A tree casts a shadow of \(15\sqrt{3}\) m when the angle of elevation of the sun is \(30^\circ\). Find the height of the tree. (Hint: \(h = d\tan30^\circ\).)
C) Match the Term with the Correct Meaning
| Term / Formula | Correct Meaning |
|---|---|
| Angle of Elevation | Angle made when looking upward |
| Angle of Depression | Angle made when looking downward |
| \(\tan\theta\) | \(\frac{\text{Opposite}}{\text{Adjacent}}\) |
| \(\sin\theta\) | \(\frac{\text{Opposite}}{\text{Hypotenuse}}\) |
| \(\cos\theta\) | \(\frac{\text{Adjacent}}{\text{Hypotenuse}}\) |
| Line of sight | Direct line from observer to object |
Aptitude Reminder
Heights and distances questions are solved by converting the situation into a right triangle. Use \(\tan\theta\) for height and ground distance, \(\sin\theta\) for height and line of sight, and \(\cos\theta\) for ground distance and line of sight.
Task: Create five questions using angle of elevation, angle of depression, shadow length, ladder length, and two-point observation.
Show Suggested Answers
Multiple Choice
-
30 m
\[ \tan45^\circ = \frac{30}{d} \]\[ 1 = \frac{30}{d} \Rightarrow d = 30 \] -
\(d\tan\theta\)
\[ \tan\theta = \frac{h}{d} \]\[ h = d\tan\theta \] -
1
\[ \tan45^\circ = 1 \] -
20 m
\[ \tan45^\circ = \frac{h}{20} \]\[ h = 20 \] -
5 m
Ladder is hypotenuse and height is opposite side.\[ \sin30^\circ = \frac{h}{10} \]\[ \frac{1}{2} = \frac{h}{10} \Rightarrow h = 5 \]
Higher-Order Problems
-
Distance \(d = 50\), angle \(30^\circ\):
\[ h = d\tan30^\circ \]\[ h = 50 \times \frac{1}{\sqrt{3}} = \frac{50}{\sqrt{3}} \]Answer = \(\frac{50}{\sqrt{3}}\) m.
-
Height \(h = 60\), angle \(60^\circ\):
\[ \tan60^\circ = \frac{60}{d} \]\[ \sqrt{3} = \frac{60}{d} \]\[ d = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]Answer = \(20\sqrt{3}\) m.
-
Ladder length \(= 20\), angle \(60^\circ\):
\[ \sin60^\circ = \frac{h}{20} \]\[ \frac{\sqrt{3}}{2} = \frac{h}{20} \]\[ h = 10\sqrt{3} \]Answer = \(10\sqrt{3}\) m.
-
Let nearer distance be \(x\), height be \(h\).
\[ \tan60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]Farther distance is \(x+30\):\[ \tan30^\circ = \frac{h}{x+30} \]\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x+30} \]\[ x+30 = 3x \Rightarrow x=15 \]\[ h = 15\sqrt{3} \]Answer = \(15\sqrt{3}\) m.
-
Shadow length \(= 15\sqrt{3}\), angle \(30^\circ\):
\[ h = d\tan30^\circ \]\[ h = 15\sqrt{3} \times \frac{1}{\sqrt{3}} = 15 \]Answer = 15 m.
Concept Matching
- Angle of Elevation → Angle made when looking upward
- Angle of Depression → Angle made when looking downward
- \(\tan\theta\) → \(\frac{\text{Opposite}}{\text{Adjacent}}\)
- \(\sin\theta\) → \(\frac{\text{Opposite}}{\text{Hypotenuse}}\)
- \(\cos\theta\) → \(\frac{\text{Adjacent}}{\text{Hypotenuse}}\)
- Line of sight → Direct line from observer to object
Clue Explanation
Heights and distances problems mainly use right triangles. The most common formula is \(\tan\theta = \frac{\text{height}}{\text{distance}}\).
Exam tips
- Always draw a right triangle first.
- Use tangent for height and ground distance.
- Use sine when line of sight is given with height.
- Use cosine when line of sight is given with ground distance.
- Convert angle of depression into angle of elevation.
- Remember standard values for \(30^\circ\), \(45^\circ\), and \(60^\circ\).