A vertical pole casts a shadow exactly equal to its own height. What is the angle of elevation of the sun?

45°. If the shadow equals the height, = height shadow = 1, so = 45^ .

From a point 30 m from the base of a tower on level ground, the angle of elevation of the top is 45°. How tall is the tower?

30 m. 45^ = h30 = 1, so h = 30 m.

From a point 20 m from the base of a tower, the angle of elevation of the top is 60°. Find the height. (Take 3 = 1.73.)

34.6 m. h = 20 60^ = 20 3 20 1.73 = 34.6 m.

As an observer walks 40 m towards a tower, the angle of elevation of its top rises from 30° to 60°. Find the height of the tower. (Take 3 = 1.73.)

34.6 m. The far distance is h 3 and the near distance is h 3; their difference is 2h 3 = 40, so h = 20 3 34.6 m.

Two poles of equal height stand on opposite sides of an 80 m wide road. From a point on the road between them the angles of elevation of the tops are 60° and 30°. Find the height of the poles. (Take 3 = 1.73.)

34.6 m. If the point is x m from the first pole, h = x 3 and h = 80 - x 3. So 3x = 80 - x, giving x = 20 and h = 20 3 34.6 m.

AI-guided practice

Quantitative Aptitude · Geometry & Mensuration

Heights and Distances — Questions & Answers

Heights-and-distances questions use right-triangle trigonometry with the standard angles. Draw the triangle, label the angle of elevation or depression, and pick the ratio that connects the known and unknown sides.

5 solved questions
·
Step-by-step AI walkthroughs
·
Key formulas included

Key formulas

$\tan\theta = \dfrac{\text{opposite (height)}}{\text{adjacent (base)}}$
$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ , $\;\tan 45^\circ = 1$ , $\;\tan 60^\circ = \sqrt{3}$
Angle of elevation from the ground equals the angle of depression from the top (alternate angles)

5 solved Heights and Distances questions

Try each question first — tap an option to check your answer, or open Learn step by step for a guided walkthrough.

01
A vertical pole casts a shadow exactly equal to its own height. What is the angle of elevation of the sun?
5 steps · reveal one at a time
1. Step 1: Visualize the setup
  Imagine a right-angled triangle formed by the pole (vertical side), its shadow (horizontal side on the ground), and the sun's ray connecting the top of the pole to the end of the shadow (hypotenuse).
2. Step 2: Choose the right formula
  The angle of elevation is the angle at the base of the pole. We know the opposite side (pole's height) and the adjacent side (shadow's length). The trigonometric ratio that relates these is the tangent: $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ .
3. Step 3: Apply the given facts
  The question states that the pole's height is exactly equal to its shadow's length. So, if the height is 'H' and the shadow is 'S', then $H = S$ . This means $\frac{\text{height}}{\text{shadow}} = \frac{H}{S} = 1$ .
4. Step 4: Calculate the angle
  Since $\tan\theta = \frac{\text{height}}{\text{shadow}}$ , and we found that $\frac{\text{height}}{\text{shadow}} = 1$ , we have $\tan\theta = 1$ . To find $\theta$ , we calculate the inverse tangent: $\theta = \arctan(1)$ .
5. Step 5: The answer
  So the answer is (B) 45°. The angle whose tangent is 1 is $45^\circ$ . This means the sun is at an elevation of $45^\circ$ .
02
From a point 30 m from the base of a tower on level ground, the angle of elevation of the top is 45°. How tall is the tower?
4 steps · reveal one at a time
1. Step 1: Visualize the scenario
  Imagine a right-angled triangle. The tower's height is the side opposite the angle of elevation, and the distance from the base is the side adjacent to it. We are given the angle of elevation as $45^\circ$ and the adjacent side as $30$ m.
2. Step 2: Apply trigonometry
  To find the height ( $h$ , the opposite side) when we know the angle and the adjacent side, we use the tangent function. The formula is $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ . So, we can write $\tan 45^\circ = \frac{h}{30}$ .
3. Step 3: Calculate the height
  We know that the value of $\tan 45^\circ$ is $1$ . Substitute this into our equation: $1 = \frac{h}{30}$ . To find $h$ , multiply both sides by $30$ . This gives us $h = 1 \times 30 = 30$ m.
4. Step 4: The answer
  So the answer is (C) 30 m, because the height of the tower is $30$ m based on our calculation using the tangent function.
03
From a point 20 m from the base of a tower, the angle of elevation of the top is 60°. Find the height. (Take $\sqrt{3} = 1.73$ .)
5 steps · reveal one at a time
1. Step 1: Visualize the setup
  Imagine the tower, the ground, and the point. This forms a right-angled triangle. The height of the tower is the opposite side, the distance from the base is the adjacent side, and the angle of elevation is given.
2. Step 2: Choose trig ratio
  We know the adjacent side (20 m) and the angle (60°), and we need to find the opposite side (height, $h$ ). The trigonometric ratio that relates these is tangent: $\tan(\theta) = \text{opposite} / \text{adjacent}$ .
3. Step 3: Calculate the height
  Substitute the values into the formula: $\tan(60^\circ) = h / 20$ . Rearrange to find $h = 20 \times \tan(60^\circ)$ . We know that $\tan(60^\circ) = \sqrt{3}$ . So, $h = 20\sqrt{3}$ .
4. Step 4: Substitute and solve
  Using the given approximation $\sqrt{3} = 1.73$ , calculate the height: $h = 20 \times 1.73 = 34.6$ m. This is the height of the tower.
5. Step 5: The answer
  So the answer is (D) 34.6 m, as our calculation directly matches this value.
04
As an observer walks 40 m towards a tower, the angle of elevation of its top rises from 30° to 60°. Find the height of the tower. (Take $\sqrt{3} = 1.73$ .)
4 steps · reveal one at a time
1. Step 1: Set up the problem
  Let the height of the tower be $h$ . When the angle of elevation is $30^\circ$ , the distance from the observer to the tower is $x_1$ . When the angle is $60^\circ$ , the distance is $x_2$ . We know $x_1 - x_2 = 40$ m.
2. Step 2: Relate distances to height
  Using trigonometry, for the $30^\circ$ angle, $\tan(30^\circ) = \frac{h}{x_1}$ , so $x_1 = \frac{h}{\tan(30^\circ)} = h\sqrt{3}$ . For the $60^\circ$ angle, $\tan(60^\circ) = \frac{h}{x_2}$ , so $x_2 = \frac{h}{\tan(60^\circ)} = \frac{h}{\sqrt{3}}$ .
3. Step 3: Calculate the height
  The difference in distances is $x_1 - x_2 = h\sqrt{3} - \frac{h}{\sqrt{3}} = 40$ . This simplifies to $\frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} = 40$ . Solving for $h$ , we get $h = \frac{40\sqrt{3}}{2} = 20\sqrt{3}$ .
4. Step 4: The answer
  So the answer is (B) 34.6 m. Substituting the given value $\sqrt{3} = 1.73$ , we find $h = 20 \times 1.73 = 34.6$ m.
05
Two poles of equal height stand on opposite sides of an 80 m wide road. From a point on the road between them the angles of elevation of the tops are 60° and 30°. Find the height of the poles. (Take $\sqrt{3} = 1.73$ .)
5 steps · reveal one at a time
1. Step 1: Set up the problem
  Let $h$ be the height of the poles. Let the point on the road be $x$ meters from the first pole. This means it is $(80 - x)$ meters from the second pole. We have two right-angled triangles formed.
2. Step 2: Formulate equations
  Using trigonometry, for the first pole, $\tan 60^\circ = h/x$ . Since $\tan 60^\circ = \sqrt{3}$ , we get $h = x\sqrt{3}$ . For the second pole, $\tan 30^\circ = h/(80-x)$ . Since $\tan 30^\circ = 1/\sqrt{3}$ , we get $h = (80-x)/\sqrt{3}$ .
3. Step 3: Solve for x
  Equate the two expressions for $h$ : $x\sqrt{3} = (80-x)/\sqrt{3}$ . Multiply both sides by $\sqrt{3}$ to get $3x = 80 - x$ . Adding $x$ to both sides gives $4x = 80$ , so $x = 20$ m.
4. Step 4: Calculate the height
  Now substitute $x = 20$ into $h = x\sqrt{3}$ . This gives $h = 20\sqrt{3}$ . Using the given approximation $\sqrt{3} = 1.73$ , we calculate $h = 20 \times 1.73 = 34.6$ m.
5. Step 5: The answer
  So the answer is (C) 34.6 m, which matches our calculated height for the poles.

Free to start

Get exam-ready on Heights and Distances

Generate fresh heights and distances practice sets in Test Studio, get step-by-step AI explanations the moment you slip, and close the gaps in Socrative Mode.

Start practising free Browse all topics

Step-by-step AI solutions
Fresh practice sets on demand
Guided Socrative Mode drills

Key formulas

5 solved Heights and Distances questions

Get exam-ready on Heights and Distances

More in Geometry & Mensuration