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CBSE Class 10  >  Class 10 Test  >  Online MCQ Tests  >  Test: Introduction to Heights & Distances - Class 10 MCQ

Introduction to Heights & Distances - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Introduction to Heights & Distances (20 Questions)

You can prepare effectively for Class 10 Online MCQ Tests for Class 10 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Introduction to Heights & Distances". These 20 questions have been designed by the experts with the latest curriculum of Class 10 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 20

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Test: Introduction to Heights & Distances - Question 1
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When the sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?

Detailed Solution: Question 1

Test: Introduction to Heights & Distances - Question 2
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The fig., shows the observation of point C from point A. The angle of depression from A is:

​

Detailed Solution: Question 2

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Test: Introduction to Heights & Distances - Question 3
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From the given figure, find h. 

​

Detailed Solution: Question 3

Answer: b) 25√3 m

Why:
Triangle OAB is right-angled at A. For the angle at O (60°):

  • Opposite side = AB = h

  • Adjacent side = OA = 25 m

In a right triangle, tan(θ) = opposite / adjacent.
So, tan 60° = h / 25.

But tan 60° = √3 ⇒ h = 25 × √3 m.

Therefore, h = 25√3 m.

Test: Introduction to Heights & Distances - Question 4
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The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called:​

Detailed Solution: Question 4

The term angle of elevation denotes the angle from the horizontal upward to an object.  An observer’s line of sight would be above the horizontal. And The term angle of depression denotes the angle from the horizontal downward to an object.  An observer’s line of sight would be below the horizontal. So The angle is called Angle of elevation.

Test: Introduction to Heights & Distances - Question 5
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A tower stands vertically on the ground from a point on the ground which is 25 m away from the foot of tower if the height of tower is 25√3 metres find the angle of elevation.​

Test: Introduction to Heights & Distances - Question 6
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The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:

Detailed Solution: Question 6


Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let ∠QPR = θ
tan θ = QR/PQ = 1 (since QR = PQ)
⇒ θ = 45°

i,e., required angle of elevation = 45°

Test: Introduction to Heights & Distances - Question 7
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If the length of a shadow cast by a pole is √3 times the length of the pole, then the angle of elevation of the sun is​

Detailed Solution: Question 7

Let AB be the tree and AC be its shadow.


Let ∠ACB = θ

Test: Introduction to Heights & Distances - Question 8
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Let AC be the rope whose length is 20 m, and AB be the vertical pole of height h m and the angle of elevation of A at point C on the ground is 30°.

Detailed Solution: Question 8


Fif. 9.11
i.e., ∠ACB = 30°
In right triangle ABC, we have

Hence, the height of the pole is 10 m.

Test: Introduction to Heights & Distances - Question 9
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A 20 m long ladder touches the wall at a height of 10 m. The angle which the ladder makes with the horizontal is

Test: Introduction to Heights & Distances - Question 10
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The ratio of the length of a rod and its shadow is 1:1/√3. what is the angle of elevation of the source of light

Detailed Solution: Question 10

Test: Introduction to Heights & Distances - Question 11
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If sun’s elevation is 60° then a pole of height 6 m will cast a shadow of length 

Test: Introduction to Heights & Distances - Question 12
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The angle of elevation from a point 30 metre from the base of tree as level ground to the top of the tree is 60°. The height of the tree is : 

Detailed Solution: Question 12

Test: Introduction to Heights & Distances - Question 13
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Tree is broken by the wind the top struck the ground at 30degree at a distance of 30m. away from the root. Find the height of the tree.

Detailed Solution: Question 13

Let BD be the whole height of the tree

The tree breaks at point C and makes ΔABC as given into the figure such that

AC = CD

In rt. ΔABC 

        

Test: Introduction to Heights & Distances - Question 14
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Find AB in the given figure

​

Detailed Solution: Question 14

Use tanθ formula in both the triangles ABC and ABD.
tan 60 =h/x
tan 30 =h/60+x

Test: Introduction to Heights & Distances - Question 15
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Guddi was standing on a road near a mall. She was 1000m away from the mall and able to see the top of the mall from the road in such a way that top of the tree, which is in between her and the mall, was exactly in line of sight with the top of the mall. The tree height is 10m and it is 20m away from Guddi. How tall is the mall?

Detailed Solution: Question 15

  • Guddi is 1000m away from the mall, with a tree of height 10m situated 20m from her.
  • The distance from the tree to the mall is 1000m - 20m = 980m.
  • Using similar triangles:
  • Height of Tree / Distance to Tree = Height of Mall / Distance to Mall
  • 10 / 20 = H / 1000
  • Solving for H: H = (10 * 1000) / 20 = 500m
  • Therefore, the correct height of the mall is 500m.

Test: Introduction to Heights & Distances - Question 16
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A tower stands vertically on the ground. From a point C on the ground, which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45 degrees. The height of the tower is

Detailed Solution: Question 16

Angle of elevation =45 degrees
Base=20 m
Height of the tower=perpendicular
We have tan 45=P/B
1=P/20
P=20 m
Hence the height of the tower  is 20 metres.

Test: Introduction to Heights & Distances - Question 17
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The angle of elevation from a point 30 feet from the base of a pole, of height h, as level ground to the top of the pole is 45o. Which equation can be used to find the height of the pole.

Test: Introduction to Heights & Distances - Question 18
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A tree casts a shadow 4 m long on the ground, when the angle of elevation of the sun is 45°. The height of the tree is:​

Detailed Solution: Question 18

In a triangle with one angle being 90 degrees (which the tree makes with the ground) and the other being 45 degrees (the angle of elevation), the 3rd angle is bound to be 45 degrees (180 - 90 - 45 = 45). 

We also know that sides opposite to equal angles are equal.

Hence, the height of the tree will also be 4m.

Test: Introduction to Heights & Distances - Question 19
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The angle of elevation of the sun when the length of the shadow of the tree is √3 times the height of the tree is​

Test: Introduction to Heights & Distances - Question 20
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A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45o. Then the height (in meters) of the tower is​

Detailed Solution: Question 20


A point on the ground which is 25 m away from the foot of the tower i. BC= 25 m

Let the height of the tower be x

The angle of elevation of the tower is found to be 45 degree.i.e.∠ACB=45°

In ΔABC

Using trigonometric ratios

Hence the height of the tower is 25 m.

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About this Class 10 Practice Test

This test contains 20 MCQs on Introduction to Heights & Distances with detailed solutions for each question. It is part of the Online MCQ Tests for Class 10 course on EduRev, designed to align with the current Class 10 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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