Class 10 Exam  >  Class 10 Tests  >  Test: Applications of Heights & Distances - Class 10 MCQ

Test: Applications of Heights & Distances - Class 10 MCQ


Test Description

10 Questions MCQ Test - Test: Applications of Heights & Distances

Test: Applications of Heights & Distances for Class 10 2024 is part of Class 10 preparation. The Test: Applications of Heights & Distances questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Applications of Heights & Distances MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Applications of Heights & Distances below.
Solutions of Test: Applications of Heights & Distances questions in English are available as part of our course for Class 10 & Test: Applications of Heights & Distances solutions in Hindi for Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Test: Applications of Heights & Distances | 10 questions in 10 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study for Class 10 Exam | Download free PDF with solutions
Test: Applications of Heights & Distances - Question 1

If the angles of depression from the top of a tower of height 40 m to the top and bottom of a tree are 45° and 60° respectively, then the height of the tree is

Detailed Solution for Test: Applications of Heights & Distances - Question 1

A tower of height 40 m is given, and the angles of depression from the top of the tower to:

  • The top of the tree = 45°
  • The bottom of the tree = 60°

We need to find the height of the tree (h).

Step 1: Use trigonometry for the angles of depression
Let:

  • The distance between the base of the tower and the base of the tree = d
  • The height of the tree = h

From the 45° angle (to the top of the tree):

The formula is:

tan(45°) = (Height of the tower - Height of the tree) / Distance (d)

Since tan(45°) = 1:

1 = (40 - h) / d

d = 40 - h (1)

From the 60° angle (to the bottom of the tree):

The formula is:

tan(60°) = Height of the tower / Distance (d)

Since tan(60°) = √3:

√3 = 40 / d

d = 40 / √3 (2)

Step 2: Solve the equations

Equate d from equations (1) and (2):

40 - h = 40 / √3

Rearrange to solve for h:

h = 40 - (40 / √3)

Rationalize the denominator:

h = 40 - (40√3 / 3)

h = (120 / 3) - (40√3 / 3)

h = (40 (3 - √3)) / 3

The height of the tree is:

b) (40 / 3) (3 - √3)

Test: Applications of Heights & Distances - Question 2

A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and angle of depression of the base of the hill as 30°. What is the height of the hill?

Detailed Solution for Test: Applications of Heights & Distances - Question 2

 

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Applications of Heights & Distances - Question 3

A vertical tower is 20 m high. A man at some distance from the tower knows that the cosine of the angle of the elevation of the top of tower is 0.5. He is standing from the foot of the tower at a distance of:

Detailed Solution for Test: Applications of Heights & Distances - Question 3

  • Base =distance from the man to the foot of the tower
  • Hypotenuse =line of sight (distance from man to top of tower)

In a right triangle, the height (20 m), base (x), and hypotenuse are related by the Pythagoras theorem:

Let the base be x and the hypotenuse be 2x.



Test: Applications of Heights & Distances - Question 4

A tower stands vertically on the ground. From a point on the ground 30 m away from the foot of the tower, the angle of elevation of the top of the tower is 45o. The height of the tower will be

Detailed Solution for Test: Applications of Heights & Distances - Question 4

Test: Applications of Heights & Distances - Question 5

The horizontal distance between two towers is 140 m. The angles of depression of the first tower, when seen from the top of the second tower is 30°. If the height of the first tower is 60 m. Find the height of the second tower.

Detailed Solution for Test: Applications of Heights & Distances - Question 5

The second tower is smaller than the first tower so let the height of the first tower is h So a part of it is equal to 60m so the remaining height is h-60 m
Tan 30=perpendicular/base

Test: Applications of Heights & Distances - Question 6

A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and angle of depression of the base of the hill as 30°. What is the height of the hill?

Test: Applications of Heights & Distances - Question 7

An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from his eyes is 45°. The height of the tower is​

Detailed Solution for Test: Applications of Heights & Distances - Question 7

To solve for the height of the tower, we use the tangent function. The angle of elevation is 45°, and the horizontal distance from the observer to the tower is 28.5 m.

Let the height of the tower be h. The observer's eye level is 1.5 m, so the difference in height between the top of the tower and the observer's eyes is h−1.5h.

Test: Applications of Heights & Distances - Question 8

The shadow of a flag pole is 30 metres. If the altitude of the sun is at 30° then the height of the flag pole is

Detailed Solution for Test: Applications of Heights & Distances - Question 8

Test: Applications of Heights & Distances - Question 9

If the angle of elevation of a cloud from a point 100 metres above a lake is 30o and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake is​

Detailed Solution for Test: Applications of Heights & Distances - Question 9

To find the height of the cloud above the lake, let the height of the cloud be h m and the horizontal distance from the observer to the point below the cloud be x. The observer is 100 m above the lake.

For the angle of elevation 30 degrees, we use the formula:

Test: Applications of Heights & Distances - Question 10

A tree is broken by wind and its upper part touches the ground at a point 10 metres from the foot of the tree and makes an angle of 45° with the ground. The entire length of the tree is​

Detailed Solution for Test: Applications of Heights & Distances - Question 10

Let the entire length of the tree be h metres and let the vertical part of the tree still standing be y metres.

So, the length of the tree inclined on the ground is (h−y) m.

In triangle ABC, we have




Information about Test: Applications of Heights & Distances Page
In this test you can find the Exam questions for Test: Applications of Heights & Distances solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Applications of Heights & Distances, EduRev gives you an ample number of Online tests for practice

Top Courses for Class 10

Download as PDF

Top Courses for Class 10