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UNIT V: Trigonometry

UNIT V: Trigonometry relates to CBSE/Class 10/Mathematics

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UNIT V: Trigonometry Questions

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Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Some Applications of Trigonometry

Sadika

Let's denote the height of the hill as hh meters. We can set up a right triangle to represent the situation. From the given information: The angle of elevation of the top of the hill from the foot of the tower is 60 degrees. The angle of elevation of the top of the tower from the foot of the hill is... read more

Let's denote the height of the hill as hh meters. We can set up a right triangle to represent the situation.

From the given information:

  1. The angle of elevation of the top of the hill from the foot of the tower is 60 degrees.
  2. The angle of elevation of the top of the tower from the foot of the hill is 30 degrees.
  3. The height of the tower is 50 meters.

In the right triangle formed by the hill, tower, and the ground, the height of the hill, the height of the tower, and the distance from the foot of the tower to the foot of the hill form the sides of the triangle.

Using trigonometric ratios, we can set up equations based on the given angles and the known side lengths:

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Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Some Applications of Trigonometry

Sadika

Given: The angles of depression from the top and bottom of the 8 m tall building are 30° and 45°, respectively. We can form two right triangles to represent the situation. For the top of the 8 m tall building: tan⁡(30∘)=8dtan(30∘)=d8 For the bottom of the 8 m tall building: tan⁡(45∘)=8dtan(45∘)=d8 We... read more

Given:

  1. The angles of depression from the top and bottom of the 8 m tall building are 30° and 45°, respectively.

We can form two right triangles to represent the situation.

For the top of the 8 m tall building: tan⁡(30∘)=8dtan(30)=d8

For the bottom of the 8 m tall building: tan⁡(45∘)=8dtan(45)=d8

We can solve these two equations simultaneously to find the values of hh and dd.

First, let's solve for dd using either equation (they are the same):

tan⁡(30∘)=8dtan(30)=d8

13=8d3

1=d8

d=83d=83

Now, using the value of dd, we can find the height of the multi-storeyed building using either equation:

tan⁡(30∘)=hdtan(30)=dh

13=h833

1=83

h

h=8h=8

So, the height of the multi-storeyed building is 8 meters, and the distance between the two buildings is 8383

meters.

 

 

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Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Some Applications of Trigonometry

Sadika

Let's denote: h as the height of the tower. x as the horizontal distance between the tower and the initial position of the car. v as the speed of the car. Given: read more

Let's denote:

  • h as the height of the tower.
  • x as the horizontal distance between the tower and the initial position of the car.
  • v as the speed of the car.

Given:

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Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry

Sadika

Given: We need to compute the value of: read more

Given:

We need to compute the value of:

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