A tower stands vertically on the ground from a point on the ground which is 30 m away

Solution:

We have to find the height of the tower.

Let us consider the height of the tower as AB, the distance between the foot of the tower to the point on the ground as BC.

In ΔABC, trigonometric ratio involving AB, BC and ∠C is tan θ.

tan C = AB/BC

tan 30° = AB/30

1/√3 = AB/30

AB = 30/√3

= (30 × √3) / (√3 × √3)

= (30√3)/3

= 10√3

Height of tower AB = 10√3 m.

☛ Check: NCERT Solutions Class 10 Maths Chapter 9

Video Solution:

Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 4

Summary:

If the angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower, is 30°, then the height of the tower is 10√3 m.

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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 45°. Then the height (in meters) of the tower is

• A. `25sqrt2`

• B. `25sqrt3`

• C. 25

• D. 12.5

Let AB be the tower and C be the point on the ground 25 m away from the foot of the

tower such that ∠ACB = 45°.

In right ΔABC:

`tan 45 ^@=(AB)/(BC)`

`rArr l=(AB)/25m`

⇒ AB = 25 m

Thus, the height of the tower is 25 m.

The correct answer is C.

Concept: Heights and Distances

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