Heights And Distances

Chapter 9

Class X

Exercise 9.1

Q.No. 1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30⁰

Solution:

From question it is clear that
AC = 20 m.
<ACB = 30⁰ = θ 
AB = ?

We know that

Sinθ = AB/AC

Now, putting the value of θ, AB, AC.

Sin30⁰  = AB/20
1/2 = AB/20
∴ AB = 10m Ans.

Q.No. 2. A tree breaks due to storm and broken parts bends so that the top of the tree touches the ground making an angle 30⁰ with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

From question it is clear that

Height of the tree = AB + AC
θ = 30⁰
BC = 8 m.

From figure 

Cos 30⁰ = BC/AC
√3/2 = 8/AC
∴ AC = 16/√3

Again, 

Tan 30⁰ = AB/BC
1/√3 = AB/8

∴ AB = 8/√3

Height of the tree = AB + AC 

= 8/√3 + 16/√3

= 24/√3 
= 8√3
= 8x1.732
= 13.856 m Ans.

Q.No. 3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30⁰ to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60⁰ to ground. What should be the length of the slide in each case?

Solution:

From question it is clear that
For first slide
AB = 1.5 m
α = 30⁰
Length of slide = AC = ?

For second slide
PQ = 3 m
β = 60⁰
Length of slide = PR = ?

For first slide
According to application of Trigonometry
Length of slide = AC = AB/Sin30⁰
∴ AC = 1.5x2 = 3 m

Now, for second slide
Length of slide PR = PQ/Sin60⁰
∴ PR = 3x2/√3
PR = 2√3
PR = 2x1.732
PR = 3.464 m 

Therefore length of slides are
3m and 3.464 m Ans.

Q.No. 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30⁰. Find the height of the tower.

Solution:

From question it is clear that
Let the height of the tower  = AB = ?
BC = 30 m
Angle of elevation = 30⁰

According to application of Trigonometry
AB = BCxTan30⁰
∴ AB = 30/√3
AB = 10√3
Therefore, height of the tower = 17.32 m Ans.

Q.No. 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60⁰. Find the length of the string, assuming that there is no slack in the string.

Solution:

Let the length of the string be AC = x which represents the hypotenuse.

The inclination represents the angle of elevation = 60⁰

The height of the kite from ground = AB = 60 m.

The concerned right angle triangle is drawn below

From figure it is clear that

Sin60⁰ = AB/AC

Now, putting the value of AB and AC.

√3/2 = 60/x

∴ x = 120/√3

= 40√3 m Ans.

Q.No. 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30⁰ to 60⁰ as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let the distance he walked towards the building be CD = x m.

The angle of elevation for Triangle ABC = 30⁰

The angle of elevation for Triangle ABD = 60⁰

Height of the building from the eye of the boy = AB = 30 - 1.5 = 28.5 m.

The concerned right angle triangle is drawn below

Consider in right angle triangle ABC
Tan30⁰ = AB/BC
Now, putting the value of BC
BC = 28.5.√3

Now consider in right angle triangle ABD
Tan60⁰ = AB/BD
BD = AB/Tan60⁰
∴ BD = 28.5/√3 = 9.5.√3 m.
Therefore,
CD = BC - BD = 28.5.√3  - 9.5.√3 = 19√3 m Ans.