Some Applications of Trigonometry

Mathematics Class Tenth

10th-Math-home


NCERT Exercise 9.1 solution Q 1 to 4

Question: (1) A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of the vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30^0 (see figure).

10 math height and distance ncert  ex 9.1 question 1

Solution:

Here, Δ BAC is a right angle, in which

∠ B = 90o

AC = Hypotenuse

AB = Perpendicular

And, BC = Base

Here, Angle of elevation, ∠ BCA = 30o

And, length of the rope, AC = 20 m

Thus, Height of the pole AB = ?

We know that, sin θ = p/h

Where, θ is the angle of elevation = 30o

h = hypotenuse = 20 m

And p = perpendicular = AB

∴ Sin 30o = AB/AC

1/2 = AB/20 m

After cross multiplication

⇒ 2 AB = 20 m

⇒ AB = 20 m/2

⇒ AB = 10 m

Thus, height of the pole = 10 m Answer

Question: (2) A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30o 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:

10 math height and distance ncert  ex 9.1 question 2

Let, AD is the given tree, which is broken from point C and bends as CB. This touches the ground at point B.

Given, Angle between the broken part of tree and ground, ∠ ABC = 30o

And length of broken part of tree, CB = 20 m

Here, ∠ A = 90o

Thus, CB = hypotenuse ( h )

AC = perpendicular ( p )

And AB = Base ( b ) = 8 m

And, ∠ ABC = 30o

Thus, AD = height of tree = ?

And, distance

We know that, tan θ = p/b

⇒ tan 30o = p/8 m

1/3 = p/8 m

By cross multiplication

⇒ √3 p = 8 m

∴ p = 8/3

Hence, AC = p = 8/3

Again, cos θ = b/h

⇒ cos 30o = 8 m/h

3/2 = 8 m/h

After cross multiplication

⇒ √3 h = 8 × 2 m

⇒ √3 h = 16 m

∴ h = 16 m/3

Hence, BC = h = 16 m/3

Now, height of the tree = AC + CD

= AC + CB

[∵ CD = CB, As it is broken part of tree]

= 8 m/3 + 16 m/3

= 8 + 16/3 m

= 24/3 m × 3/3 m

= 24 √3/3 m

= 8 √3 m

Thus, height of the tree 8 √3 m Answer

Question: (3) A contractor plans to install two slides for the children to play in a park. For 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 30o to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, inclined at an angle of 60o to the ground. What should be the length of the slide in each case?

Solution:

10 math height and distance ncert  ex 9.1 question 3

10 math height and distance ncert  ex 9.1 question 3a

Let, ABC is the slide for children below 5 years and DEF is the slide for elder children.

In Δ ABC ,

AB = height of the slide = p = 1.5 m

∠ ABC = 30o

Thus, CB (h) = ?

We know that, sin θ = p/h

⇒ sin 30o = 1.5 m/h

1/2 = 1.5 m/h

After cross multiplication

⇒ h = 1.5 m × 2

⇒ h = 3 m

⇒ CB = h = 3 m

In Δ DEF

DF = perpendicular ( p ) = 3 m

∠ DEF = 30o

Thus, EF = hypotenuse ( h ) = ?

We know that, sin θ = p/h

⇒ sin 60o = 3 m/h

3/2 = 3 m/h

After cross multiplication

⇒ √3 h = 3 m × 2

⇒ √3 h = 6 m

∴ h = 6 m/3

⇒ h = 6 m/3 × 3/3

⇒ h = 6√3/3 m

⇒ h = 2√3 m

⇒ EF = h = 2√3 m

Thus, length of slide for children below five years = 3 m and length of slide for elder children = 2√3 m Answer

Question: (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 30o. Find the height of the tower.

Solution:

10 math height and distance ncert  ex 9.1 question 4

Here, let AC is the given tower.

Here, ∠ A is the right angle.

Thus, AC = perpendicular (p)

AB = Base ( b )

And CB is the hypotenuse ( h )

Here, given,

AB = b = 30 m

And angle of elevation, ∠ ABC = 30o

Thus, AC = p = ?

We know that, tan θ = p/b

⇒ tan 30o = p/30 m

1/3 = p/30 m

After cross multiplication, we get

3 p = 30 m

∴ p = 30 m/3

⇒ p = 30 m/3 × 3/3

⇒ p = 30 √3/3 m

⇒ p = 10 √3 m

Thus, height of the tower = 10√3 m Answer

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