Some Applications of Trigonometry

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).

Solution:

Here, Δ BAC is a right angle, in which

∠ B = 90^o

AC = Hypotenuse

AB = Perpendicular

And, BC = Base

Here, Angle of elevation, ∠ BCA = 30^o

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 = 30^o

h = hypotenuse = 20 m

And p = perpendicular = AB

∴ Sin 30^o = ^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 30^o 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:

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 = 30^o

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

Here, ∠ A = 90^o

Thus, CB = hypotenuse ( h )

AC = perpendicular ( p )

And AB = Base ( b ) = 8 m

And, ∠ ABC = 30^o

Thus, AD = height of tree = ?

And, distance

We know that, tan θ = ^p/_b

⇒ tan 30^o = ^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 30^o = ^{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 30^o 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 60^o to the ground. What should be the length of the slide in each case?

Solution:

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 = 30^o

Thus, CB (h) = ?

We know that, sin θ = ^p/_h

⇒ sin 30^o = ^{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 = 30^o

Thus, EF = hypotenuse ( h ) = ?

We know that, sin θ = ^p/_h

⇒ sin 60^o = ^{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 30^o. Find the height of the tower.

Solution:

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 = 30^o

Thus, AC = p = ?

We know that, tan θ = ^p/_b

⇒ tan 30^o = ^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

MCQs Test

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