Some Applications of Trigonometry

NCERT Exercise 9.1 solution Q 5 to 8

Question: (5) A kite is flying at a height of 60m 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 60^o. Find the length of the string, assuming that there is no slack in the string.

Solution:

Let, F is a kite flying in the sky and EF is the string tied with the kite.

Thus, DEF so formed is a right angle.

Here, ∠ D = 90^o

Thus, DF = perpendicular ( p )

EF = hypotenuse ( h )

And, DE = base ( b )

Now, given,

DE = height of the kite from ground = perpendicular ( p ) = 60m

And, Angle of elevation = ∠ DEF = 60^o

Thus, length of string (EF) = hypotenuse ( h ) = ?

We know that, sin θ = ^p/_h

⇒ sin 60^o = ^{60 m}/_h

⇒ ^√3/₂ = ^{60 m}/_h

After cross multiplication, we get

√3 h = 2 × 60 m

∴ h = ¹²⁰/_√3 m

⇒ h = ¹²⁰/_√3 × ^√3/_√3 m

⇒ h = ^{120 √3}/₃ m

⇒ h = 40 √3 m

Thus, length of the string of the kite = 40 √3 m Answer

Question: (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 30^o to 60^o as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let, boy AF is standing at point A or F in front of building BE.

In this case the angle of elevation from the top of the building is 30^o. Let him walk towards the building and reach a point G or D, the angle of elevation from the top of the building became 60^o

In the given condition, A triangle CAB is formed.

In the Δ CAB , ∠ C = 90^o

Thus, AB = hypotenuse ( h )

And, AC = Base ( b )

And CB = perpendicular ( p )

Now, given,

EB = height of building = 30 m

AF = height of boy = 1.5 m

Angle of elevation CAB = 30^o

And angle of elevation CDB = 60^o

Thus, AD = ?

Since, CE || AF and CE = AF

And given, AF = 1.5 m

Thus, CE = 1.5 m

Now, BC = BE – EC

⇒ BC = 30 m –  1.5 m = 28.5 m

In Δ CDB

Now, we know that, tan θ = ^p/_b

= tan θ = ^BC/_CD

= tan 60^o = ^{28.5 m}/_CD

⇒ √3 = ^{28.5 m}/_CD

After cross multiplication, we get

√3 CD = 28.5 m

∴ CD = ^{28.5 m}/_√3 m

Now, in Δ CAB

We know that, tan θ = ^p/_b

⇒ tan 30^o = ^CB/_CA

⇒ ¹/_√3 = ^{28.5 m}/_CA

After cross multiplication, we get

CA = 28.5 √3 m

Now, AD = CA – CD

⇒ AD = 28.5 √3 – ^28.5/_√3 m

⇒ AD = ^{28.5 √3 √3 – 28.5}/_√3 m

⇒ AD = ^{28.5 × 3 – 28.5}/_√3 m

⇒ AD = ^{28.5(3 – 1)}/_√3 m

⇒ AD = ^{28.5 × 2}/_√3 m

⇒ AD = ⁵⁷/_√3 m

By multiplying with ^√3/_√3

⇒ AD = ⁵⁷/_√3 × ^√3/_√3 m

⇒ AD = ^57√3/₃ m

⇒ AD = 19 √3 m

Thus, distance walked by boy towards building = 19 √3 m Answer

Question: (7) From a point on the ground, the angle of elevation of the bottom and the top of a transmission tower fixed at the top of a 20m high building are 45^o and 60^o respectively. Find the height of the tower.

Solution:

Let CD is a transmission tower fixed at a building AD.

And the angle of elevation from top point C of the transmission tower to a point of ground B is 60^o and And the angle of elevation from the bottom of the tower D at point B on the ground is 45^o

Thus, according to the given condition in question, two triangles ABC and ABD are formed, in which ∠ A is equal to 90^o. Thus, these triangles are right-angle triangles.

Thus, according to the question.

Height of building, AD = 20 m

∠ ABD = 45^o

And ∠ ABC = 60^o

Then, height of transmission tower, DC = ?

Now, in Δ ABD

∠ A = 90^o

Thus, AD = perpendicular ( p )

AB = Base ( b )

And, BD = hypotenuse ( h )

As given, ∠ ABD = 45^o

Now, we know that, tan θ = ^p/_b

⇒ tan 45^o = ^AD/_AB

⇒ 1 = ^{20 m}/_AB

[∵ tan 45^o = 1 ]

After cross multiplication, we get

AB = 20 m

Now, in Δ ABC

AC = perpendicular ( p ) and

AB = base ( b )

As given in question, ∠ ABC = 60^o

We know that, tan θ = ^p/_b

⇒ tan 60^o = ^AC/_AB

⇒ √3 = ^AC/₂₀ m

[∵ AB = 20 m as calculated]

After cross multiplication, we get

AC = 20 √3

⇒ AD + DC = 20 √3

[∵ AC = AD + DC]

⇒ 20 m + DC = 20 √3

[∵ As given in question, AD = 20m]

⇒ DC = 20 √3 – 20 m

⇒ DC = 20 (√3 – 1) m

⇒ DC = 20 (1.732 – 1) m

⇒ DC = 20 × 0.732 = 14.64 m

Thus, height of the given transmission tower, 20 √3 m or   14.64 m Answer

Question: (8) A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60^o and from the same point the angle of elevation of the top of the pedestal is 45^o. Find the height of the pedestal.

Solution:

Let, A statue CD stands on the pedestal AD.

The angle of elevation of the top of the statue is ABC = 60^o and angle of elevation of the top of the pedestal is ABD = 45^o

Thus, height of pedestal AD = ?

Thus, in the condition given in the question, two triangles ABD and ABC are formed.

Given, Height of statue CD = 1.6 m

∠ ABD = 45^o

And ∠ ABC = 60^o

Thus, height of pedestal AD = ?

Now, in Δ ABD

∠ DAB = 90^o

AB = base ( b )

AD = perpendicular ( p )

And DB = hypotenuse ( h )

Now, we know that, tan θ = ^p/_b

∴ tan 45^o  = ^AD/_AB

⇒ 1 = ^AD/_AB

After cross multiplication

⇒ AB = AD

Now, in Δ ABC

AC = perpendicular ( p )

AB = base ( b )

CD = 1.6 m

And ∠ ABC = 60^o

Now, we know that, tan θ = ^p/_b

⇒ tan 60^o = ^AC/_AB

⇒ √3 = ^AC/_AB

After cross multiplication, we get

√3 AB = AC

⇒ √3 AD = AC

[∵ AB = AD]

⇒ √3 AD = AD + CD

[∵ AC = AD + CD]

⇒ √3 AD = AD + 1.6 m

[∵ given CD = 1.6 m ]

⇒ √3 AD – AD = 1.6 m

⇒ AD (√3 – 1) = 1.6 m

∴ AD = ^{1.6 m}/_{√3 – 1}

⇒ AD = ^{1.6 m}/_{√3 – 1} × ^{√3 + 1}/_{√3 + 1}

⇒ AD = ^{1.6 (√3 + 1)}/_{(√3)² – 1²} m

⇒ AD = ^{1.6 (√3 + 1)}/_{3 – 1} m

⇒ AD = ^{1.6 (√3 + 1)}/₂ m

⇒ AD = 0.8 (√3 + 1) m

Thus, height of the pedestal = 0.8 (√3 + 1) m Answer

MCQs Test

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