Some Applications of Trigonometry

NCERT Exercise 9.1 solution Q 9 to 12

Question: (9) The angle of elevation of the top of a building from the foot of the tower is 30^o and the angle of elevation of the tower from the foot of the building is 60^o. If the tower is 50 m high, find the height of the building.

Solution:

Let, AD is the given building and BC is the given tower.

The angle of elevation of the top of the building D from the foot of the tower B is equal to 30^o

And the angle of elevation of the tower from the foot of building A is equal to 60^o

Now, given,

Height of the tower BC = 50 m

And angle ABD = 30^o

And angle BAC = 60^o

Thus, height of the building, AD = ?

Now, in Δ ABC

CB = perpendicular ( p )

And AB = base ( b )

Now, we know that, tan θ = ^p/_b

⇒ tan 60^o = ^CB/_AB

⇒ √3 = ^CB/_AB

After cross multiplication, we get,

√3 AB = CB

⇒ √3 AB = 50 m

[ ∵ CB =50 m ]

⇒ AB = ⁵⁰/_√3 m

Now, in Δ ABD

AB = base ( b )

And AD = perpendicular ( p )

Now, we know that, tan θ = ^p/_b

⇒ tan 30^o = ^AD/_AB

⇒ ¹/_√3 = ^AD/_AB

After cross multiplication, we get

AB = √3 AD

⇒ √3 AD = AB

⇒ √3 AD = ⁵⁰/_√3 m

∴ AD = ^50/_√3/_√3 m

⇒ AD = ⁵⁰/_{√3 × √3} m

⇒ AD = ⁵⁰/₃ m

⇒ AD = 16 ²/₃ m

Thus, height of the building = 16 ²/₃ m Answer

Question: (10) Two poles of equal heights are standing opposite each other on either side of the road, which 80m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60^o and 30^o, respectively. Find the height of the poles and the distances of the point from the poles.

Solution:

Let, AD and BC are two poles of similar height and standing opposite to each other of the road AB

Let, At point O between road the angle of elevation of the top of the pole AD = 30^o and the angle of elevation of the pole BC = 60^o

Given,

Width of road AB = 80 m

Height of pole AD = height of pole BC

i.e. AD = BC

Thus, AD = BC = ?

And, AO = ? And OB = ?

Now, in Δ AOD

∠ AOD = 30^o

AD = perpendicular ( p )

And, AO = Base ( b )

Now, we know that, tan θ = ^p/_b

⇒ tan 30^o = ^AD/_AO

⇒ ¹/_√3 = ^AD/_AO

After cross multiplication, we get

AO = √3 AD -----(i)

Now, in Δ OBC

∠ BOD = 60^o

BC = perpendicular ( p )

And OB = base ( b )

Now, we know that, tan θ = ^p/_b

⇒ tan 60^o = ^BC/_OB

⇒ √3 = ^BC/_OB

After cross multiplication, we get,

√3 OB = BC

⇒ √3 OB = AD

[∵ AD = OB]

⇒ OB = ^AD/_√3 ------(ii)

Now, by adding equation (i) and (ii), we get

AO + OB = √3 AD + ^AD/_√3

⇒ 80 m = ^{√3 × √3 AD + AD}/_√3

[∵ AO + OB = 80 m. As given in question]

⇒ 80 m = ^{3AD + AD}/_√3

⇒ 80 m = ^{4 AD}/_√3

After cross multiplication, we get

80 √3 m = 4 AD

⇒ AD = ^{80 √3}/₄ m

⇒ AD = 20 √3 m

Now, after substituting the value of AD in equation (i), we get

√3 × 20 √3 m = AO

⇒ AO = 3 × 20 m

⇒ AO = 60 m

Now, since, AO + OB = 80m

∴ 60 m + OB = 80 m

∴ OB = 80 m –  60 m = 20 m

Thus, height of the pole = 20 √3 m and distances of point of angle of elevation are 60 m and 20 m respectively. Answer

Question: (11) A TV tower stands vertically on a bank of the canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60^o. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30^o (see figure). Find the height of the tower and the width of the canal.

Solution:

Given, DC = 20 m

∠ ADB = 30^o

And, ∠ ACB = 60^o

Thus, AB = ?,

And CB = ?

In Δ ACB

AB = perpendicular ( p )

And CB = base ( b )

Now, we know that, tan θ = ^p/_b

⇒ tan 60^o = ^AB/_CB

⇒ √3 = ^AB/_CB

After cross multiplication, we get

√3 CB = AB

⇒ CB = ^AB/_√3 -----(i)

Now, in Δ ADB

AB = perpendicular ( p )

And, DB = base ( b )

We know, that, tan θ = ^p/_b

⇒ tan 30^o = ^AB/_DB

⇒ ¹/_√3 = ^AB/_DB

After cross multiplication, we get

DB = √3 AB

⇒ 20 m + CB = √3 AB

[∵ DB = 20 m + CB]

After substituting the value of CB from equation (i), we get

20 m + ^AB/_√3 = √3 AB

⇒ 20 m = √3 AB – ^AB/_√3

⇒ 20 m = ^{√3 × √3 × AB –  AB}/_√3

⇒ 20 m = ^{3 AB – AB}/_√3

⇒ ^2AB/_√3 = 20 m

After cross multiplication

⇒ 2 AB = 20 √3 m

∴ AB = ^{20 √3}/₂ m

⇒ AB = 10 √3 m -----(ii)

Now by substituting the value of AB from equation (ii) in equation (i), we get

CB = ^{10 √3}/_√3 m

⇒ CB = 10 m

Thus, height of the tower is 20 √3 m and width of the canal is 10 m Answer

Question: (12) From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^o and the angle of depression of its foot is 45^o. Determine the height of the tower.

Solution:

Let, AE is the building and BD is the cable tower.

As given, height of the building AE = 7 m

Angle of elevation from the top of the tower to the top of the building, ∠ DEC = 60^o

And, angle of depression of the foot of the cable tower to the top of the building, ∠ CEB = 45^o

Since AB||EC thus, ∠ BEC = ∠ ABE = 45^o

Thus, height of cable tower, BD = ?

Now, AE = BC = 7m

And, AB = EC

Now, in Δ ABE

AE = perpendicular ( p ) = 7 m

And AB = base ( b )

And ∠ BCA = 45^o

Now, We know that tan θ = ^p/_b

⇒ tan 45^o = ^AB/_BC

⇒ 1 = ^{7 m}/_BC

By cross multiplication

⇒ BC = 7 m -----------(i)

Now, in Δ ADE

AD = BC = 7 m (from equation (i))

ED = perpendicular ( p )

AD = base ( b )

We know that, tan θ = ^p/_b

⇒ tan 60^o = ^ED/_AD

⇒ √3 = ^ED/₇ m

By cross multiplication, we get

ED = 7 √3 -----(ii)

Now, CE = height of tower = ED + CD

= 7 √3 m + 7 m

[∵ CD = AB = 7m]

⇒ CE = 7(√3 + 1) m

Thus, height of the tower = 7(√3 + 1) m Answer

MCQs Test

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