Some Applications of Trigonometry

Mathematics Class Tenth

10th-Math-home


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 60o. Find the length of the string, assuming that there is no slack in the string.

Solution:

10 math height and distance ncert  ex 9.1 question 5

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 = 90o

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 = 60o

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

We know that, sin θ = p/h

⇒ sin 60o = 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 30o to 60o as he walks towards the building. Find the distance he walked towards the building.

Solution:

10 math height and distance ncert  ex 9.1 question 6

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

In the given condition, A triangle CAB is formed.

In the Δ CAB , ∠ C = 90o

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 = 30o

And angle of elevation CDB = 60o

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 60o = 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 30o = 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 √328.5/3 m

⇒ AD = 28.5 √33 – 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 45o and 60o respectively. Find the height of the tower.

Solution:

10 math height and distance ncert  ex 9.1 question 7

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 60o and And the angle of elevation from the bottom of the tower D at point B on the ground is 45o

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

Thus, according to the question.

Height of building, AD = 20 m

∠ ABD = 45o

And ∠ ABC = 60o

Then, height of transmission tower, DC = ?

Now, in Δ ABD

∠ A = 90o

Thus, AD = perpendicular ( p )

AB = Base ( b )

And, BD = hypotenuse ( h )

As given, ∠ ABD = 45o

Now, we know that, tan θ = p/b

⇒ tan 45o = AD/AB

⇒ 1 = 20 m/AB

[∵ tan 45o = 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 = 60o

We know that, tan θ = p/b

⇒ tan 60o = 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 60o and from the same point the angle of elevation of the top of the pedestal is 45o. Find the height of the pedestal.

Solution:

10 math height and distance ncert  ex 9.1 question 8

Let, A statue CD stands on the pedestal AD.

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

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 = 45o

And ∠ ABC = 60o

Thus, height of pedestal AD = ?

Now, in Δ ABD

∠ DAB = 90o

AB = base ( b )

AD = perpendicular ( p )

And DB = hypotenuse ( h )

Now, we know that, tan θ = p/b

∴ tan 45o  = 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 = 60o

Now, we know that, tan θ = p/b

⇒ tan 60o = 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

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