In this page we have *Class 10 Maths NCERT solutions for Some Applications Of Trigonometry* for
Exercise 9.1 on pages 203,204 and 205. Hope you like them and do not forget to like , social_share
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(i) The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level, i.e., the case when we raise our head to look at the object.

(iii) The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level, i.e., the case when we lower our head to look at the object.

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

Let AB be the vertical pole AC be 20 m long rope tied to point C.

In ΔABC,

sin 30° = Perpendicular/Base=AB/AC

⇒ 1/2 = AB/20

⇒ AB = 20/2

⇒ AB = 10

The height of the pole is 10 m.

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° 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.

Let us draw the situation below

Let AC be the broken part of the tree.

AB +AC is the tree height

In right ΔABC,

cos 30° = Base/Hypotenuse =BC/AC

⇒ √3/2 = 8/AC

⇒ AC = 16/√3

Also,

tan 30° = AB/BC

⇒ 1/√3 = AB/8

⇒ AB = 8/√3

Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3

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

Let us take small slide first

AB is the height and AC is the length of the slide

In right ΔABC,

sin 30° = Perpendicular/hypotenuse=AB/AC

⇒ 1/2 = 1.5/AC

⇒ AC = 3m

Now take the case of elder children slide

PQ is the height

PR is the length of slide

In r ΔPQR,

sin 60° = Perpendicular/hypotenuse= PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3 m

Hence, length of the slides are 3 m and 2√3 m respectively.

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°. Find the height of the tower.

The situation is depicted in the below figure

AB is height of the tower

BC =30 m (given)

In right ΔABC,

tan 30° = Perpendicular/base=AB/BC

⇒ 1/√3 = AB/30

⇒ AB = 10√3

Thus, the height of the tower is 10√3 m.

A kite is flying at a height of 60 m 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°. Find the length of the string, assuming that there is no slack in the string.

The situation is depicted in the below figure

Let BC be the height of the kite from the ground,

AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.

In ΔABC,

sin 60° =Perpendicular/base= BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

Thus, the length of the string from the ground is 40√3 m.

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° to 60° as he walks towards the building. Find the distance he walked towards the building

The situation is depicted in the below figure

Given

DY=1.5m

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

so XY is the distance he walked towards the building.

Height of the building = AZ=30 m

also, XY = CD.

AB = AZ - BZ = (30 - 1.5) = 28.5 m

In right ΔABD,

tan 30° = AB/BD

⇒ 1/√3 = 28.5/BD

⇒ BD = 28.5√3 m

also,

In right ΔABC,

tan 60° = AB/BC

⇒ √3 = 28.5/BC

⇒ BC = 28.5/√3 = 28.5√3/3 m

∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.

Thus, the distance boy walked towards the building is 57/√3 m.

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

The situation is depicted in the below figure

D is the point on Ground

BC be the 20 m high building..

Height of transmission tower = AB = AC - BC

In right ΔBCD,

tan 45° = BC/CD

⇒ 1 = 20/CD

⇒ CD = 20 m

also,

In right ΔACD,

tan 60° = AC/CD

⇒ √3 = AC/20

⇒ AC = 20√3 m

Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.

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° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

The situation is depicted in the below figure

Let AB be the height of statue.

D is the point on the ground from where the elevation is taken.

Height of pedestal = BC = AC - AB

In right ΔBCD,

tan 45° = BC/CD

⇒ 1 = BC/CD

⇒ BC = CD.

also,

In right ΔACD,

tan 60° = AC/CD

⇒ √3 = AB+BC/CD

⇒ √3CD = 1.6 m + BC

⇒ √3BC = 1.6 m + BC

⇒ √3BC - BC = 1.6 m

⇒ BC(√3-1) = 1.6 m

⇒ BC = 1.6/(√3-1) m

⇒ BC = 0.8(√3+1) m

Thus, the height of the pedestal is 0.8(√3+1) m.

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Let CD be the tower whose height is equal to 50 m

Let AB be the building and height of the building is x

BC be the distance between the foots of the building and the tower.

In right Δ BCD,

$ \tan 60 = \frac {CD}{BC}$

$ \sqrt {3} = \frac {50}{BC}$

$BC = \frac {50}{\sqrt {3}}$

Also,

In right Δ ABC,

$ \tan 30 = \frac {AB}{BC}$

$ \frac {1}{\sqrt {3}} = \frac {x}{BC}$

Substituting the value of BC from above

$x = \frac {50}{3}$ m

Hence the height of the building is 50/3 m.

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

Let AB and CD be the two poles of equal height h

O is the point on the road from where elevation of the top of the poles are 60° and 30°, respectively

Let OB=x and OD= 80 -x

Now

In right ΔOAB,

$ \tan 60 = \frac {AB}{OB}$

$ \sqrt {3} = \frac {h/x}$

or $h = x \sqrt {3} $ --(1)

Also,

In right Δ OCD,

$\tan 30 = \frac {CD}{OB}$

$\frac {1}{\sqrt {3}} = \frac {h}{80-x}$

Substituting the value from h

$\frac {1}{\sqrt {3}} = \frac {x \sqrt {3}}{80-x}$

80-x = 3x

x = 20

Putting the value of x in equation (1)

$h= 20 \sqrt {3}$

Also,

OD= 80 -20 = 60 m

Thus, the height of the poles are $20 \sqrt {3}$ m and distance from the point of elevation are 20 m and 60 m respectively.

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° . Find the height of the tower and the width of the canal.

In right Δ ABD,

$ \tan 30 = \frac {AB}{BD}$

$ \frac {1}{\sqrt {3}} = \frac {AB}{20+BC}$

$AB = \frac {20+BC}{\sqrt {3}} $... (A)

Also,

In right Δ ABC,

$ \tan 60 = \frac {AB}{BC}$

$ \sqrt {3} = \frac {AB}{BC}$

$ AB = \sqrt {3} BC$ ... (B)

From equation (A) and (B)

$ \sqrt {3} BC = \frac {20+BC}{\sqrt {3}}$

BC = 10 m

Putting the value of BC in eqn (B)

AB = $10 \sqrt {3}$ m

Hence, the height of the tower $10 \sqrt {3}$ m and the width of the canal is 10 m.

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Now

Height of Tower = CD = CE + ED= CE + AB

Now

In right Δ AEC,

$ \tan 60 =\frac {CE}{AE}$

or

$CE= AE \sqrt {3}$

ALso

In right Δ ABD

$\tan 45 = \frac {AB}{BD}$

or BD = AB = 7 cm

Now AE = BD = 7 cm

So, $CE = 7 \sqrt {3}$

So height of tower

$=7 \sqrt {3} + 7$ m

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Now

In right Δ ABC,

$\tan 45 =\frac {75}{BC}$

or BC = 75 m

ALso

In right Δ ABD,

$\tan 30 = \frac {75}{BC + CD}$

$ \frac {1}{\sqrt {3}} = \frac {75}{CD + 75}$

or

CD = $ 75 (\sqrt {3} -1)$ m

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Now

In right Δ OAC,

$\tan 60 = \frac {AC}{OC}$

$OC = \frac {87}{\sqrt {3}}$

Also

In right Δ OBD,

$\tan 30 = \frac {BD}{OD}$

$\frac {1}{\sqrt {3}}= \frac {87}{OC + CD}$

$OC + CD= 87 \sqrt {3}$

$ CD = 87 \sqrt {3} -\frac {87}{\sqrt {3}}$

$CD = 58 \sqrt {3}$ m

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

In right Δ ABC,

$\tan 60 = \frac {AB}{BC}$

$BC = \frac {AB}{\sqrt {3}}$

Also

In right Δ ABD,

$\tan 30 = \frac {AB}{BD}$

$\frac {1}{\sqrt {3}}=\frac {AB}{BD}$

$BD = AB \sqrt {3}$

Now

$CD = BD - BC = AB \sqrt {3} - \frac {AB}{\sqrt {3}}= \frac {2AB}{\sqrt {3}}$

We can observe that $BC =\frac {1}{2} CD$

Now Time taken to travel is 6 sec,So time taken to travel BC = 6/2 = 3sec

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m

Let AB be the tower and C and D are points. Let complementary angles are x and 90 -x

In right Δ ABC,

$\tan x = \frac {AB}{4}$ -(1)

ALso

In right Δ ABD,

$\tan (90-x) = \frac {AB}{9}$

$ \cot x = \frac {AB}{9}$

$ \tan x = \frac {9}{AB}$ --(2)

From (1) and (2)

$\frac {AB}{4}= \frac {9}{AB}$

$AB^2 = 36$

$AB = 6$

- NCERT Solutions for Class 10th Maths Some Applications Of Trigonometry Exercise 9.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

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