# Some Applications of Trigonometry Chapter 9 NCERT Solutions For Class 10 CBSE Mathematics

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## Solution:

Length of the rope is 20 m and angle made by the rope with the ground level is 30°.

Given: AC = 20 m and angle C = 30°

To Find: Height of the pole

Let AB be the vertical pole

In right ΔABC, using sine formula

sin 30° = AB/AC

Using value of sin 30 degrees is ½, we have

1/2 = AB/20

AB = 20/2

AB = 10

Therefore, the height of the pole is 10 m.

## Solution:

Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°

BC = 8 m

To Find: Height of the tree, which is AB

From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right ΔABC,

Using Cosine and tangent angles,

cos 30° = BC/AC

We know that, cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

Also,

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.

## Solution:

As per contractor’s plan

Let, ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at

60° with length PR.

To Find: AC and PR

In right ΔABC,

sin 30° = AB/AC

1/2 = 1.5/AC

AC = 3

Also,

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3

Hence, length of the slide for below 5 = 3 m and

Length of the slide for elders children = 2√3 m

## Solution:

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

To Find: AB (height of the tower)

In right ABC

tan 30° = AB/BC

1/√3 = AB/30

⇒ AB = 10√3

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

## Solution:

Draw a figure, based on given instruction,

Let BC = Height of the kite from the ground, BC = 60 m

AC = Inclined length of the string from the ground and

A is the point where string of the kite is tied.

To Find: Length of the string from the ground i.e. the value of AC

From the above figure,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

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

## Solution:

Let the boy initially stand at point Y with inclination 30° and then he approaches the building to the

point X with inclination 60°.

To Find: The distance boy walked towards the building i.e. XY

From figure,

XY = CD.

Height of the building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

Measure of AB is 28.5 m

In right ΔABD,

tan 30° = AB/BD

1/√3 = 28.5/BD

BD = 28.5√3 m

Again,

In right ΔABC,

tan 60° = AB/BC

√3 = 28.5/BC

BC = 28.5/√3 = 28.5√3/3

Therefore, the length of BC is 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.

## Solution:

Let BC be the 20 m high building.

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

Height of transmission tower = AB = AC – BC

To Find: AB, Height of the tower

From figure, In right ΔBCD,

tan 45° = BC/CD

1 = 20/CD

CD = 20

Again,

In right ΔACD,

tan 60° = AC/CD

√3 = AC/20

AC = 20√3

Now, AB = AC – BC = (20√3-20) = 20(√3-1)

Height of transmission tower = 20(√3 – 1) m.

## Solution:

Let AB be the height of statue.

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

To Find: Height of pedestal = BC = AC-AB

From figure,

In right triangle BCD,

tan 45° = BC/CD

1 = BC/CD

BC = CD …..(1)

Again,

In right ΔACD,

√3 = ( AB+BC)/CD

√3CD = 1.6 + BC

√3BC = 1.6 + BC (using equation (1)

√3BC – BC = 1.6

BC(√3-1) = 1.6

BC = 1.6/(√3-1) m

BC = 0.8(√3+1)

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

## Solution:

Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30degree and 60 degree from the tower and the building respectively.

In right ΔBCD,

tan 60° = CD/BC

√3 = 50/BC

BC = 50/√3 …(1)

Again,

In right ΔABC,

tan 30° = AB/BC

⇒ 1/√3 = AB/BC

Use result obtained in equation (1)

AB = 50/3

Thus, the height of the building is 50/3 m.

## Solution:

Let AB and CD be the poles of equal height.

O is the point between them from where the height of elevation taken. BD is the distance between the poles.

As per above figure, AB = CD,

OB + OD = 80 m

Now,

In right ΔCDO,

tan 30° = CD/OD

1/√3 = CD/OD

CD = OD/√3 … (1)

Again,

In right ΔABO,

tan 60° = AB/OB

√3 = AB/(80-OD)

AB = √3(80-OD)

AB = CD (Given)

√3(80-OD) = OD/√3 (Using equation (1))

3(80-OD) = OD

240 – 3 OD = OD

4 OD = 240

OD = 60

Putting the value of OD in equation (1)

CD = OD/√3

CD = 60/√3

CD = 20√3 m

Also,

OB + OD = 80 m

⇒ OB = (80-60) m = 20 m

Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and

60 m respectively.

## 11. 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 joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Solution: Given, AB is the height of the tower.

DC = 20 m (given)

As per given diagram, In right ΔABD,

tan 30° = AB/BD

1/√3 = AB/(20+BC)

AB = (20+BC)/√3 … (i)

Again,

In right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

AB = √3 BC … (ii)

From equation (i) and (ii)

√3 BC = (20+BC)/√3

3 BC = 20 + BC

2 BC = 20

BC = 10

Putting the value of BC in equation (ii)

AB = 10√3

This implies, the height of the tower is 10√3 m and the width of the canal is 10 m.

## Solution:

Let AB be the building of height 7 m and EC be the height of the tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°.

EC = DE + CD

Also, CD = AB = 7 m. and BC = AD

To Find: EC = Height of the tower

Design a figure based on given instructions:

In right ΔABC,

tan 45° = AB/BC

1= 7/BC

BC = 7

√3 = DE/7

⇒ DE = 7√3 m

Now: EC = DE + CD

= (7√3 + 7) = 7(√3+1)

Therefore, Height of the tower is 7(√3+1) m.

## Solution:

Let AB be the lighthouse of height 75 m. Let C and D be the positions of the ships.

30° and 45° are the angles of depression from the lighthouse.

To Find: CD = distance between two ships

Step 1: From right triangle ABC,

tan 45° = AB/BC

1= 75/BC

BC = 75 m

Step 2: Form right triangle ABD,

tan 30° = AB/BD

1/√3 = 75/BD

BD = 75√3

Step 3: To find measure of CD, use results obtained in step 1 and step 2.

CD = BD – BC = (75√3 – 75) = 75(√3-1)

The distance between the two ships is 75(√3-1) m.

14. 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° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Solution:

Let the initial position of the balloon be A and final position be B.

Height of balloon above the girl height = 88.2 m – 1.2 m = 87 m.

To Find: Distance travelled by the balloon = DE = CE – CD

Step 1: In right ΔBEC,

tan 30° = BE/CE

1/√3= 87/CE

CE = 87√3

Step 2:

√3= 87/CD

CD = 87/√3 = 29√3

Step 3:

DE = CE – CD = (87√3 – 29√3) = 29√3(3 – 1) = 58√3

Distance travelled by the balloon = 58√3 m.

## Solution:

Let AB be the tower.

D is the initial and C is the final position of the car respectively.

Since man is standing at the top of the tower so, Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

Step 1: In right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

BC = AB/√3

AB = √3 BC

Step 2:

In right ΔABD,

tan 30° = AB/BD

1/√3 = AB/BD

AB = BD/√3

Step 3: Form step 1 and Step 2, we have

√3 BC = BD/√3 (Since LHS are same, so RHS are also same)

3 BC = BD

3 BC = BC + CD

2BC = CD

or BC = CD/2

Here, distance of BC is half of CD. Thus, the time taken is also half.

Time taken by car to travel distance CD = 6 sec. Time taken by car to travel BC = 6/2 = 3 sec.

## Solution:

Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As per

question,

In right ΔABC,

tan x = AB/BC

tan x = AB/4

AB = 4 tan x … (i)

Again, from right ΔABD,

tan (90°-x) = AB/BD

cot x = AB/9

AB = 9 cot x … (ii)

Multiplying equation (i) and (ii)

AB2 = 9 cot x × 4 tan x

⇒ AB2 = 36 (because cot x = 1/tan x

⇒ AB = ± 6

Since height cannot be negative. Therefore, the height of the tower is 6 m.

Hence Proved.