Class 10 - Maths - Applications of Trigonometry

**Exercise 9.1**

**Question 1:**

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^{0} (see Fig. 9.11).

Answer:

It can be observed from the figure that AB is the pole.

In ∆ABC,

AB/AC = sin 30^{0}

=> AB/20 = 1/2

=> AB = 20/2

=> AB = 10

Therefore, the height of the pole is 10 m.

**Question 2:**

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^{0} 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.

Answer:

Let AD is the broken part of the tree.

So total length of the tree = AB + AD

Again AD = AC

So, total length of the tree = AB + AC

Now in triangle ABC,

cos 30^{0} = BC/AC

=> √3/2 = 8/AC

=> AC = 16/√3

Again tan 30^{0} = AB/BC

=> 1/√3 = AB/8

=> AB = 8/√3

So, height of tree = AB + AC

= 8/√3 + 16/√3

= 24/√3 m

**Question 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^{0} to the ground,

whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60^{0} to the ground.

What should be the length of the slide in each case?

Answer:

It can be observed that AC and PR are the slides for younger and elder children respectively.

In ∆ABC,

AB/AC = sin 30^{0}

=> 1.5/AC = 1/2

=> AC = 1.5 * 2

=> AC = 3 m

In ∆PQR,

PQ/PR = sin 60^{0}

=> 3/PR = √3/2

=> √3 * PR = 3 * 2

=> √3 * PR = 6

=> PR = 6/√3

=> PR = (2 * √3 * √3)/ √3

=> PR = 2√3

Therefore, the lengths of these slides are 3 m and 2√3 m.

**Question 4:**

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^{0}.

Find the height of the tower.

Answer:

Let AB be the tower and the angle of elevation from point C (on ground) is 30°.

In ∆ABC,

AB/BC = tan 30^{0}

=> AB/30 = 1/√3

=> AB = 30/√3

=> AB = (10 * √3 * √3)/ √3

=> AB = 10√3

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

**Question 5:**

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^{0}. Find the length of the string, assuming that there is no slack in the string.

Answer:

Let K be the kite and the string is tied to point P on the ground.

In ∆KLP,

KL/KP = sin 60^{0}

=> 60/KP = √3/2

=> KP/60 = 2/√3

=> KP = (60 * 2)/ √3

=> KP = (6 * 10 * 2)/ √3

=> KP = (2 * √3 * √3 * 10 * 2)/ √3

=> KP = 2 * √3 * 10 * 2

=> KP = 40√3

Hence, the length of the string 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^{0} to 60^{0} as he walks towards the building. Find the distance he walked towards the building.

Answer:

Let the boy was standing at point S initially. He walked towards the building and reached

at point T. It can be observed that

PR = PQ − RQ

= (30 − 1.5) m = 28.5 m = 57/2 m

In ∆PAR,

PR/AR = tan 30^{0}

=> 57/2AR = 1/√3

=> AR = 57√3/2 m

In ∆PRB,

PR/BR = tan 60^{0}

=> 57/2BR = √3

=> BR = 57/2√3

=> BR = (19 * √3 * √3)/2√3

=> BR = 19√3/2

Now, ST = AB

= AR – BR = 57√3/2 - 19√3/2

= 38√3/2

= 19√3/2

Hence, he walked 19√3 m towards the building.

**Question 7:**

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^{0} and 60^{0} respectively. Find the height of the tower.

Answer:

Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

In ∆BCD,

BC/CD = tan 45^{0}

=> 20/CD = 1

=> CD = 20 m

In ∆ACD,

AC/CD = tan 60^{0}

=> (AB + BC)/CD = √3

=> (AB + 20)/20 = √3

=> AB + 20 = 20√3

=> AB = 20√3 – 20

=> AB = 20(√3 - 1)

Therefore, the height of the transmission tower is 20(√3 − 1) m.

**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 tatue is 60^{0}

and from the same point the angle of elevation of the top of the pedestal is 45^{0}. Find the height of the pedestal.

Answer:

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

In ∆BCD,

BC/CD = tan 45^{0}

=> BC/CD = 1

=> BC = CD ……………1

In ∆ACD,

(AB + BC)/CD = tan 60^{0}

=> (AB + BC)/BC = √3 ………..From equation 1

=> 1.6 + BC = √3BC

=> BC√3 – BC = 1.6

=> BC(√3 - 1) = 1.6

=> BC = 1.6/(√3 - 1)

=> BC = {1.6/(√3 - 1)} * {(√3 + 1)/ (√3 + 1)}

=> BC = {1.6 * (√3 + 1)}/{(√3 - 1) * (√3 + 1)}

=> BC = {1.6 * (√3 + 1)}/{(√3)^{2} – 1^{2}}

=> BC = {1.6 * (√3 + 1)}/(3 - 1)

=> BC = {1.6 * (√3 + 1)}/2

=> BC = 0.8 * (√3 + 1)

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

**Question 9:**

The angle of elevation of the top of a building from the foot of the tower is 30^{0} and the angle of elevation of the top of the tower

from the foot of the building is 60^{0}. If the tower is 50 m high, find the height of the building.

Answer:

Let AB be the building and CD be the tower.

In ∆CDB,

CD/BD = tan 60^{0}

=> 50/BD = √3

=> BD = 50/√3

In ∆ABD,

AB/BD = tan 30^{0}

=> AB = BD * (1/√3)

=> AB = (50/√3) * (1/√3)

=> AB = 50/3

=> AB = 16⅔

Therefore, the height of the building is 16⅔ m.

**Question 10:**

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^{0} and 30^{0}, respectively.

Find the height of the poles and the distances of the point from the poles.

Answer:

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ∆ABO,

AB/BO = tan 60^{0}

=> AB/BO = √3

=> BO = AB/√3

In ∆CDO,

CD/DO = tan 30^{0}

=> CD/DO = 1/√3

=> CD/(80 – BO) = 1/√3

=> CD * √3 = 80 – BO

=> CD * √3 = 80 – AB/√3

=> CD * √3 + AB/√3 = 80

Since the poles are of equal heights,

So, CD = AB

=> CD * √3 + CD/√3 = 80

=> CD(√3 + 1/√3) = 80

=> CD{(3 + 1)/√3} = 80

=> 4CD/√3 = 80

=> CD/√3 = 80/4

=> CD/√3 = 20

=> CD = 20√3

Now, BO = AB/√3 = CD/√3 = 20√3/√3 = 20

DO = BD − BO = (80 − 20) m = 60 m

Therefore, the height of poles is 20√3 m and the point is 20 m and 60 m far from these poles.

**Question 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^{0}. 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^{0}

(see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer:

In ∆ABC,

AB/BC = tan 60^{0}

=> AB/BC = √3

=> BC = AB/√3

In ∆ABD,

AB/BD = tan 30^{0}

=> AB/(BC + CD) = 1/√3

=> AB/(AB/√3 + 20) = 1/√3

=> AB√3/(AB + 20√3) = 1/√3

=> AB * √3 * √3 = AB + 20√3

=> 3AB = AB + 20√3

=> 3AB - AB + 20√3

=> 2AB = 20√3

=> AB = 20√3/2

=> AB = 10√3

Now, BC = AB/√3 = 10√3/√3 = 10 m

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

**Question 12:**

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^{0} and the angle of depression of its foot is 45^{0}.

Determine the height of the tower.

Answer:

Let AB be a building and CD be a cable tower.

In ∆ABD,

AB/BD = tan 45^{0}

=> 7/BD = 1

=> BD = 7

In ∆ACE,

AC = BD = 7 m

CE/AE = tan 60^{0}

=> CE/7 = √3

=> CE = 7√3

Now, CD = CE + ED

= 7√3 + 7

= 7(√3 + 1)

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

**Question 13:**

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30^{0} and 45^{0}.

If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:

Let AB be the lighthouse and the two ships be at point C and D respectively.

In ∆ABC,

AB/BC = tan 45^{0}

=> 75/BC = 1

=> BC = 75

In ∆ABD,

AB/BD = tan 30^{0}

=> 75/(BC + CD) = 1/√3

=> 75/(75 + CD) = 1/√3

=> 75√3 = 75 + CD

=> CD = 75√3 – 75

=> CD = 75(√3 – 1)

Therefore, the distance between the two ships is 75(√3 − 1) m.

**Question 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^{0}. After some time,

the angle of elevation reduces to 30^{0} (see Fig. 9.13).

Find the distance travelled by the balloon during the interval.

Answer:

Let the initial position A of balloon change to B after some time and CD be the girl.

In ∆ACE,

AE/CE = tan 60^{0}

=> (AF - EF)/CE = √3

=> (88.2 – 1.2)/CE = √3

=> 87/CE = √3

=> CE * √3 = 87

=> CE = 87/√3

=> CE = (29 * √3 * √3)/ √3

=> CE = 29√3

In ∆BCG,

BG/CG = tan 30^{0}

=> (88.2 – 1.2)/CE = 1/√3

=> 87/CE = 1/√3

=> CE = 87√3

Distance travelled by balloon = EG

= CG – CE

= 87√3 - 29√3

= 58√3 m

**Question 15:**

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^{0},

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^{0}. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In ∆ADB,

AB/DB = tan 60^{0}

=> AB/DB = √3

=> DB = AB/√3

In ∆ABC,

AB/BC = tan 30

=> AB/(BD + DC) = 1/√3

=> AB√3 = BD + DC

=> AB√3 = AB/√3 + DC

=> DC = AB√3 – AB/√3

=> DC = AB(√3 – 1/√3)

=> DC = AB{(3 – 1)/√3}

=> DC = 2AB√3

Time taken by the car to travel distance DC = (i.e. 2AB/√3) 6 seconds.

Time taken by the car to travel distance DB (i.e. 2AB/√3) = 6/(2AB/√3) * (AB/√3)

= 6/2

= 3 seconds

**Question 16:**

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.

Answer:

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower

respectively.

The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ∆AQR,

AQ/QR = tan θ

=> AQ/4 = tan θ ……….1

In ∆AQS,

AQ/SQ = tan (90^{0} – θ)

=> AQ/9 = cot θ ……….2

On multiplying equations 1 and 2, we obtain

(AQ/4) * (AQ/9) = tan θ * cot θ

=> AQ^{2}/36 = tan θ * 1/tan θ

=> AQ^{2}/36 = 1

=> AQ^{2} = 36

=> AQ = ±√36

=> AQ = ±6

Since height cannot be negative.

Therefore, the height of the tower is 6 m.

.