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 300 (see Fig. 9.11). Answer:

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

In ∆ABC,

AB/AC = sin 300

=> 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 300 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 300 = BC/AC

=> √3/2 = 8/AC

=> AC = 16/√3

Again tan 300 = 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 300 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 600 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 300

=> 1.5/AC = 1/2

=> AC = 1.5 * 2

=> AC = 3 m

In ∆PQR,

PQ/PR = sin 600

=> 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 300.

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 300

=> 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 600. 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 600

=> 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 300 to 600 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 300

=> 57/2AR = 1/√3

=> AR = 57√3/2 m

In ∆PRB,

PR/BR = tan 600

=> 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 450 and 600 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 450

=> 20/CD = 1

=> CD = 20 m

In ∆ACD,

AC/CD = tan 600

=> (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 600

and from the same point the angle of elevation of the top of the pedestal is 450. 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 450

=> BC/CD = 1

=> BC = CD ……………1

In ∆ACD,

(AB + BC)/CD = tan 600

=> (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 – 12}

=> 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 300 and the angle of elevation of the top of the tower

from the foot of the building is 600. 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 600

=> 50/BD = √3

=> BD = 50/√3

In ∆ABD,

AB/BD = tan 300

=> 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 600 and 300, 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 600

=> AB/BO = √3

=> BO = AB/√3

In ∆CDO,

CD/DO = tan 300

=> 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 600. 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 300

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

In ∆ABC,

AB/BC = tan 600

=> AB/BC = √3

=> BC = AB/√3

In ∆ABD,

AB/BD = tan 300

=> 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 600 and the angle of depression of its foot is 450.

Determine the height of the tower.

Answer:

Let AB be a building and CD be a cable tower. In ∆ABD,

AB/BD = tan 450

=> 7/BD = 1

=> BD = 7

In ∆ACE,

AC = BD = 7 m

CE/AE = tan 600

=> 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 300 and 450.

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 450

=> 75/BC = 1

=> BC = 75

In ∆ABD,

AB/BD = tan 300

=> 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 600. After some time,

the angle of elevation reduces to 300 (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 600

=> (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 300

=> (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 300,

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 600. 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 600

=> 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 (900 – θ)

=> AQ/9 = cot θ   ……….2

On multiplying equations 1 and 2, we obtain

(AQ/4) * (AQ/9) = tan θ * cot θ

=> AQ2/36 = tan θ * 1/tan θ

=> AQ2/36 = 1

=> AQ2 = 36

=> AQ = ±√36

=> AQ = ±6

Since height cannot be negative.

Therefore, the height of the tower is 6 m.

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