Exercise 9.1

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° (see figure ).

Answer:

Let AB be the vertical pole Ac be 20 m  long rope tied to point C.
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.

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° 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 AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In  right ΔABC,
cos 30° = 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

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?

Answer:

There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides.
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
A/q,
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,

In  right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.

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°. Find the height of the tower.

Answer:

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.

In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.

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

Answer:

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.
A/q,
In  right ΔABC,
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.

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.

Answer:

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ – BZ = (30 – 1.5) = 28.5 m
A/q,
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.

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° and 60° respectively. Find the height of the tower.

Answer:

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
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.

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 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer:

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.

9. 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.

Answer:

Let AB be the building and CD be the tower

In ΔCDB,

CB/BD = tan 60º

50/BD = √3

BD = √3/50

In ΔABD,

(AB)/(BD) = tan 30º

AB = 50/√3 x 1/√3 = 50/3 = 16 2/3

Therefore, the height of the building is 16 2/3 m.

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° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

In right triangle PRQ,

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

In ΔABO

(AB)/(BO) = tan 60º

AB/BO = √3

BO = AB/√3

In ΔCDO,

(CD)/(DO) = tan 30º

CD/80 – BO = 1/√3

CD x √3 = 80 – BO

CD x √3 = 80 – AB/√3

CD x √3 + AB/√3 = 80

Since the poles are of equal heights,

CD = AB

CD[√3 + 1/√3] = 80

CD[3+1/√3] = 80

CD = 20√3

BO = AB/√3 = CD/√3 = [20√3/√3]m = 20m

DO = BD − BO = (80 − 20) m = 60 m

Hence the heights of the poles are 20√3 m each and the distances of the point from poles 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 figure). Find the height of the tower and the width of the canal.

Answer:

In ΔABC,

AB/BC = tan 60º

AB/BC = √3

BC = AB/ √3

In ΔABD,

AB/BD = tan 30º

AB/BC+CD = 1/√3

[AB/(AB/√3) + 20] = 1/√3

[AB x √3/AB + 20 x √3] = 1/√3

3AB = AB + 20√3 =

2AB = 20√3

AB = 10√3m

BC = AB/√3 = {10√3/√3}m = 10m

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

12. 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.

Answer:

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

In ΔABD,

AB/BD = tan 45º

7/BD = 1

BD = 7 m

In ΔACE,

AE = BD = 7 m

CE/AE = tan 60º

CE/7 = √3

CE = 7 x √3

CD = CE + ED = [7 x √3 + 7]m = 7[√3 + 1]m

Hence height of the tower is 7[√3 + 1]m.

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° and 45°. 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 ΔABC,

AB/BC = tan 45º

75/BC = 1

BC = 75 m

In ΔABD,

AB/BD = tan 30º

75/BC+CD = 1/√3

75/75+CD = 1/√3

75 x √3 = 75 + CD

75[√3 – 1]m = CD

Hence 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 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.

Answer:

In right triangle ABC,

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

In ΔACE,

AE/CE = tan 60º

(AF – EF)/(CE) = tan 60º

88.2 – 1.2/CE = √3

87/CE = √3

CE = 87/√3 = 87x√3m

In ΔBCG,

(BG)/(CG) = tan 30º

88.2 – 1.2/CG = 1/√3

87/CG = 1/√3

CG = 87x√3 m

Distance travelled by balloon = EG = CG − CE

= (87x√3 – 29x√3 )m

= 58√3m

Hence the distance travelled by the balloon during the interval is 58√3 m.

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°, 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.

Answer:

Let AB be the tower.

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

In ΔADB,

AB/DB = tan 60º

AB/DB = √3

DB = AB/√3

In ΔABC,

AB/BC = tan 30º

AB/BD + DC = 1/√3

AB √3 = BD + DC

AB √3 = AB/√3 + DC

DC = AB √3 – AB/√3 = AB(√3 – 1/√3) = 2AB/√3

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

Time taken by the car to travel a distance DB (i.e. AB/√3) = [6/2AB/√3] x [AB/√3] = 6/2 = 3 seconds.

Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.

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 ΔAQR,

AQ/QR = tanΘ

AQ/4 = tanΘ    … 1

In ΔAQS,

AQ/SQ = tan(90 – Θ)

AQ/9 = cot Θ …2

On multiplying equations (1) & (2)

(AQ/4)(AQ/9) = (tanΘ).(cot Θ)

AQ²/36 = 1

AQ² = 36

AQ = √36

AQ = ±6

However, height cannot be negative.

Therefore, the height of the tower is 6 m