﻿ Class 10 NCERT Math Solution
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TOPICS
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° (see Figure).

Solution :-
```
Here, AB is the Pole.
Length of rope (AC) = 20 m
Height of the pole (AB) = ?

In Δ ABC,
AB/AC = sin 30°
AB/20 = 1/2
AB = 20/2
AB = 10 m
Therefor, 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° 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.

Solution :-
```
Let AC was a original tree.
Length of Break part of tree (A'C) = 8 m
Total Height of the tree (AC) = ?

In Δ A'BC,
BC/A'C = tan 30°
BC/8 = 1/√3
BC = 8/√3 m

A'C/A'B = cos 30°
8/A'B = √3/2
A'B = 16/√3 m

Total Heght of tree (AC) = A'B + BC = 16/√3 + 8/√3 = 24/√3 = 8√3 m
Therefor, Height of the tree is 8√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° 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° to the ground. What should be the length of the slide in each case?

Solution :-
```  Let AC and PR be the slides of Younger and Elder Children respectively.
AB = 1.5 m, PQ = 3 m

In Δ ABC,
AB/AC = sin 30°
1.5/AC = 1/2
AC = 3.0 m
Therefore, length of younger slides is 3 m.

In Δ PQR,
PQ/PR = sin 60°
3/PR = √3/2
PR = 6/√3 = 2√3 m
Therefore, length of elder slides is 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°. Find the height of the tower.

Solution :-
```
Let AB the tower and Distance BC = 30 m
Angle of elevation = 30°
Height of the tower (AB) = ?

In Δ ABC,
AB/BC = tan 30°
AB/30 = 1/√3
AB = 30/√3 = 10√3 m
Therefore, the height of 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°. Find the length of the string, assuming that there is no slack in the string.

Solution :-
```
Let K is a Kite.
Height of kite from the ground (LK) = 60m
Length of string (KP) = ?

In Δ KLP,
KL/KP = sin 60°
60/KP = √3/2
KP = 120/√3 = 40√3 m
Therefore, the Length of 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° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution :-
```
Let a boy was standing at point S initially.
He walked towards the building and reached at T point.
Height of the Boy = 1.5 m
Height of the building (PQ) = 30 m
So, PR = PQ - RQ = 30 - 1.5 = 28.5 = 57/2 m
Distance from walked towards the building (ST) = ?

In Δ PAR,
PR/AR = tan 30°
57/2AR = 1/√3
AR = (57√3)/2 m

In Δ PRB,
PR/BR = tan 60°
57/2BR = √3
BR = 57/(2√3) = (19√3)/2 m

Now, ST = AB
= AR - BR
= (57√3)/2 - (19√3)/2
= (38√3)/2
= 19√3 m
Therefore, the Distance from walked towards the building is 19√3 m.
```

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

Solution :-
```
Let AB be the Transmission Tower.
Height of building (BC) = 20 m
AC = AB + BC
Height of Transmission Tower (AB) = ?

In Δ BCD,
BC/CD = tan 45°
20/CD = 1
CD = 20 m

In Δ ACD,
AC/CD = tan 60°
(AB + BC)/BC = √3
AB + BC = √3 BC
√3 BC - BC = AB
BC (√3 - 1) = AB
20(√3 - 1) = AB
AB = 20(√3 - 1) m
Therefore, the height of 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 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.

Solution :-
```
Let the height of pedestal is BC.
Height of statue (AB) = 1.6 m
AC = AB + BC
Height of pedestal (BC) = ?

In Δ BCD,
BC/CD = tan 45°
BC/CD = 1
BC = CD

In Δ ACD,
AC/CD = tan 60°
(AB + BC)/BC = √3
1.6 + BC = √3 BC
√3 BC - BC = 1.6
BC (√3 - 1) = 1.6
BC = 1.6/(√3 - 1)  By rationalising of Denominator
BC = 1.6(√3 + 1)/[(√3 - 1)(√3 + 1)]
BC = 1.6(√3 + 1)/2
BC = 0.8(√3 + 1) m
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° 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.

Solution :-
```
Let AB be a Building.
Height of tower (CD) = 50 m
Height of the Building (AB) = ?

In Δ CDB,
CD/BD = tan 60°
50/BD = √3
BD = 50/√3 m

In Δ ABD,
AB/BD = tan 30°
AB/BD = 1/√3
AB = BD/√3
AB = 50/(√3 x √3)
AB = 50/3 m

Therefore, the height of the building is 50/3 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° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Solution :-
```
Let AB and CD are be the two equal poles and BO and Do distances of the point from the poles.
Distance between both poles (BD) = 80 m
Height of poles = ?
Distances of the point from the poles = ?

In Δ ABO,
AB/BO = tan 60°
AB/BO = √3
BO = AB/√3

In Δ CDO,
CD/DO = tan 30°
CD/DO = 1/√3
CD/(80 - BO) = 1/√3
CD √3 = 80 - BO
CD √3 = 80 - AB/√3
CD √3 + AB/√3 = 80
By equal Poles,
CD = AB
CD √3 + CD/√3 = 80
CD(√3 + 1/√3) = 80
CD(4/√3) = 80
CD = (80√3)/4
CD = 20√3 m
or AB = 20√3 m
BO = AB/√3 = (20√3)√3 = 20 m
DO = BD - BO = 80 - 20 = 60 m
Therefore, the height of Poles are 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°. 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° (see Figure). Find the height of the tower and the width of the canal.

Solution :-
```
Let AB be a TV Tower and BC be a Canal.
The height of the tower (AB) and the width of the canal (BC) = ?

In Δ ABC,
AB/BC = tan 60°
AB/BC = √3
BC = AB/√3

In Δ ABD,
AB/BD = tan 30°
AB/(BC + CD) = 1/√3
AB/(AB/√3 + 20) = 1/√3
(√3 AB)/(AB + 20√3) = 1/√3
3AB = AB + 20√3
3AB - AB = 20√3
2AB = 20√3
AB = (20√3)/2
AB = 10√3 m

BC = AB/√3 = (10√3)/√3 = 10 m
Therefore, the height of tower is 10√3 m and width of 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° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution :-
```
Let AB be a building and CD be a cable tower.
Height of building (AB) = 7 m
Height of tower = ?

In Δ ABD,
AB/BD = tan 45°
7/BD = 1
BD = 7 m

In Δ ACE,
AE = BD = 7m
CE/AE = tan 60°
CE/7 = √3
CE = 7√3 m

CD = CE + ED = 7√3 + 7 = 7(√3 + 1) m
Therefore, the height of cable tower is 7(√3 + 1) m.
Therefore, the height of tower is 10√3 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° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution :-
```
Let AB be a light house and C and D are be the two ships.
Height of light house (AB) = 75 m
The distance between the two ships (CD) = ?

In Δ ABC,
AB/BC = tan 45°
75/BC = 1
BC = 75 m

In Δ ABD,
AB/BD = tan 30°
75/(BC + CD) = 1/√3
BC + CD = 75√3
75 + CD = 75√3
CD = 75√3 - 75
CD = 75(√3 - 1) m

Therefore, The distance between the two ships 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°. After some time, the angle of elevation reduces to 30° (see Figure). Find the distance travelled by the balloon during the interval.

Solution :-
```
Let A and B are be the initial point and final point of baloon from the ground.
Height of baloon from the ground (BH) or (AF) = 88.2 m
Height of girl (CD) = 1.2m
Distance travelled by the baloon (EG) = ?

In Δ ACE,
AE/CE = tan 60°
(AF - EF)/CE = √3
(88.2 - 1.2)/CE = √3
87/CE = √3
CE = 87/√3 = 29√3 m

In Δ BCG,
BG/CG = tan 30°
(BH - GH)/CG = 1/√3
(88.2 - 1.2)/CG = 1/√3
87/CG = 1/√3
CG = 87√3 m

Therefore, Distance travelled by baloon (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°, 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.

Solution :-
```
Let AB be a tower and C and D are be the initial and final positions of car.
Time taken = ?

In Δ ABD,
AB/BD = tan 60°
AB/BD = √3
BD = AB/√3

In Δ ABC,
AB/BC = tan 30°
AB/(BD + CD) = 1/√3
AB √3 = BD + CD
AB √3 = AB/√3 + CD
CD = AB √3 - AB/√3
CD = 2AB/√3

Therefore, Time taken by the car to travel distance CD (2AB/√3) = 6 seconds
Time taken by the car to travel distance DB (AB/√3) = 6/(2AB/√3) x (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.

Solution :-
```
Let AQ be a tower and Base RQ and SQ are be the 4m and 9m respectively.
The angles are complementary. If one angle is θ and another is (90° - θ)

In Δ AQR,
AQ/QR = tan θ
AQ/4 = tan θ  .....(1)

In Δ AQS,
AQ/QS = tan(90 - θ)
AQ/9 = cot θ  .....(2)

Multipying eq (1) and (2)
AQ/4 x AQ/9 = tan θ . cot θ
AQ/36 = tan θ . 1/tan θ
AQ²/36 = 1
AQ² = 36
AQ = √36
AQ = 6
Therefore, the height of tower is 6 m (Hence Proved).
```
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