Chapter 9: Some Applications of Trigonometry
Exercise 9.1
Q1
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 \( 30^{\circ} \).
Solution:
Let AB be the vertical pole and AC be the rope.
Length of rope (AC) = 20 m.
Angle of elevation \( \angle C = 30^{\circ} \).
We need to find the height of the pole (AB).
In right \( \Delta ABC \):
\[ \sin 30^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} \]
\[ \frac{1}{2} = \frac{AB}{20} \]
\[ AB = \frac{20}{2} = 10 \text{ m} \]
Answer: The height of the pole is 10 m.
Q2
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle \( 30^{\circ} \) 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 the original tree be \( AC \). It breaks at point \( B \), and the top part \( AB \) touches the ground at point \( D \).
So, \( AB = BD \) (Broken part).
Distance from foot to top touching ground (\( CD \)) = 8 m.
Angle of elevation \( \angle D = 30^{\circ} \).
Total height of tree = \( BC + BD \).
1. Find BC:
\[ \tan 30^{\circ} = \frac{BC}{CD} \]
\[ \frac{1}{\sqrt{3}} = \frac{BC}{8} \Rightarrow BC = \frac{8}{\sqrt{3}} \text{ m} \]
2. Find BD:
\[ \cos 30^{\circ} = \frac{CD}{BD} \]
\[ \frac{\sqrt{3}}{2} = \frac{8}{BD} \Rightarrow BD = \frac{16}{\sqrt{3}} \text{ m} \]
3. Total Height:
\[ \text{Height} = BC + BD = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} \]
Rationalizing the denominator:
\[ \frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m} \]
Answer: The height of the tree is \( 8\sqrt{3} \) m.
Q3
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 \( 30^{\circ} \) 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 \( 60^{\circ} \) to the ground. What should be the length of the slide in each case?
Solution:
Case 1: For younger children
Height (\( AB \)) = 1.5 m. Angle (\( \theta \)) = \( 30^{\circ} \). Let length of slide be \( AC \).
\[ \sin 30^{\circ} = \frac{AB}{AC} \]
\[ \frac{1}{2} = \frac{1.5}{AC} \Rightarrow AC = 3 \text{ m} \]
Case 2: For elder children
Height (\( PQ \)) = 3 m. Angle (\( \theta \)) = \( 60^{\circ} \). Let length of slide be \( PR \).
\[ \sin 60^{\circ} = \frac{PQ}{PR} \]
\[ \frac{\sqrt{3}}{2} = \frac{3}{PR} \]
\[ PR = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \text{ m} \]
Answer: Length of slide for younger children is 3 m, and for elder children is \( 2\sqrt{3} \) m.
Q4
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 \( 30^{\circ} \). Find the height of the tower.
Solution:
Let the height of the tower be \( h \) and the distance from the foot be \( d = 30 \) m.
In the right triangle formed:
\[ \tan 30^{\circ} = \frac{\text{Height}}{\text{Distance}} = \frac{h}{30} \]
\[ \frac{1}{\sqrt{3}} = \frac{h}{30} \]
\[ h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m} \]
Answer: The height of the tower is \( 10\sqrt{3} \) m.
Q5
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 \( 60^{\circ} \). Find the length of the string, assuming that there is no slack in the string.
Solution:
Let the length of the string be \( l \) and the height be \( h = 60 \) m.
Angle of elevation = \( 60^{\circ} \).
\[ \sin 60^{\circ} = \frac{\text{Height}}{\text{String Length}} = \frac{60}{l} \]
\[ \frac{\sqrt{3}}{2} = \frac{60}{l} \]
\[ l = \frac{120}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \text{ m} \]
Answer: The length of the string is \( 40\sqrt{3} \) m.
Q6
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 \( 30^{\circ} \) to \( 60^{\circ} \) as he walks towards the building. Find the distance he walked towards the building.
Solution:
Height of building = 30 m.
Height of boy = 1.5 m.
Height of building above eye level (\( AB \)) = \( 30 - 1.5 = 28.5 \) m.
Let the initial distance be \( x \) (at \( 30^{\circ} \)) and the final distance be \( y \) (at \( 60^{\circ} \)). The distance walked is \( x - y \).
1. At \( 60^{\circ} \):
\[ \tan 60^{\circ} = \frac{28.5}{y} \Rightarrow \sqrt{3} = \frac{28.5}{y} \Rightarrow y = \frac{28.5}{\sqrt{3}} \]
2. At \( 30^{\circ} \):
\[ \tan 30^{\circ} = \frac{28.5}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3} \]
3. Distance walked:
\[ x - y = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} \]
\[ = 28.5 \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = 28.5 \left( \frac{3-1}{\sqrt{3}} \right) \]
\[ = \frac{28.5 \times 2}{\sqrt{3}} = \frac{57}{\sqrt{3}} = \frac{57\sqrt{3}}{3} = 19\sqrt{3} \text{ m} \]
Answer: The distance walked is \( 19\sqrt{3} \) m.
Q7
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 \( 45^{\circ} \) and \( 60^{\circ} \) respectively. Find the height of the tower.
Solution:
Let height of building (\( BC \)) = 20 m.
Let height of tower (\( AB \)) = \( h \) m.
Let point on ground be \( D \) at distance \( x \) from the building base.
1. Triangle BCD (Bottom of tower, \( 45^{\circ} \)):
\[ \tan 45^{\circ} = \frac{BC}{CD} \Rightarrow 1 = \frac{20}{x} \Rightarrow x = 20 \text{ m} \]
2. Triangle ACD (Top of tower, \( 60^{\circ} \)):
\[ \tan 60^{\circ} = \frac{AC}{CD} = \frac{20 + h}{x} \]
\[ \sqrt{3} = \frac{20 + h}{20} \]
\[ 20\sqrt{3} = 20 + h \]
\[ h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \text{ m} \]
Answer: Height of the tower is \( 20(\sqrt{3} - 1) \) m.
Q8
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 \( 60^{\circ} \) and from the same point the angle of elevation of the top of the pedestal is \( 45^{\circ} \). Find the height of the pedestal.
Solution:
Let height of pedestal be \( h \) m.
Height of statue = 1.6 m. Total height = \( h + 1.6 \) m.
Let the distance of point from the base be \( x \).
1. For pedestal top (\( 45^{\circ} \)):
\[ \tan 45^{\circ} = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h \]
2. For statue top (\( 60^{\circ} \)):
\[ \tan 60^{\circ} = \frac{h + 1.6}{x} \]
\[ \sqrt{3} = \frac{h + 1.6}{h} \quad (\text{Since } x = h) \]
\[ h\sqrt{3} = h + 1.6 \]
\[ h(\sqrt{3} - 1) = 1.6 \]
\[ h = \frac{1.6}{\sqrt{3} - 1} \]
Rationalizing:
\[ h = \frac{1.6(\sqrt{3} + 1)}{3 - 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \text{ m} \]
Answer: Height of the pedestal is \( 0.8(\sqrt{3} + 1) \) m.
Q9
The angle of elevation of the top of a building from the foot of the tower is \( 30^{\circ} \) and the angle of elevation of the top of the tower from the foot of the building is \( 60^{\circ} \). If the tower is 50 m high, find the height of the building.
Solution:
Height of tower (\( AB \)) = 50 m. Let height of building (\( CD \)) = \( h \) m.
Distance between them = \( x \).
1. From foot of building to top of tower (\( 60^{\circ} \)):
\[ \tan 60^{\circ} = \frac{50}{x} \Rightarrow \sqrt{3} = \frac{50}{x} \Rightarrow x = \frac{50}{\sqrt{3}} \]
2. From foot of tower to top of building (\( 30^{\circ} \)):
\[ \tan 30^{\circ} = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow x = h\sqrt{3} \]
3. Equate \( x \):
\[ h\sqrt{3} = \frac{50}{\sqrt{3}} \]
\[ 3h = 50 \Rightarrow h = \frac{50}{3} = 16\frac{2}{3} \text{ m} \]
Answer: Height of the building is \( 16\frac{2}{3} \) m.
Q10
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 \( 60^{\circ} \) and \( 30^{\circ} \), respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
Let height of poles be \( h \).
Total width = 80 m. Let point be at distance \( x \) from one pole, so distance from other is \( 80 - x \).
1. First Pole (\( 60^{\circ} \)):
\[ \tan 60^{\circ} = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]
2. Second Pole (\( 30^{\circ} \)):
\[ \tan 30^{\circ} = \frac{h}{80 - x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \Rightarrow h = \frac{80 - x}{\sqrt{3}} \]
3. Equate \( h \):
\[ x\sqrt{3} = \frac{80 - x}{\sqrt{3}} \]
\[ 3x = 80 - x \Rightarrow 4x = 80 \Rightarrow x = 20 \text{ m} \]
Distance from other pole = \( 80 - 20 = 60 \text{ m} \).
Height \( h = 20\sqrt{3} \) m.
Answer: Height of poles is \( 20\sqrt{3} \) m. Distances are 20 m and 60 m.
Q11
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 \( 60^{\circ} \). 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 \( 30^{\circ} \). Find the height of the tower and the width of the canal.
Solution:
Let height of tower be \( h \) and width of canal be \( x \).
1. At canal bank (\( 60^{\circ} \)):
\[ \tan 60^{\circ} = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]
2. At 20 m away (\( 30^{\circ} \)):
\[ \tan 30^{\circ} = \frac{h}{x + 20} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 20} \Rightarrow h = \frac{x + 20}{\sqrt{3}} \]
3. Equate \( h \):
\[ x\sqrt{3} = \frac{x + 20}{\sqrt{3}} \]
\[ 3x = x + 20 \Rightarrow 2x = 20 \Rightarrow x = 10 \text{ m} \]
Height \( h = 10\sqrt{3} \) m.
Answer: Height of tower is \( 10\sqrt{3} \) m, width of canal is 10 m.
Q12
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is \( 60^{\circ} \) and the angle of depression of its foot is \( 45^{\circ} \). Determine the height of the tower.
Solution:
Height of building (\( AB \)) = 7 m.
Let distance between building and tower be \( x \).
Angle of depression of foot is \( 45^{\circ} \), so angle of elevation from foot of tower to top of building is \( 45^{\circ} \).
\[ \tan 45^{\circ} = \frac{7}{x} \Rightarrow 1 = \frac{7}{x} \Rightarrow x = 7 \text{ m} \]
Now consider the top part of the tower above the building level. Let this height be \( h_1 \).
\[ \tan 60^{\circ} = \frac{h_1}{x} = \frac{h_1}{7} \]
\[ h_1 = 7\sqrt{3} \text{ m} \]
Total height of tower = Building height + \( h_1 \)
\[ = 7 + 7\sqrt{3} = 7(\sqrt{3} + 1) \text{ m} \]
Answer: \( 7(\sqrt{3} + 1) \) m.
Q13
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are \( 30^{\circ} \) and \( 45^{\circ} \). If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
Height of lighthouse = 75 m.
Let distance to first ship (angle \( 45^{\circ} \)) be \( x \) and second ship (angle \( 30^{\circ} \)) be \( y \).
1. For ship at \( 45^{\circ} \):
\[ \tan 45^{\circ} = \frac{75}{x} \Rightarrow 1 = \frac{75}{x} \Rightarrow x = 75 \text{ m} \]
2. For ship at \( 30^{\circ} \):
\[ \tan 30^{\circ} = \frac{75}{y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{y} \Rightarrow y = 75\sqrt{3} \text{ m} \]
Distance between ships = \( y - x = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) \) m.
Answer: \( 75(\sqrt{3} - 1) \) m.
Q14
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 \( 60^{\circ} \). After some time, the angle of elevation reduces to \( 30^{\circ} \). Find the distance travelled by the balloon during the interval.
Solution:
Height of balloon from ground = 88.2 m. Height of girl = 1.2 m.
Effective vertical height (\( h \)) = \( 88.2 - 1.2 = 87 \) m.
1. Initial Position (\( 60^{\circ} \)):
Let horizontal distance be \( x \).
\[ \tan 60^{\circ} = \frac{87}{x} \Rightarrow x = \frac{87}{\sqrt{3}} = 29\sqrt{3} \text{ m} \]
2. Final Position (\( 30^{\circ} \)):
Let horizontal distance be \( y \).
\[ \tan 30^{\circ} = \frac{87}{y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{87}{y} \Rightarrow y = 87\sqrt{3} \text{ m} \]
Distance Travelled:
\[ y - x = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3} \text{ m} \]
Answer: \( 58\sqrt{3} \) m.
Q15
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 \( 30^{\circ} \), 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 \( 60^{\circ} \). Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let height of tower be \( h \).
Let distance at \( 30^{\circ} \) be \( x \) and at \( 60^{\circ} \) be \( y \).
1. At \( 30^{\circ} \):
\[ \tan 30^{\circ} = \frac{h}{x} \Rightarrow x = h\sqrt{3} \]
2. At \( 60^{\circ} \):
\[ \tan 60^{\circ} = \frac{h}{y} \Rightarrow y = \frac{h}{\sqrt{3}} \]
Distance travelled = \( x - y = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = h \left( \frac{2}{\sqrt{3}} \right) \).
Time taken for this distance = 6 seconds.
Speed \( v = \frac{\text{Distance}}{\text{Time}} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}} \).
Time to cover remaining distance \( y \):
\[ t = \frac{y}{v} = \frac{h/\sqrt{3}}{h/3\sqrt{3}} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} = 3 \text{ seconds} \]
Answer: 3 seconds.