CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 09

NCERT 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 \( 30^\circ \).

Answer:
Let AB be the pole and AC the rope, \( AC = 20 \) m and \( \angle ACB = 30^\circ \). Let \( AB = x \) m.
In right \( \triangle ABC \): \( \sin 30^\circ = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{x}{20} \Rightarrow x = 10 \).
The height of the pole is 10 m.
In simple words: The pole is the side opposite the \( 30^\circ \) angle and the rope is the hypotenuse, so use sine.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 30^\circ \) 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.

Answer:
Let the tree break at P, so PQ is the standing part and PR the broken part, with \( QR = 8 \) m and \( \angle PRQ = 30^\circ \).
In right \( \triangle PQR \): \( \cos 30^\circ = \frac{QR}{PR} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{PR} \Rightarrow PR = \frac{16}{\sqrt{3}} \) m.
\( \tan 30^\circ = \frac{PQ}{QR} \Rightarrow \frac{1}{\sqrt{3}} = \frac{PQ}{8} \Rightarrow PQ = \frac{8}{\sqrt{3}} \) m.
Height of the tree \( = PQ + PR = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \) m, about 13.86 m.
In simple words: Height of the tree = standing part + broken part.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 30^\circ \) 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?

Answer:
Let AM be the slide for the younger children, \( AB = 1.5 \) m, \( \angle AMB = 30^\circ \), and PN the slide for the elder children, \( PQ = 3 \) m, \( \angle PNQ = 60^\circ \).
In right \( \triangle ABM \): \( \sin 30^\circ = \frac{AB}{AM} \Rightarrow \frac{1}{2} = \frac{1.5}{AM} \Rightarrow AM = 3 \) m.
In right \( \triangle PQN \): \( \sin 60^\circ = \frac{PQ}{PN} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{PN} \Rightarrow PN = \frac{6}{\sqrt{3}} = 2\sqrt{3} \) m.
The slide for children below 5 years is 3 m long and the slide for elder children is \( 2\sqrt{3} \) m (about 3.46 m) long.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 30^\circ \). Find the height of the tower.

Answer:
Let AB be the tower and C the point, \( BC = 30 \) m and \( \angle ACB = 30^\circ \).
\( \tan 30^\circ = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{30} \Rightarrow AB = \frac{30}{\sqrt{3}} = 10\sqrt{3} \) m.
The height of the tower is \( 10\sqrt{3} \) m (about 17.3 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 60^\circ \). Find the length of the string, assuming that there is no slack in the string.

Answer:
Let C be the kite, B the point below it and A the point where the string is tied: \( BC = 60 \) m, \( \angle CAB = 60^\circ \).
\( \sin 60^\circ = \frac{BC}{AC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{AC} \Rightarrow AC = \frac{120}{\sqrt{3}} = 40\sqrt{3} \) m.
The length of the string is \( 40\sqrt{3} \) m (about 69.3 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 30^\circ \) to \( 60^\circ \) as he walks towards the building. Find the distance he walked towards the building.

Answer:
The eye level is 1.5 m above the ground, so the building rises \( 30 - 1.5 = 28.5 \) m above his eyes. Let the boy walk from M to L; the eye positions are D and C, and \( AB = 28.5 \) m.
In right \( \triangle ABD \): \( \tan 30^\circ = \frac{28.5}{BD} \Rightarrow BD = 28.5\sqrt{3} \) m.
In right \( \triangle ABC \): \( \tan 60^\circ = \frac{28.5}{BC} \Rightarrow BC = \frac{28.5}{\sqrt{3}} = \frac{28.5\sqrt{3}}{3} \) m.
Distance walked \( = CD = BD - BC = 28.5\sqrt{3} - \frac{28.5\sqrt{3}}{3} = 28.5\sqrt{3} \times \frac{2}{3} = 19\sqrt{3} \) m, about 32.9 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 45^\circ \) and \( 60^\circ \) respectively. Find the height of the tower.

Answer:
Let BC be the building (20 m) and AB the tower of height h m; D is the point on the ground.
In right \( \triangle BCD \): \( \tan 45^\circ = \frac{BC}{CD} \Rightarrow 1 = \frac{20}{CD} \Rightarrow CD = 20 \) m.
In right \( \triangle ACD \): \( \tan 60^\circ = \frac{AC}{CD} \Rightarrow \sqrt{3} = \frac{h + 20}{20} \Rightarrow h = 20\sqrt{3} - 20 \).
The height of the tower is \( 20(\sqrt{3} - 1) \) m (about 14.64 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 60^\circ \) and from the same point the angle of elevation of the top of the pedestal is \( 45^\circ \). Find the height of the pedestal.

Answer:
Let BC be the pedestal of height h m and AB the statue, \( AB = 1.6 \) m; D is the point on the ground.
In right \( \triangle BCD \): \( \tan 45^\circ = \frac{BC}{CD} \Rightarrow CD = BC = h \).
In right \( \triangle ACD \): \( \tan 60^\circ = \frac{AB + BC}{CD} \Rightarrow \sqrt{3} = \frac{1.6 + h}{h} \Rightarrow h(\sqrt{3} - 1) = 1.6 \).
\( h = \frac{1.6}{\sqrt{3} - 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \).
The height of the pedestal is \( 0.8(\sqrt{3} + 1) \) m (about 2.19 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 9. 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.

Answer:
Let PQ be the tower (50 m) and AB the building, with P and B the feet of the tower and of the building; \( \angle QBP = 60^\circ \) and \( \angle APB = 30^\circ \).
In right \( \triangle BPQ \): \( \tan 60^\circ = \frac{PQ}{BP} \Rightarrow \sqrt{3} = \frac{50}{BP} \Rightarrow BP = \frac{50}{\sqrt{3}} \) m.
In right \( \triangle ABP \): \( \tan 30^\circ = \frac{AB}{BP} \Rightarrow AB = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = \frac{50}{3} \) m.
The height of the building is \( \frac{50}{3} \) m (about 16.67 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 60^\circ \) and \( 30^\circ \) respectively. Find the height of the poles and the distances of the point from the poles. [CBSE 2022]

Answer:
Let AB and CD be the poles, each of height h m, O the point on the road with \( BD = 80 \) m, \( \angle AOB = 60^\circ \), \( \angle COD = 30^\circ \). Let \( OD = x \), so \( BO = 80 - x \).
In \( \triangle ABO \): \( \tan 60^\circ = \frac{h}{80 - x} \Rightarrow h = \sqrt{3}(80 - x) \) ...(i)
In \( \triangle CDO \): \( \tan 30^\circ = \frac{h}{x} \Rightarrow h = \frac{x}{\sqrt{3}} \) ...(ii)
From (i) and (ii): \( \sqrt{3}(80 - x) = \frac{x}{\sqrt{3}} \Rightarrow 240 - 3x = x \Rightarrow x = 60 \). Then \( h = \frac{60}{\sqrt{3}} = 20\sqrt{3} \) and \( BO = 20 \).
Height of each pole \( = \) \( 20\sqrt{3} \) m (about 34.6 m); the point is 20 m from the first pole and 60 m from the second pole.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 60^\circ \). 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.

Answer:
Let PQ be the tower, Q its foot, C the point opposite the tower and D the second point, \( DC = 20 \) m. Let \( QC = x \) m.
In \( \triangle PQC \): \( \tan 60^\circ = \frac{PQ}{x} \Rightarrow PQ = \sqrt{3}x \) ...(i)
In \( \triangle PQD \): \( \tan 30^\circ = \frac{PQ}{20 + x} \Rightarrow PQ = \frac{20 + x}{\sqrt{3}} \) ...(ii)
From (i) and (ii): \( \sqrt{3}x = \frac{20 + x}{\sqrt{3}} \Rightarrow 3x = 20 + x \Rightarrow x = 10 \). Then \( PQ = 10\sqrt{3} \).
The height of the tower is \( 10\sqrt{3} \) m (about 17.3 m) and the width of the canal is 10 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 12. 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.

Answer:
Let PQ be the building (7 m) and EC the tower; the horizontal through P meets EC at D, so \( CD = PQ = 7 \) m.
In right \( \triangle PQC \): \( \tan 45^\circ = \frac{PQ}{QC} \Rightarrow QC = 7 \) m, so \( PD = QC = 7 \) m.
In right \( \triangle PDE \): \( \tan 60^\circ = \frac{DE}{PD} \Rightarrow DE = 7\sqrt{3} \) m.
Height of the tower \( EC = DE + CD = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \) m, about 19.12 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 30^\circ \) 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.

Answer:
Let PQ be the lighthouse (75 m) and C, D the positions of the ships, with \( \angle PCQ = 45^\circ \) and \( \angle PDQ = 30^\circ \) (alternate angles).
In \( \triangle PQC \): \( \tan 45^\circ = \frac{75}{QC} \Rightarrow QC = 75 \) m.
In \( \triangle PQD \): \( \tan 30^\circ = \frac{75}{QD} \Rightarrow QD = 75\sqrt{3} \) m.
Distance between the ships \( CD = QD - QC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) \) m, about 54.9 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 60^\circ \). After some time, the angle of elevation reduces to \( 30^\circ \). Find the distance travelled by the balloon during the interval.

Answer:
Height of the balloon above the girl's eyes \( = 88.2 - 1.2 = 87 \) m. Let C be the girl's eyes, P the first and Q the second position of the balloon, and D, E the feet of the perpendiculars from P and Q on the horizontal line through C. So \( PD = QE = 87 \) m.
In right \( \triangle QEC \) (angle \( 30^\circ \)): \( \tan 30^\circ = \frac{87}{CE} \Rightarrow CE = 87\sqrt{3} \) m.
In right \( \triangle PDC \) (angle \( 60^\circ \)): \( \tan 60^\circ = \frac{87}{CD} \Rightarrow CD = \frac{87}{\sqrt{3}} = 29\sqrt{3} \) m.
Distance travelled \( DE = CE - CD = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3} \) m, about 100.5 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

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 \( 30^\circ \), 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.

Answer:
Let PQ be the tower and let the car move from D to C in 6 seconds, with \( \angle PDQ = 30^\circ \) and \( \angle PCQ = 60^\circ \) (alternate angles).
In \( \triangle PQC \): \( \tan 60^\circ = \frac{PQ}{QC} \Rightarrow PQ = \sqrt{3}\,QC \). In \( \triangle PQD \): \( \tan 30^\circ = \frac{PQ}{QD} \Rightarrow QD = \sqrt{3}\,PQ = 3\,QC \).
So \( CD = QD - QC = 2\,QC \). The car covers \( CD = 2\,QC \) in 6 seconds, i.e. \( QC \) in 3 seconds, at a uniform speed.
The car takes 3 seconds to reach the foot of the tower from C.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angle first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.