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Some Applications of Trigonometry: 4 marks Questions (CBSE Class 10)

25 different 4 marks questions on Some Applications of Trigonometry from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (30)2 marks (3)3 marks (20)4 marks (25)5 marks (73)

Q374 marksCase StudySome Applications of Trigonometry CBSE 2026 · Standard 30/1/1

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure).
Distance between the base of the tower and point 'O' is 6 m. From point 'O', the angle of elevation of the top of the section 'B' is $3 0^{\circ}$ and the angle of elevation of the top of section 'A' is $6 0^{\circ}$ .
Based on the above information, answer the following questions :
(i) Find the length of the wire from the point 'O' to the top of section 'B'. (1)
(ii) Find the length of the wire from the point 'O' to the top of section 'A'. (1)
(iii) (a) Find the distance AB. (2)
OR
(iii) (b) Find the area of $△ O P B$ . (2)

Diagram for CBSE 2026 Class 10 Maths question 37

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Answer: (i)

43

m (ii) 12 m (iii) (a)

43

m OR (b)

63

^{2}

P is the base of the tower, $O P = 6$ m, $∠ B O P = 3 0^{\circ}$ , $∠ A O P = 6 0^{\circ}$ , $∠ O P A = 9 0^{\circ}$ .
(i) $cos 3 0^{\circ} = \frac{O P}{O B}$ , so $O B = \frac{6}{3 /2} = \frac{12}{3} = 43$ m.
(ii) $cos 6 0^{\circ} = \frac{O P}{O A}$ , so $O A = \frac{6}{1/2} = 12$ m.
(iii) (a) $A P = 6 tan 6 0^{\circ} = 63$ m and $B P = 6 tan 3 0^{\circ} = \frac{6}{3} = 23$ m.
$A B = 63 - 23 = 43$ m.
(iii) (b) Area of $△ O P B = \frac{1}{2} \times O P \times B P = \frac{1}{2} \times 6 \times 23 = 63$ m $^{2}$ .

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q374 marksCase StudySome Applications of Trigonometry CBSE 2026 · Standard 30/2/1

Tejas is standing at the top of a building and observes a car at an angle of depression of $3 0^{\circ}$ as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to $6 0^{\circ}$ , and at that moment, the car is 25 m away from the building.
Based on the information given above, answer the following questions :
(i) What is the height of the building ? (1)
(ii) What is the distance between the two positions of the car ? (1)
(iii) (a) What would be the total time taken by the car to reach the foot of the building from the starting point ? (2)
OR
(iii) (b) What is the distance of the observer from the car when it makes an angle of $6 0^{\circ}$ ? (2)

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Answer: (i)

253

m (ii) 50 m (iii) (a) 9 seconds OR (b) 50 m

In the figure, AB is the building, C and D are the two positions of the car, CB = 25 m.
(i) In $△ A B C$ , $tan 6 0^{\circ} = \frac{A B}{C B}$ , so $A B = 253$ m.
(ii) In $△ A B D$ , $tan 3 0^{\circ} = \frac{A B}{D B}$ , so $D B = 253 \times 3 = 75$ m. DC = 75 - 25 = 50 m.
(iii)(a) Speed $= \frac{50}{6} = \frac{25}{3}$ m/s. Time for the remaining 25 m $= 25 \div \frac{25}{3} = 3$ s.
Total time = 6 + 3 = 9 seconds.
(iii)(b) $cos 6 0^{\circ} = \frac{C B}{A C}$ , so $A C = \frac{25}{1/2} = 50$ m.

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q384 marksCase StudySome Applications of Trigonometry CBSE 2026 · Standard 30/5/1

Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is 54 metres away from the water tank.
From a window (W) of the building, the angle of elevation of top of the tank is $4 5^{\circ}$ and angle of depression of its foot is $3 0^{\circ}$ .
(i) Write a relation between d (the height of window) and y. (1)
(ii) Determine the value of h. (1)
(iii) (a) Determine height of the water tank. (2)
OR (iii) (b) Find the value of $x$ and height of the window above ground level. (2)

Diagram for CBSE 2026 Class 10 Maths question 38

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Answer: (i) y = 2d (ii) h = 54 m (iii) (a)

54 + 183

m ≈ 85.18 m OR (iii) (b)

x = 542

m, window height

= 183

m ≈ 31.18 m

XW = AC = 54 m, $∠ B W X = 4 5^{\circ}$ , $∠ X W A = 3 0^{\circ}$ , AX = WC = d.
(i) In right $△ A X W$ : $sin 3 0^{\circ} = \frac{A X}{A W} = \frac{d}{y}$ , so $y = 2 d$ .
(ii) In right $△ B X W$ : $tan 4 5^{\circ} = \frac{h}{54}$ , so h = 54 m.
(iii) (a) $tan 3 0^{\circ} = \frac{d}{54}$ , so $d = \frac{54}{3} = 183$ m.
Height of tank AB $= h + d = 54 + 183 \approx 54 + 31.18 = 85.18$ m
(iii) (b) $cos 4 5^{\circ} = \frac{54}{x}$ , so $x = 542$ m (≈ 76.37 m).
Height of window $= d = 183 \approx 31.18$ m

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q364 marksCase StudySome Applications of Trigonometry CBSE 2025 · Basic 430/1/1

An injured bird was found on the roof of a building. The building is 15 m high. A fireman was called to rescue the bird. The fireman used an adjustable ladder to reach the roof. He placed the ladder in such a way that the ladder makes an angle of $6 0^{\circ}$ with the ground in order to reach the roof.
Based on the above information, answer the following questions :
(i) Find the length of the ladder used by the fireman to reach the roof. (1)
(ii) Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building. (1)
(iii) In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of $3 0^{\circ}$ with the ground.
(a) Draw a neat diagram to represent the above situation and hence find the width of the road between the building and the wall. (2)
OR
(b) Find the length of the ladder used by the fireman in this case. (2)

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Answer: (i)

103

m (ii)

53

m (iii) (a)

153

m OR (b) 30 m

(i) $sin 6 0^{\circ} = \frac{15}{l}$ , so $l = \frac{15 \times 2}{3} = 103$ m.
(ii) $tan 6 0^{\circ} = \frac{15}{d}$ , so $d = \frac{15}{3} = 53$ m.
(iii) (a) Diagram: vertical building AB = 15 m, foot of the wall C on the ground, ladder AC with $∠ A C B = 3 0^{\circ}$ .
$tan 3 0^{\circ} = \frac{A B}{B C}$ , so $B C = 153$ m; the road is $153$ m wide.
(iii) (b) $sin 3 0^{\circ} = \frac{15}{A C}$ , so $A C = 30$ m.

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q364 marksCase StudySome Applications of Trigonometry CBSE 2025 · Basic 430/2/1

Kite festival is a popular festival in India which takes place during Makar Sankranti. The festival is celebrated by people flying kites from their rooftops. Reena and Ravi are also flying kites to enjoy the festival. The height of Reena’s kite is 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground, and the inclination of the string with the ground is $3 0^{\circ}$ . Ravi is flying a kite from a 10 m high building. His kite is also flying 60 m above the ground and the length of the string used by Ravi is same as that of Reena’s. $θ$ is the angle of elevation of Ravi’s kite from a point on the rooftop.
Based on the above information, answer the following questions :
(i) Find the length of string used by Reena. (1)
(ii) Find the value of $sin θ$ . (1)
(iii) (a) If $θ$ changes to $6 0^{\circ}$ , without changing the length of the string, what will be the height of Ravi’s kite above the ground ? (Use $3 = 1.7$ ) (2)
OR (b) What would have been the height of Ravi’s kite above the ground, if the string had an inclination of $3 0^{\circ}$ with the ground, assuming that the length of the string does not change ? (2)

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Answer: (i) 120 m (ii)

sin θ = \frac{5}{12}

(iii) (a) 112 m OR (b) 70 m

(i) For Reena: $sin 3 0^{\circ} = \frac{60}{L} \Rightarrow \frac{1}{2} = \frac{60}{L} \Rightarrow L = 120$ m.
(ii) Ravi’s kite is $60 - 10 = 50$ m above the rooftop and his string is 120 m long, so $sin θ = \frac{50}{120} = \frac{5}{12}$ .
(iii) (a) With $θ = 6 0^{\circ}$ : height above rooftop $= 120 sin 6 0^{\circ} = 120 \times \frac{3}{2} = 603 = 60 \times 1.7 = 102$ m.
Height above ground $= 102 + 10 = 112$ m.
OR (b) With inclination $3 0^{\circ}$ : height above rooftop $= 120 sin 3 0^{\circ} = 60$ m.
Height above ground $= 60 + 10 = 70$ m.

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q374 marksCase StudySome Applications of Trigonometry CBSE 2025 · Basic 430/3/1

Two motorboats A and B are waiting at the opposite banks of a river in order to reach the opposite side. From a point P on the bridge, 20 m above the river, the angles of depression of the boats are $3 0^{\circ}$ and $4 5^{\circ}$ respectively, as shown in the figure given below. Both the boats leave at the same time at the speed of 10 m/s and 5 m/s, respectively
Based on the above information, answer the following questions :
(i) Find the distance travelled by boat A to reach point D in the river, vertically below the point P. (Use $3 = 1.73$ ) (1)
(ii) What is the width of the river ? (1)
(iii) (a) Which boat will reach point D first, and how much earlier, than the other boat ? (2)
OR
(b) What is the distance between the two boats after 3 seconds ? (2)

Diagram for CBSE 2025 Class 10 Maths question 37

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Answer: (i) 34.6 m (ii) 54.6 m (iii) (a) Boat A, 0.54 s earlier OR (b) 9.6 m

PD = 20 m; angle of depression of A is $3 0^{\circ}$ , of B is $4 5^{\circ}$ , so $∠ P A D = 3 0^{\circ}$ and $∠ P B D = 4 5^{\circ}$ .
(i) $tan 3 0^{\circ} = \frac{P D}{A D} \Rightarrow A D = 203 = 20 \times 1.73 = 34.6$ m.
(ii) $tan 4 5^{\circ} = \frac{P D}{D B} \Rightarrow D B = 20$ m. Width $= A D + D B = 34.6 + 20 = 54.6$ m.
(iii) (a) Time for A $= \frac{34.6}{10} = 3.46$ s; time for B $= \frac{20}{5} = 4$ s. Boat A reaches D first, $4 - 3.46 = 0.54$ s earlier.
(iii) (b) In 3 s, A travels 30 m and B travels 15 m towards each other. Distance between them $= 54.6 - 30 - 15 = 9.6$ m.

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q384 marksCase StudySome Applications of Trigonometry CBSE 2025 · Standard 30/1/1

Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be $6 0^{\circ}$ . Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be $4 5^{\circ}$ .
Based on the above given information, answer the following questions :
(i) If CD is h metres, find the distance BD in terms of 'h'. (1)
(ii) Find distance BC in terms of 'h'. (1)
(iii) (a) Find the height CE of the lighthouse [Use $3 = 1.73$ ] (2)
OR
(iii) (b) Find distance AE, if AC = 100 m. (2)

Diagram for CBSE 2025 Class 10 Maths question 38

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Answer: (i)

B D = h

m (ii)

B C = h 2

m (iii) (a)

C E = 20 (3 + 3) = 94.6

m OR (iii) (b)

A E = 50

$A$ is the original position, $B$ the deck ( $A B = 40$ m), $C E$ the lighthouse, $B D ∥ A E$ , $D E = 40$ m, $C D = h$ .
(i) In right $△ B D C$ : $tan 4 5^{\circ} = \frac{C D}{B D}$ , so $B D = h$ m.
(ii) $B C = B D^{2} + C D^{2} = h^{2} + h^{2} = h 2$ m.
(iii) (a) $A E = B D = h$ and $C E = h + 40$ . In right $△ A E C$ : $tan 6 0^{\circ} = \frac{h + 40}{h} = 3$ .
$h (3 - 1) = 40$ , so $h = \frac{40}{3 - 1} = 20 (3 + 1) = 20 \times 2.73 = 54.6$ m.
$C E = h + 40 = 54.6 + 40 = 94.6$ m.
(iii) (b) In right $△ A E C$ : $cos 6 0^{\circ} = \frac{A E}{A C}$ , so $A E = 100 \times \frac{1}{2} = 50$ m.

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q374 marksCase StudySome Applications of Trigonometry CBSE 2025 · Standard 30/3/1

A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $3 0^{\circ}$ and $4 5^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is 50 metres.
Based on the information given above, answer the following questions :
(i) Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is $4 5^{\circ}$ . (1)
(ii) Find the measures of $∠ P B A$ and $∠ QB A$ . (1)
(iii) (a) Find the distance travelled by the ship between points P and Q. (2)
OR
(b) If the ship continues moving towards the shore and takes 10 minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A. (2)

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Answer: (i) 50 m (ii)

∠ P B A = 6 0^{\circ}

∠ QB A = 4 5^{\circ}

(iii) (a)

50 (3 - 1)

\approx 36.6

m OR (b) 0.3 km/h

(i) In right $△ Q A B$ , $∠ B Q A = 4 5^{\circ}$ , so $tan 4 5^{\circ} = \frac{A B}{A Q}$ , $A Q = A B = 50$ m.
(ii) $∠ P B A = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ and $∠ QB A = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$ .
(iii)(a) In right $△ P A B$ , $∠ B P A = 3 0^{\circ}$ , so $tan 3 0^{\circ} = \frac{50}{A P}$ , $A P = 503$ m.
(iii)(a) $P Q = A P - A Q = 503 - 50 = 50 (3 - 1) \approx 36.6$ m.
(iii)(b) QA = 50 m = 0.05 km; time = 10 min = $\frac{1}{6}$ h.
(iii)(b) Speed $= 0.05 \div \frac{1}{6} = 0.3$ km/h.

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q384 marksCase StudySome Applications of Trigonometry CBSE 2025 · Standard 30/4/1

A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident.
The ambulance was travelling at the speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be $3 0^{\circ}$ at a particular instant.
Based on above information, answer the following questions :
(i) Represent the above situation with the help of a diagram. (1)
(ii) Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use $3 = 1.73$ ) (1)
(iii) (a) Find the time (in seconds) in which the angle of depression changes from $3 0^{\circ}$ to $4 5^{\circ}$ . (2)
OR (iii) (b) How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is $6 0^{\circ}$ from the drone ? (2)

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Answer: (i) Right triangle QPA with QP = 100 m vertical, ambulance at A on the highway, angle of depression

3 0^{\circ}

at Q (ii) 173 m (iii) (a) 4.38 s OR (iii) (b)

23 \approx 3.46

(i) Draw QP = 100 m vertical with P on the highway; ambulance at A on the highway; $∠$ of depression from Q to A $= 3 0^{\circ}$ , so $∠ Q A P = 3 0^{\circ}$ .
(ii) $tan 3 0^{\circ} = \frac{QP}{P A} \Rightarrow P A = 1003 = 173$ m
Speed $= 60$ km/h $= \frac{60 \times 1000}{3600} = \frac{50}{3}$ m/s
(iii)(a) At $4 5^{\circ}$ : distance from P $= 100$ m. Distance covered $= 173 - 100 = 73$ m
Time $= 73 \div \frac{50}{3} = \frac{219}{50} = 4.38$ s
(iii)(b) At T: $P T = \frac{100}{t a n 6 0 ^{\circ}} = \frac{100}{3}$ m
Time $= \frac{100}{3} \times \frac{3}{50} = 23 \approx 3.46$ s

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q384 marksCase StudySome Applications of Trigonometry CBSE 2025 · Standard 30/5/1

Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the aircraft doors and the ground. Larger aircraft have door sills 5 to 20 feet (1 foot = 30 cm) high. Stairs facilitate safe boarding and de-boarding.
An aircraft has a door sill at a height of 15 feet above the ground. A stair car is placed at a horizontal distance of 15 feet from the plane.
Based on given information, answer the questions given in part (i) and (ii).
(i) Find the angle at which stairs are inclined to reach the door sill 15 feet high above the ground. (1)
(ii) Find the length of stairs used to reach the door sill. (1)
Further, answer any one of the following questions :
(iii) (a) If the 20 feet long stairs is inclined at an angle of $6 0^{\circ}$ to reach the door sill, then find the height of the door sill above the ground. (use $3 = 1.732$ ) (2)
OR (iii) (b) What should be the shortest possible length of stairs to reach the door sill of the plane 20 feet above the ground, if the angle of elevation cannot exceed $3 0^{\circ}$ ? Also, find the horizontal distance of base of stair car from the plane. (2)

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Answer: (i)

4 5^{\circ}

(ii)

152

feet (iii)(a)

103 = 17.32

feet OR (iii)(b) 40 feet; horizontal distance

203 \approx 34.64

feet

(i) $tan θ = \frac{15}{15} = 1 \Rightarrow θ = 4 5^{\circ}$
(ii) Length $= 1 5^{2} + 1 5^{2} = 152$ feet
(iii)(a) Height $= 20 sin 6 0^{\circ} = 20 \times \frac{3}{2} = 103 = 17.32$ feet
(iii)(b) Length $= \frac{20}{s i n θ}$ is least when $θ$ is largest, i.e. $θ = 3 0^{\circ}$ : length $= \frac{20}{1/2} = 40$ feet
Horizontal distance $= 20 cot 3 0^{\circ} = 203 \approx 34.64$ feet

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q384 marksCase StudySome Applications of Trigonometry CBSE 2025 · Standard 30/6/1

The Statue of Unity situated in Gujarat is the world's largest Statue which stands over a 58 m high base. As part of the project, a student constructed an inclinometer and wishes to find the height of Statue of Unity using it.
He noted following observations from two places :
Situation – I :
The angle of elevation of the top of Statue from Place A which is $803$ m away from the base of the Statue is found to be $6 0^{\circ}$ .
Situation – II :
The angle of elevation of the top of Statue from a Place B which is 40 m above the ground is found to be $3 0^{\circ}$ and entire height of the Statue including the base is found to be 240 m.
Based on given information, answer the following questions :
(i) Represent the Situation – I with the help of a diagram. (1)
(ii) Represent the Situation – II with the help of a diagram. (1)
(iii) (a) Calculate the height of Statue excluding the base and also find the height including the base with the help of Situation – I. (2)
OR (iii) (b) Find the horizontal distance of point B (Situation – II) from the Statue and the value of $tan α$ , where $α$ is the angle of elevation of top of base of the Statue from point B. (2)

Show answer & solution

Answer: (i) Right triangle: vertical side = total height (statue + base), horizontal side

803

m from A, angle

6 0^{\circ}

at A (ii) B at 40 m above ground; horizontal line from B to statue; angle

3 0^{\circ}

to the top; height of top above B's level

= 240 - 40 = 200

m (iii) (a) 182 m excluding base, 240 m including base OR (iii) (b)

2003

tan α = \frac{3 3}{100}

(i) Draw the statue with base as a vertical line PQ (Q at ground), A on the ground with $A Q = 803$ m and $∠ P A Q = 6 0^{\circ}$ .
(ii) Draw B at height 40 m above the ground, a horizontal line from B meeting the statue's vertical line at M, and $∠ P B M = 3 0^{\circ}$ ; $P M = 240 - 40 = 200$ m.
(iii) (a) $tan 6 0^{\circ} = \frac{P Q}{80 3} \Rightarrow P Q = 803 \times 3 = 240$ m (including base)
Height of statue excluding base $= 240 - 58 = 182$ m
(iii) (b) $tan 3 0^{\circ} = \frac{200}{B M} \Rightarrow B M = 2003$ m
Top of base is $58 - 40 = 18$ m above B's level.
$tan α = \frac{18}{200 3} = \frac{9}{100 3} = \frac{3 3}{100}$

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q384 marksCase StudySome Applications of Trigonometry CBSE 2024 · Basic 430/5/1

Two poles of different heights stand on level ground and at a distance of 40 m. Both poles are supported by wires attached from the top of each pole to the bottom of the other. A coupling is placed at point C, where the two wires cross (as shown in the figure).
Based on the above information, answer the following questions :
(i) Find the height of pole AB. (1)
(ii) Find the height of pole PQ. (1)
(iii) (a) If the angle of elevation of the top of the pole PQ from the top of the pole AB is $3 0^{\circ}$ , find the distance BQ. (2)
OR
(b) If the coupling is at a height of 20 m from the ground, how far down the wire from the smaller pole AB is the coupling ? (2)

Diagram for CBSE 2024 Class 10 Maths question 38

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Answer: (i)

\frac{40 3}{3}

m (ii)

403

m (iii) (a)

\frac{80 3}{3}

m OR (b)

\frac{40 3}{3}

m (along wire AQ from A)

In the figure, $∠ Q A P = 6 0^{\circ}$ , $∠ B P A = 3 0^{\circ}$ and AP = 40 m.
(i) In right $△ B A P$ : $tan 3 0^{\circ} = \frac{A B}{40}$ , so $A B = \frac{40}{3} = \frac{40 3}{3}$ m.
(ii) In right $△ QP A$ : $tan 6 0^{\circ} = \frac{P Q}{40}$ , so $P Q = 403$ m.
(iii) (a) The horizontal distance from B to PQ is 40 m. With elevation $3 0^{\circ}$ , $cos 3 0^{\circ} = \frac{40}{B Q}$ .
$B Q = \frac{40 \times 2}{3} = \frac{80 3}{3}$ m.
(iii) (b) C lies on wire AQ, which makes $6 0^{\circ}$ with the ground at A (the foot of pole AB).
$sin 6 0^{\circ} = \frac{20}{A C}$ , so $A C = \frac{40}{3} = \frac{40 3}{3}$ m.

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q374 marksCase StudySome Applications of Trigonometry CBSE 2023 · Standard 30/6/1

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O.
Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the Section B is $3 0^{\circ}$ and the angle of elevation of the top of Section A is $4 5^{\circ}$ .
Based on the above information, answer the following questions :
(i) Find the length of the wire from the point O to the top of Section B. (1)
(ii) Find the distance AB. (2)
OR Find the area of $△ O P B$ .
(iii) Find the height of the Section A from the base of the tower. (1)

Diagram for CBSE 2023 Class 10 Maths question 37

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Answer: (i)

243

cm (ii) AB =

12 (3 - 3)

cm; OR:

2163

^{2}

(iii) 36 cm

(i) In right $△ O P B$ , $cos 3 0^{\circ} = \frac{O P}{O B}$ , so $O B = \frac{36}{\frac{3}{2}} = \frac{72}{3} = 243$ cm.
(ii) $P B = 36 tan 3 0^{\circ} = \frac{36}{3} = 123$ cm and $P A = 36 tan 4 5^{\circ} = 36$ cm.
AB = PA – PB = $36 - 123 = 12 (3 - 3)$ cm.
OR: Area of $△ O P B = \frac{1}{2} \times O P \times P B = \frac{1}{2} \times 36 \times 123 = 2163$ cm $^{2}$ .
(iii) PA = $36 tan 4 5^{\circ}$ = 36 cm.

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q144 marksCase StudySome Applications of Trigonometry CBSE 2022 · Basic 430/1/1

The following TV Tower was built in 1988 and is located in Pitampura, Delhi. It has an observation deck. Observe the picture given below :
The TV Tower stands vertically on the ground. From a point ‘A’ on the ground, the angle of elevation of top of the tower (point ‘B’) is $6 0^{\circ}$ . There is a point ‘C’ on the tower which is 78 m (approx.) above the ground.
The angle of elevation of the point C from point A is found to be $3 0^{\circ}$ .
(a) Draw a well-labelled figure, based on the information given above. (2)
(b) Find the height of the tower and the distance of the tower from point A. (2)

Show answer & solution

Answer: (a) Right triangle with tower BD vertical, C on BD with CD = 78 m,

∠ C A D = 3 0^{\circ}

∠ B A D = 6 0^{\circ}

(b) Height = 234 m; distance =

783 \approx 135.1

(a) Let D be the foot of the tower. Draw vertical BD, mark C on BD with CD = 78 m, A on the ground; join AC and AB with $∠ C A D = 3 0^{\circ}$ and $∠ B A D = 6 0^{\circ}$
(b) In right $△ A D C$ : $tan 3 0^{\circ} = \frac{C D}{A D}$ , so $A D = 783$ m $\approx 135.1$ m
In right $△ A D B$ : $tan 6 0^{\circ} = \frac{B D}{A D}$ , so $B D = 783 \times 3 = 234$ m
Height of tower = 234 m; distance from A = $783$ m

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q134 marksCase StudySome Applications of Trigonometry CBSE 2022 · Basic 430/2/1

Qutub Minar, located in South Delhi, India, was built in the year 1193. It is 72 m high tower. Working on a school project, Charu and Daljeet visited the monument. They used trigonometry to find their distance from the tower. Observe the picture given below. Points C and D represent their positions on the ground in line with the base of tower, the angles of elevation of top of the tower (Point A) are $6 0^{\circ}$ and $4 5^{\circ}$ from points C and D respectively.
(1) Based on above information, draw a well-labelled diagram. (1)
(2) Find the distances CD, BC and BD. (use $3$ = 1.73) (3)

Show answer & solution

Answer: (1) Vertical tower AB = 72 m with C and D on the ground in line with B;

∠ A C B = 6 0^{\circ}

∠ A D B = 4 5^{\circ}

(2) BC = 41.52 m, BD = 72 m, CD = 30.48 m

(1) Draw AB vertical (72 m), B on the ground line BCD with C nearer to B; mark $∠ A C B = 6 0^{\circ}$ and $∠ A D B = 4 5^{\circ}$ .
(2) In $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow B C = \frac{72}{3} = 243 = 41.52$ m.
In $△ A B D$ : $tan 4 5^{\circ} = \frac{A B}{B D} \Rightarrow B D = 72$ m.
CD = BD − BC = 72 − 41.52 = 30.48 m.

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q124 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/3/1

The angles of depression of the top and bottom of a 6 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $3 = 1.73$ )

Show answer & solution

Answer: Height of multi-storeyed building

= (9 + 33)

= 14.19

m; distance between the buildings

= 14.19

Let the multi-storeyed building be H m tall and the buildings be d m apart.
Angle of depression of the bottom is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{H}{d} \Rightarrow d = H$
Angle of depression of the top is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{H - 6}{d} \Rightarrow d = 3 (H - 6)$
$H = 3 H - 63 \Rightarrow H = \frac{6 3}{3 - 1} = \frac{6 3 ( 3 + 1 )}{2} = 9 + 33$
$H = 9 + 3 (1.73) = 14.19$ m, and $d = H = 14.19$ m

Also asked in: 2022 Basic 430/3/2

Q114 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/3/3

A statue, 1.8 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of top of the statue is $6 0^{\circ}$ and from the same point the angle of elevation of top of the pedestal is $4 5^{\circ}$ . Find the height of the pedestal. (Use $3 = 1.73$ )

Show answer & solution

Answer: Height of pedestal

= 0.9 (3 + 1)

\approx 2.457

Let the pedestal be h m high and the point be d m from its foot.
$tan 4 5^{\circ} = \frac{h}{d} \Rightarrow d = h$
$tan 6 0^{\circ} = \frac{h + 1.8}{d} \Rightarrow 3 h = h + 1.8$
$h = \frac{1.8}{3 - 1} = \frac{1.8 ( 3 + 1 )}{2} = 0.9 (3 + 1)$
$h = 0.9 \times 2.73 = 2.457$ m

Q114 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/4/1

As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are found to be $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use $3 = 1.732$ ]

Show answer & solution

Answer: 73.2 m

Let AB = 100 m be the lighthouse, with ships at C (angle $4 5^{\circ}$ ) and D (angle $3 0^{\circ}$ ) on the same side, D beyond C.
Angle of depression equals the angle of elevation from the ship (alternate angles).
In $△ A B C$ : $tan 4 5^{\circ} = \frac{100}{B C} \Rightarrow B C = 100$ m.
In $△ A B D$ : $tan 3 0^{\circ} = \frac{100}{B D} \Rightarrow B D = 1003$ m.
$C D = B D - B C = 100 (3 - 1) = 100 \times 0.732 = 73.2$ m.

Also asked in: 2022 Basic 430/4/2, 2022 Basic 430/4/3

Q144 marksCase StudySome Applications of Trigonometry CBSE 2022 · Standard 30/1/1

Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was built by the King of Jaisalmer and rebuilt by Gadsi Singh in 14 $^{t h}$ century. The lake has many Chhatris. One of them is shown below :
Observe the picture. From a point A $h$ m above from water level, the angle of elevation of top of Chhatri (point B) is $4 5^{\circ}$ and angle of depression of its reflection in water (point C) is $6 0^{\circ}$ . If the height of Chhatri above water level is (approximately) 10 m, then
(a) draw a well-labelled figure based on the above information; (2)
(b) find the height ( $h$ ) of the point A above water level. (Use $3 = 1.73$ ) (2)

Show answer & solution

Answer: (a) Figure: A at height

h

above water; B is 10 m above water and C is 10 m below water on the vertical through the Chhatri; elevation of B

4 5^{\circ}

, depression of C

6 0^{\circ}

. (b)

h = 20 - 103 \approx 2.7

(a) Draw the water line and the vertical line of the Chhatri meeting it at D, with B 10 m above D and its reflection C 10 m below D. A is a point $h$ m above the water; draw the horizontal AM to the vertical BC. Mark $∠ B A M = 4 5^{\circ}$ and $∠ M A C = 6 0^{\circ}$ .
(b) Let AM $= x$ . Then BM $= 10 - h$ and MC $= 10 + h$ .
$tan 4 5^{\circ} = \frac{10 - h}{x} \Rightarrow x = 10 - h$ .
$tan 6 0^{\circ} = \frac{10 + h}{x} \Rightarrow 10 + h = 3 (10 - h)$ .
$h (1 + 3) = 10 (3 - 1) \Rightarrow h = \frac{10 ( 3 - 1 )}{3 + 1} = 5 (3 - 1)^{2} = 20 - 103$
$h = 20 - 17.3 = 2.7$ m.

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q134 marksCase StudySome Applications of Trigonometry CBSE 2022 · Standard 30/2/1

Kite Festival
Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites.
The picture given below, shows three kites flying together.
In Fig. 5, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Taking AD = 50 m and BE = 60 m, find
(1) the lengths of strings used (take them straight) for kites A and B as shown in the figure. (2)
(2) the distance ‘d’ between these two kites (2)

Diagram for CBSE 2022 Class 10 Maths question 13

Show answer & solution

Answer: (1) AC = 100 m, BC =

403

\approx 69.28

m (2) d =

2037

\approx 121.66

(1) In right $△ A D C$ : $sin 3 0^{\circ} = \frac{A D}{A C} \Rightarrow A C = \frac{50}{1/2} = 100$ m.
In right $△ B E C$ : $sin 6 0^{\circ} = \frac{B E}{B C} \Rightarrow B C = \frac{60 \times 2}{3} = 403$ m.
(2) $∠ A C B = 18 0^{\circ} - 3 0^{\circ} - 6 0^{\circ} = 9 0^{\circ}$ .
In right $△ A C B$ : $d^{2} = A C^{2} + B C^{2} = 10000 + 4800 = 14800$ .
$d = 14800 = 2037$ m $\approx 121.66$ m.

Also asked in: 2022 Standard 30/2/2

Q134 marksCase StudySome Applications of Trigonometry CBSE 2022 · Standard 30/2/3

Kite Festival
Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites.
The picture given below, three kites flying together.
In Fig. 5, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Taking AD = 50 m and BE = 60 m, find
(1) the lengths of strings used (take them straight) for kites A and B as shown in the figure. (2)
(2) the distance ‘d’ between these two kites (2)

Show answer & solution

Answer: (1) AC = 100 m, BC =

403

\approx 69.28

m (2) d =

2037

\approx 121.66

(1) In right $△ A D C$ : $sin 3 0^{\circ} = \frac{A D}{A C} \Rightarrow A C = \frac{50}{1/2} = 100$ m.
In right $△ B E C$ : $sin 6 0^{\circ} = \frac{B E}{B C} \Rightarrow B C = \frac{60 \times 2}{3} = 403$ m.
(2) $∠ A C B = 18 0^{\circ} - 3 0^{\circ} - 6 0^{\circ} = 9 0^{\circ}$ .
In right $△ A C B$ : $d^{2} = A C^{2} + B C^{2} = 10000 + 4800 = 14800$ .
$d = 14800 = 2037$ m $\approx 121.66$ m.

Q124 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/3/1

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 $3 0^{\circ}$ , which is approaching the foot of the tower with a uniform speed. Ten seconds later, the angle of depression of the car is found to be $6 0^{\circ}$ . Find the time taken by the car to reach the foot of the tower from this point.

Show answer & solution

Answer: 5 seconds

Let the tower height be h. At $3 0^{\circ}$ the car is at distance $\frac{h}{t a n 3 0 ^{\circ}} = h 3$ from the foot.
At $6 0^{\circ}$ it is at distance $\frac{h}{t a n 6 0 ^{\circ}} = \frac{h}{3}$ .
Distance covered in 10 s $= h 3 - \frac{h}{3} = \frac{2 h}{3}$ , so speed $= \frac{2 h}{10 3} = \frac{h}{5 3}$ .
Time for the remaining $\frac{h}{3}$ $= \frac{h}{3} \div \frac{h}{5 3} = 5$ s.

Q114 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/3/2

The angle of elevation of an aeroplane from a point on the ground is $6 0^{\circ}$ . After a flight of 30 seconds, the angle of elevation from the same point becomes $3 0^{\circ}$ . If the aeroplane is flying at a constant height of $30003$ m, find the speed of the aeroplane.

Show answer & solution

Answer: 200 m/s (= 720 km/h)

Height $h = 30003$ m.
At $6 0^{\circ}$ : horizontal distance $= \frac{h}{t a n 6 0 ^{\circ}} = \frac{3000 3}{3} = 3000$ m.
At $3 0^{\circ}$ : horizontal distance $= \frac{h}{t a n 3 0 ^{\circ}} = 30003 \times 3 = 9000$ m.
Distance flown in 30 s $= 9000 - 3000 = 6000$ m.
Speed $= \frac{6000}{30} = 200$ m/s $= 720$ km/h.

Q114 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/3/3

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is $6 0^{\circ}$ . From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower PQ is $4 5^{\circ}$ . Find the height of the tower PQ and the distance PX. [Use $3$ = 1.73]

Show answer & solution

Answer: PQ =

20 (3 + 3)

m ≈ 94.6 m; PX =

20 (3 + 1)

m ≈ 54.6 m

Let $P Q = H$ and $P X = x$ .
From X: $tan 6 0^{\circ} = \frac{H}{x} \Rightarrow H = x 3$ .
From Y (40 m above X): $tan 4 5^{\circ} = \frac{H - 40}{x} \Rightarrow H - 40 = x$ .
$x 3 - 40 = x \Rightarrow x = \frac{40}{3 - 1} = 20 (3 + 1) \approx 20 \times 2.73 = 54.6$ m.
$H = x + 40 = 203 + 60 \approx 34.6 + 60 = 94.6$ m.

Q124 marksLong AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/4/1

From the top of an 8 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower. (Take $3 = 1.732$ ).

Show answer & solution

Answer: 21.856 m (approx.)

Let the building be AB = 8 m and the tower CD, with horizontal distance BD = $x$ .
Angle of depression of the foot is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{8}{x} \Rightarrow x = 8$ m.
Let the top of the tower be $h$ m above the top of the building. $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow h = 83$ m.
Height of tower $= 8 + 83 = 8 + 8 \times 1.732 = 21.856$ m.

Also asked in: 2022 Standard 30/4/2, 2022 Standard 30/4/3

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