CBSE Class 10 · Maths · board exam 2027
Chapter 9: Some Applications of Trigonometry
This chapter’s questions from CBSE’s real papers of 2023–2026 — the 20 asked most often, 3 with the full model answer, 16 with how the answer begins — and objective questions with the board’s key. All 76 questions of this chapter, with answers, are in the app.
- marks a year (about)
- 5
- questions a year (about)
- 2
- rank in Maths
- #8
From 4 years of papers, 2023–2026 · full chapter-wise marks
Important questions — the ones the board asked most (20)
- Q1Long answer5 marksAsked: 2025, 2026
34. The angle of elevation of the top of a building from the foot of the tower is 30 degree and the angle of elevation of the top of the tower from the foot of the building is 60 degree. If the tower is 60 m high, find the height of the building and distance between the building and the tower. (Use sqrt(3) = 1.73)
Medha Ma’am
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Model answer · heights and distances, two angles
- 1.The height of the building is 20 m and the distance between the building and the tower is 34.6 m. To find these, draw the figure. Let DC be the tower, with top D and foot C, so DC = 60 m. Let AB be the building, with top B and foot A, so AB = h m. Both stand vertically on level ground, so the angles at C and at A are right angles. Let the distance between the building and the tower be CA = x m. The angle of elevation of the top of the tower from the foot of the building is ∠DAC = 60°. The angle of elevation of the top of the building from the foot of the tower is ∠BCA = 30°.1 mark
- 2.In right Δ DCA, right-angled at C: tan 60° = DC/CA √3 = 60/x1 mark
- 3.x = 60/√3 = 60√3/3 = 20√3 x = 20 × 1.73 = 34.6 m1 mark
- 4.In right Δ BAC, right-angled at A: tan 30° = AB/CA 1/√3 = h/x = h/(20√3)1 mark
- 5.h = 20√3/√3 h = 20 m1 mark
Diagram: Draw the tower DC (D at the top, C at the foot, DC = 60 m) and the building AB (B at the top, A at the foot, AB = h m) standing upright on the horizontal line CA = x m. Join A to D and mark ∠DAC = 60°, join C to B and mark ∠BCA = 30°, and mark right angles at C and A.
- Label the figure fully: tower DC = 60 m, building AB = h m, distance CA = x m, ∠DAC = 60° and ∠BCA = 30°.
- Use 1.73 to change 20√3 into 34.6 m, and write the unit m with both answers.
Marks split as in the board’s published marking scheme
- Q2Long answer5 marksAsked: 2026
A boy standing on a horizontal plane is flying a kite with a string of length 60 m, at an angle of elevation of 30°. Another boy standing on the roof of a 20 m high building, finds the angle of elevation of same kite to be 45°. If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use sqrt(3) = 1.73)
Medha Ma’am
AskMedha’s AI teacher
Model answer · heights and distances, angle of elevation
- 1.The height of the kite from the ground is 30 m, and the first boy is 61.9 m from the base of the building. This is worked out below. Given: string BK = 60 m, angle of elevation of the kite from the first boy = 30°, height of building MD = 20 m, angle of elevation of the kite from the second boy = 45°. Let K be the kite and B the first boy on the ground. KP is perpendicular to the ground, so KP is the height of the kite. M is the second boy on the roof of the building MD, and D is the base of the building, on the other side of P from B. The horizontal line ML from M meets KP at L, so LP = MD = 20 m. Let KL = h m, BP = y m and LM = PD = x m. Then ∠KBP = 30° and ∠KML = 45°.1 mark
- 2.In rt. angled ΔKPB, KP/BK = sin 30° (h + 20)/60 = sin 30° = 1/20.5 marks
- 3.h + 20 = 30 => h = 10 m0.5 marks
- 4.Also, in ΔKPB, BP/BK = cos 30° y/60 = cos 30° = √3/20.5 marks
- 5.=> y = 30√3 = 30 × 1.73 = 51.9 m0.5 marks
- 6.In rt. angled ΔKLM, KL/LM = tan 45° 10/x = tan 45° = 10.5 marks
- 7.=> x = 10 m, so PD = LM = 10 m0.5 marks
- 8.Height of kite from ground = KP = KL + LP = 10 + 20 = 30 m0.5 marks
- 9.Distance of first boy from the base of the building = BD = BP + PD = 51.9 + 10 = 61.9 m0.5 marks
Diagram: Draw the ground line with the first boy B on one side and the building MD (20 m, D = base, M = top) on the other side, with the kite K above and between them. Draw the string BK = 60 m at 30° to the ground and KP perpendicular to the ground, and draw the horizontal ML from the roof meeting KP at L with ∠KML = 45°. Label KL = h, LP = 20 m, BP = y and LM = PD = x.
- Begin with both results, "30 m" and "61.9 m", then draw and label the figure: K, P, B, L, M, D, the 60 m string, the 20 m building and the angles 30° and 45°.
- Write the kite's height as h + 20, and finish with "10 + 20 = 30 m".
- For the distance from the building, add both parts: "51.9 + 10 = 61.9 m", and write the unit m.
Marks split as in the board’s published marking scheme
- Q3Long answer5 marksAsked: 2026
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff. From a point on the ground 30 m away from the tower, wires are attached to the top and bottom of the flagstaff making angles of elevation 60 degree and 30 degree respectively. Find the height of the tower and lengths of the wires attached. (Take sqrt(3) = 1.73)
Medha Ma’am
AskMedha’s AI teacher
Model answer · Heights and distances, flagstaff
- 1.Given: AB is the vertical tower of height h m, with A its foot on the ground and B its top. BC is the vertical flagstaff on the tower, with C its top. D is the point on the ground with AD = 30 m. Wire DB (length l₁) goes to the bottom of the flagstaff and wire DC (length l₂) goes to its top. The angle of elevation of B from D is ∠BDA = 30°, the angle of elevation of C from D is ∠CDA = 60°, and ∠BAD = 90°. (Figure drawn and labelled as described.)1 mark
- 2.In right angled ΔBAD, AB/AD = tan 30° h/30 = tan 30° = 1/√31 mark
- 3.h = 30/√3 × √3/√3 = 30√3/3 = 10√3 h = 10 × 1.73 = 17.30.5 marks
- 4.So the height of the tower is 17.3 m. Also, in right angled ΔBAD, AD/BD = cos 30° 30/l₁ = cos 30° = √3/21 mark
- 5.l₁ = 60/√3 × √3/√3 = 60√3/3 = 20√3 l₁ = 20 × 1.73 = 34.60.5 marks
- 6.So the length of the wire attached to the bottom of the flagstaff is 34.6 m. In right angled ΔCAD, AD/CD = cos 60° 30/l₂ = cos 60° = 1/20.5 marks
- 7.l₂ = 30 × 2 = 60 So the length of the wire attached to the top of the flagstaff is 60 m.0.5 marks
Diagram: Draw a vertical tower AB (A = foot, B = top) with the flagstaff BC on top of it (C = top of the flagstaff), and a point D on the ground with AD = 30 m. Join DB (label l₁) and DC (label l₂), mark ∠BDA = 30° and ∠CDA = 60°, mark the right angle at A, and label AB = h.
- Label the figure fully: A, B, C, D, AD = 30 m, the angles 30° and 60°, h, l₁ and l₂.
- Use cos 30° = √3/2 for the bottom wire and cos 60° = 1/2 for the top wire, and write which right triangle (ΔBAD or ΔCAD) each one comes from.
- Write every final answer with its unit in metres: 17.3 m, 34.6 m and 60 m.
Marks split as in the board’s published marking scheme
- Q4Long answer5 marksAsked: 2026
(a) The angle of elevation of the top of a building from a point A, on the ground, is 30°. On moving a distance of 24 m towards its base to the point B, the angle of elevation changes to 60°. Find the height of the building and distance of point A from the base of the building. (Take sqrt(3) = 1.73) OR (b) A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of 30°, who is approaching the foot of the tower with a uniform speed. 30 seconds later, the angle of depression changes to 60°. Find the time taken by his friend to reach the foot of the tower from this point.
How the model answer begins
1. Let DC be the building, with top D and base C, and let its height be h m. A and B are points on the ground in a line with C. AB = 24 m, and BC = x m. The angle of elevation of D is 30° from A and 60° from B. The figure…
The full answer from Medha Ma’am → - Q5Long answer5 marksAsked: 2026
A statue, 2 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 degree and from the same point the angle of elevation of the bottom of the statue is 45 degree. Find the height of the pedestal and its distance from the point of observation on the ground. (use sqrt(3) = 1.73)
How the model answer begins
1. The height of the pedestal is 2.73 m, and its distance from the point of observation on the ground is also 2.73 m. These values are found as follows. Figure: Draw the pedestal BD standing vertically on the ground, with…
The full answer from Medha Ma’am → - Q6Long answer5 marksAsked: 2026
A straight highway leads to the foot of a tower. The shadow of the tower is found to be 40 m longer when the sun's altitude changes from 60 degree to 30 degree. Find the height of the tower and the length of the shadow in both cases. (Take sqrt(3) = 1.73)
How the model answer begins
1. Let AB be the tower of height h metres standing on a level ground, and let BC be the length of the shadow when the Sun's altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow…
The full answer from Medha Ma’am → - Q7Long answer5 marksAsked: 2025
As observed from the top of a 70 m high lighthouse from the sea level, the angles of depression of two ships are 30 degree and 45 degree. If one ship is exactly behind the other on the same sides of the lighthouse, find the distance between the two ships. (Use sqrt(3) = 1.73)
How the model answer begins
1. Let AB be the lighthouse of height 70 m standing at the sea level point B, with A at the top. Let C be the position of the nearer ship and D be the position of the farther ship, both on the same side of the lighthouse,…
The full answer from Medha Ma’am → - Q8Long answer5 marksAsked: 2025
From a point on the ground, the angle of elevation of the top of a tree observed by a person is 60 degree. When moved back by 28 m, in the same line, the angle of elevation from another point on ground becomes 30 degree. Find the height of the tree and its distance from the initial point. (Use sqrt(3) = 1.73)
How the model answer begins
1. The height of the tree is 24.22 m and its distance from the initial point is 14 m. The working is shown below. Given: from a point on the ground, the angle of elevation of the top of the tree is 60°. From a second point…
The full answer from Medha Ma’am → - Q9Long answer5 marksAsked: 2025
A statue 3 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 30°. Find the height of the pedestal and the distance of the pedestal from the point of observation on the ground. (Take sqrt(3) = 1.73)
How the model answer begins
1. The height of the pedestal is 1.5 m and its distance from the point of observation is 2.6 m. This is found as follows. Let AB be the pedestal of height h m, with its foot A on the ground. The statue BC = 3 m stands on…
The full answer from Medha Ma’am → - Q10Long answer5 marksAsked: 2025
The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are 30 degree and 60 degree respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use sqrt(3) = 1.73)
How the model answer begins
1. The multi-storeyed building is 13.5 m high and stands 7.79 m from the 9 m building. Figure (drawn and labelled): DC is the multi-storeyed building of height h m, and AB is the 9 m tall building. AC = x m is the distance…
The full answer from Medha Ma’am → - Q11Long answer5 marksAsked: 2025
Two poles of equal heights are standing opposite each other on either side of the road which is 85 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60 degree and 30 degree respectively. Find the height of the poles and the distances of the point from the poles. (Use sqrt(3) = 1.73)
How the model answer begins
1. Each pole is 36.76 m high, and the point is 21.25 m from one pole and 63.75 m from the other. These values are found below. Given: two poles of equal height h m stand on opposite sides of a road 85 m wide. Let AB and CD…
The full answer from Medha Ma’am → - Q12Long answer5 marksAsked: 2025
A drone is flying at a height of h metres. At an instant it observes the angle of elevation of top of an industrial turbine as 60° and angle of depression of foot of the turbine as 30°. If height of turbine is 200 metres, find the value of h and the distance of drone from the turbine. (Use sqrt(3) = 1.73)
How the model answer begins
1. The drone is flying at h = 50 m, and its distance from the turbine is 86.5 m; this is found from two right triangles as shown below. In the figure drawn here, AB is the vertical industrial turbine, with top A and foot B…
The full answer from Medha Ma’am → - Q13Long answer5 marksAsked: 2024
A pole 6m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is 60 degree and the angle of depression of the point P from the top of the tower is 45 degree. Find the height of the tower and the distance of point P from the foot of the tower. (Use sqrt(3) = 1.73)
- Q14Long answer5 marksAsked: 2024
From the top of a building 60 m high, the angles of depression of the top and bottom of the vertical lamp post are observed to be 30 degree and 60 degree respectively. (i) Find the horizontal distance between the building and the lamp post. (ii) Find the distance between the tops of the building and the lamp post.
How the model answer begins
1. Let AB be the building, with A its top and B its foot, so AB = 60 m. Let CD be the vertical lamp post, with D its top and C its foot, standing on the same level ground. Let BC = x m be the horizontal distance between…
The full answer from Medha Ma’am → - Q15Long answer5 marksAsked: 2024
From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be 30 degree. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60 degree. Find the height of the tower and the distance between tower and the building.
How the model answer begins
1. Let AB represent the building of height 15 m, with A at the top and B at the bottom, and let CD represent the tower, with D at its foot and C at its top, both standing on the same level ground. A line AE is drawn from A…
The full answer from Medha Ma’am → - Q16Long answer5 marksAsked: 2024
From the top of a 45 m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be 30 degree and 60 degree. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use sqrt(3) = 1.73)
How the model answer begins
1. The distance between the two ships is 60√3 m = 103.8 m, worked out as follows. Given: AB is the light house, with A its top and B its foot, so AB = 45 m. The two ships C and D are on opposite sides of the light house,…
The full answer from Medha Ma’am → - Q17Long answer5 marksAsked: 2024
From a point on a bridge across the river, the angles of depressions of the banks on opposite sides of the river are 30° and 60° respectively. If the bridge is at a height of 4 m from the banks, find the width of the river.
How the model answer begins
1. The width of the river is (16/3)√3 m. It is found as follows. Given: a point P on the bridge is 4 m above the banks. The angles of depression of the banks on opposite sides of the river are 30° and 60°. Let A and B be…
The full answer from Medha Ma’am → - Q18Long answer5 marksAsked: 2024
From a window 15 metres high above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 30 degree and 45 degree respectively. Find the height of the opposite house. (Use sqrt(3) = 1.732)
How the model answer begins
1. The height of the opposite house is (5√3 + 15) m = 23.66 m. Here is how it is found. Let P be the window, 15 m above the ground. Let CD be the opposite house, with its foot at C and its top at D. A horizontal line from…
The full answer from Medha Ma’am → - Q19Long answer5 marksAsked: 2024
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. (Use sqrt(3) = 1.732)
How the model answer begins
1. The height of the cable tower is 7√3 + 7 = 19.124 m. It is found as follows. Given: AB is the building, 7 m high. A is its foot and B is its top. PC is the cable tower. C is its foot and P is its top. Both stand…
The full answer from Medha Ma’am → - Q20Long answer5 marksAsked: 2024
As observed from the top of a 75 m light house from the sea-level, the angles of depression of two ships are 30 degree and 45 degree. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use sqrt(3) = 1.732]
How the model answer begins
1. The distance between the two ships is 54.9 m. Given: the lighthouse AB is 75 m high, with top A and foot B at sea-level. P is the nearer ship and Q is the ship exactly behind it, with B, P and Q on one straight line.…
The full answer from Medha Ma’am →
56 more questions of this chapter, each with its model answer, in the AskMedha app.
Objective questions with the board’s key (10)
- 1. CBSE 2026
A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is 10sqrt(3) m away from the base of the tower, is :
- (A) 30°
- (B) 45°
- (C) 90°
- (D) 60° ✓
Answer: (D)
- 2. CBSE 2026
The string of a flying kite is tied to a point on the ground. The length of the string between the kite and that point on the ground is 80 m. The string makes an angle of 30 degree with the ground. The height of the kite from the ground is :
- (A) 20sqrt(3) metres
- (B) 40 metres ✓
- (C) 40sqrt(3) metres
- (D) 80sqrt(3) metres
Answer: (B)
- 3. CBSE 2026
From a point on the ground, which is 60 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be 45 degree. The height (in metres) of the tower is :
- (A) 10 sqrt(3)
- (B) 30 sqrt(3)
- (C) 60 ✓
- (D) 30
Answer: (C)
- 4. CBSE 2026
If the length of the shadow of a tower is sqrt(3) times that of its height, then altitude of the Sun is :
- (A) 45°
- (B) 30° ✓
- (C) 60°
- (D) 15°
Answer: (B)
- 5. CBSE 2026
If the length of the shadow of a tower is sqrt(3) times its height, then the angle of elevation of the sun is :
- (A) 45°
- (B) 30° ✓
- (C) 60°
- (D) 15°
Answer: (B)
- 6. CBSE 2026
A wire is attached from a point A on the ground to the top of a pole BC, making an angle of elevation as 60°. If AB = 5sqrt(3) m, then length of the wire is
- (A) 10 m
- (B) 10sqrt(3) m ✓
- (C) 15 m
- (D) 5/2sqrt(3) m
Answer: (B)
- 7. CBSE 2025
The length of the shadow of a tower when the sun's altitude changes from 30° to 60° will :
- (A) become shorter ✓
- (B) become longer
- (C) remain same
- (D) be doubled
Answer: (A)
- 8. CBSE 2025
When the angle of elevation of the sun changes from 30° to 60°, the length of the shadow of a tower :
- (A) will become shorter ✓
- (B) will become longer
- (C) will remain the same
- (D) will become double
Answer: (A)
- 9. CBSE 2025
In the given figure, the angle of elevation of point A from point C is :
- (A) 30° ✓
- (B) 45°
- (C) 60°
- (D) Cannot be determined
Answer: (A)
- 10. CBSE 2025
At any time, the length of the shadow of a tower is sqrt(3) times its height. The angle of elevation of the sun is
- (A) 45°
- (B) 30° ✓
- (C) 60°
- (D) 0°
Answer: (B)
Questions parents and students ask
- How many marks does "Some Applications of Trigonometry" carry in CBSE Class 10 Maths?
- In the CBSE papers of 2023–2026, this chapter carried about 5 marks and about 2 questions a year; by marks it ranks 8 in Maths.
- Which are the most important (VVI) questions of "Some Applications of Trigonometry"?
- The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 76 written questions of this chapter, with answers, are in the AskMedha app.
- Do the answers follow the board’s marking scheme?
- Yes. CBSE publishes its marking schemes; every model answer here is written point by point to the marks the board’s own scheme gives, and checked by a second, independent model.
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