NDA Mathematics · 27 questions · 2017 to 2026
NDA Heights and Distances previous year questions with solutions
Heights and Distances gives one or two questions in most Mathematics papers, none in some: about one mark in a hundred. Since 2022 they come mostly as linked pairs on one figure, leaning on exact ratios at 15°, 75° and 67.5°. Every question UPSC has set on it since 2017, solved. Grouped by the method each needs, newest first; open a question for its answer and worked solution.
14.1 Angles of elevation and depression: the single right triangle (4)14.2 Two angles at one point: an object standing on another (6)14.3 Two or more points in line with the foot (7)14.4 An observer above the ground: tops and bottoms of two heights (5)14.5 Problems in three dimensions: directions, spheres and plots (5)
14.1 Angles of elevation and depression: the single right triangle
1. NDA 2026 (I)
A plane is observed to be approaching the airport. It is at a distance of 10 km from the point of observation and makes an angle of elevation of 67.5°. What is the height of the plane above the ground?
Answer and solution
Answer: (c) 5√(2 + √2) km
Method. height = slant distance × sin(elevation); sin 67.5° from the half-angle formula
Height = 10 sin 67.5°.
sin2 67.5° = 1 − cos 135°2 = 1 + 1√22 = 2 + √24 ⇒ sin 67.5° = √(2 + √2)2.
Height = 10 × √(2 + √2)2 = 5√(2 + √2) km.
2. NDA 2018 (II)
A balloon is directly above one end of a bridge. The angle of depression of the other end of the bridge from the balloon is 48°. If the height of the balloon above the bridge is 122 m, then what is the length of the bridge?
Answer and solution
Answer: (b) 122 tan 42° m
Method. horizontal distance = height × cot(angle of depression); cot 48° = tan 42°
The balloon is 122 m vertically above one end A; let B be the other end. The angle of elevation of the balloon from B equals the angle of depression, 48°.
tan 48° = 122AB ⇒ AB = 122 cot 48° = 122 tan(90° - 48°) = 122 tan 42° m.
3. NDA 2018 (I)
If a flag-staff of 6 m height placed on the top of a tower throws a shadow of 2√3 m along the ground, then what is the angle that the sun makes with the ground?
Answer and solution
Answer: (a) 60°
Method. the flag-staff and its own shadow form a right triangle with the sun's rays
The tower's shadow ends where the ray past the tower's top meets the ground; the extra 2√3 m beyond it is cast by the 6 m flag-staff, and the ray past its top makes the same angle θ with the ground.
So tan θ = 62√3 = √3
θ = 60°.
4. NDA 2017 (I)
From the top of a lighthouse, 100m high, the angle of depression of a boat is tan-1(512). What is the distance between the boat and the lighthouse?
Answer and solution
Answer: (c) 240 m
Method. angle of depression equals the angle of elevation at the boat; tan = height/distance
Let the boat be d m from the foot of the lighthouse. The angle of elevation of the top from the boat equals the angle of depression, so tan θ = 512.
tan θ = 100d ⇒ 100d = 512
d = 100 × 125 = 240 m.
14.2 Two angles at one point: an object standing on another
5. NDA 2025 (I)
A man at M, standing 100 m away from the base (P) of a chimney of height 50 m, observes the angle of elevation of the highest point (Q) of the smoke to be 45°. The highest point of the chimney is at R. Further P, R and Q are in a straight line and the straight line is perpendicular to PM. What is the angle RMQ equal to?
Answer and solution
Answer: (b) tan-1(13)
Method. angle RMQ is the difference of two elevations; use tan(A − B)
PQ is vertical and perpendicular to PM, with PM = 100 m and PR = 50 m.
∠PMQ = 45° ⇒ PQ = 100 m; tan ∠PMR = 50100 = 12.
tan ∠RMQ = tan(45° - ∠PMR) = 1 − 121 + 12 = 13.
∠RMQ = tan-1(13).
6. NDA 2023 (I)
Consider the following for the next two (02) items that follow:
A flagstaff 20 m long standing on a pillar 10 m high subtends an angle tan-1(0.5) at a point P on the ground. Let θ be the angle subtended by the pillar at this point P.
A flagstaff 20 m long standing on a pillar 10 m high subtends an angle tan-1(0.5) at a point P on the ground. Let θ be the angle subtended by the pillar at this point P.
If x is the distance of P from bottom of the pillar, then consider the following statements:
1. x can take two values which are in the ratio 1 : 3
2. x can be equal to height of the flagstaff
Which of the statements given above is/are correct?
1. x can take two values which are in the ratio 1 : 3
2. x can be equal to height of the flagstaff
Which of the statements given above is/are correct?
Answer and solution
Answer: (a) 1 only
Method. tan of the difference of two elevations gives a quadratic in x
From P, the pillar's top is at elevation θ with tan θ = 10x; the flagstaff's top at β with tan β = 30x.
tan(β − θ) = 30x − 10x1 + 300x2 = 20xx2 + 300 = 12
x2 - 40x + 300 = 0 ⇒ (x − 10)(x − 30) = 0 ⇒ x = 10 or 30.
Statement 1: 10 : 30 = 1 : 3, true. Statement 2: x = 20 is not a root (it gives 400700 = 47 ≠ 12), false.
So 1 only.
7. NDA 2023 (I)
What is a possible value of tan θ?
Answer and solution
Answer: (c) 13
Method. tan θ = 10x for the two admissible x
From the previous item, x = 10 or 30 m.
tan θ = 10x = 1 or 13.
Of the options, only 13 occurs.
8. NDA 2022 (I)
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane the angles of elevation of the bottom and top of the flagstaff are θ and 2θ respectively. What is the height of the tower?
Answer and solution
Answer: (c) h cos 2θ
Method. sine rule in the triangle formed by the observer, the tower's top and the flagstaff's top
Let O be the point, B the top of the tower (bottom of the flagstaff), C the top of the flagstaff, with BC = h.
∠BOC = 2θ − θ = θ. The exterior angle gives ∠OBC = 90° + θ, so ∠OCB = 180° - θ − (90° + θ) = 90° - 2θ.
Sine rule: OBsin(90° - 2θ) = hsin θ ⇒ OB = h cos 2θsin θ.
Height of tower = OB sin θ = h cos 2θ.
9. NDA 2021 (I)
A ladder 6 m long reaches a point 6 m below the top of a vertical flagstaff. From the foot of the ladder, the elevation of the top of the flagstaff is 75°. What is the height of the flagstaff?
Answer and solution
Answer: (d) (6 + 3√3) m
Method. coordinates: H = d tan 75°, ladder length 6 m to the point H − 6
Let the foot of the ladder be d from the flagstaff of height H, so H = d tan 75°; write t = tan 75° = 2 + √3.
d2 + (H − 6)2 = 36 ⇒ d2(1 + t2) − 12td = 0 ⇒ d = 12t1 + t2 = 6 sin 150° = 3 m.
H = 3 tan 75° = 3(2 + √3) = (6 + 3√3) m.
Shortcut. Ladder FL = LT = 6 (T the staff top), so triangle FLT is isosceles with equal angles at F and T; ∠FTL = 90° - 75° = 15° = ∠TFL, so the ladder rises at 75° - 15° = 60°: H = 6 + 6 sin 60° = (6 + 3√3) m.
10. NDA 2019 (II)
A ladder 9 m long reaches a point 9 m below the top of a vertical flagstaff. From the foot of the ladder, the elevation of the flagstaff is 60°. What is the height of the flagstaff?
Answer and solution
Answer: (c) 13.5 m
Method. coordinates: elevation fixes height = d tan 60°, ladder length fixes the rest
Let the foot of the ladder be at horizontal distance d from the flagstaff of height H. Then H = d tan 60° = √3 d.
The ladder reaches height H − 9 on the flagstaff: d2 + (H − 9)2 = 92.
d2 + (√3 d − 9)2 = 81 ⇒ 4d2 - 18√3 d = 0 ⇒ d = 9√32.
H = √3 × 9√32 = 272 = 13.5 m.
Shortcut. Ladder FL = LT = 9 (T the staff top), so triangle FLT is isosceles with equal angles at F and T; ∠FTL = 90° - 60° = 30° = ∠TFL, so the ladder rises at 60° - 30° = 30°: H = 9 + 9 sin 30° = 13.5 m.
14.3 Two or more points in line with the foot
11. NDA 2025 (I)
Consider the following for the two (02) items that follow:
The top (M) of a tower is observed from three points P, Q and R lying in a horizontal straight line which passes directly along the foot (N) of the tower. The angles of elevations of M from P, Q and R are 30°, 45° and 60° respectively. Let PQ = a and QR = b.
The top (M) of a tower is observed from three points P, Q and R lying in a horizontal straight line which passes directly along the foot (N) of the tower. The angles of elevations of M from P, Q and R are 30°, 45° and 60° respectively. Let PQ = a and QR = b.
What is PN equal to?
Answer and solution
Answer: (b) (3 + √32)a
Method. each distance from the foot is h cot(elevation); express h through a
Let MN = h. With elevations 30°, 45°, 60°, P is farthest and R nearest, all on one side of N:
PN = h√3, QN = h, RN = h√3.
a = PQ = h(√3 − 1) ⇒ h = a√3 − 1 = (√3 + 1)a2.
PN = √3 h = (3 + √3)a2.
12. NDA 2025 (I)
What is MN equal to?
Answer and solution
Answer: (a) (3 + √32)b
Method. QR = h − h√3; solve for h
From the previous item, QN = h and RN = h√3.
b = QR = h(1 − 1√3) = h(√3 − 1)√3
h = √3 b√3 − 1 = √3(√3 + 1)b2 = (3 + √32)b.
So MN = (3 + √32)b.
13. NDA 2022 (II)
Consider the following for the next three (03) items that follow:
There are two points P and Q due south of a leaning tower, which leans towards north. P is at a distance x and Q is at a distance y from the foot of the tower (x > y). The angles of elevation of the top of the tower from P and Q are 15° and 75° respectively.
There are two points P and Q due south of a leaning tower, which leans towards north. P is at a distance x and Q is at a distance y from the foot of the tower (x > y). The angles of elevation of the top of the tower from P and Q are 15° and 75° respectively.
At what height is the top of the tower above the ground level?
Answer and solution
Answer: (a) x − y2√3
Method. coordinates along the north–south line: the top's horizontal offset cancels on subtracting
Take the foot O as origin, north positive; P is at -x, Q at -y, and the top T at (a, H), a the northward lean.
From P: tan 15° = Ha + x ⇒ a + x = H cot 15° = H(2 + √3).
From Q: tan 75° = Ha + y ⇒ a + y = H cot 75° = H(2 - √3).
Subtracting: x − y = 2√3 H ⇒ H = x − y2√3.
14. NDA 2022 (II)
If θ is the inclination of the tower to the horizontal, then what is cot θ equal to?
Answer and solution
Answer: (d) 2 − √3(x + y)x − y
Method. cot θ = horizontal lean ÷ height of the top
From the same equations, adding: 2a + x + y = 4H ⇒ a = 2H − x + y2.
The tower OT makes angle θ with the horizontal, so cot θ = aH = 2 − x + y2H.
With 2H = x − y√3: cot θ = 2 − √3(x + y)x − y.
15. NDA 2022 (II)
What is the length of the tower?
Answer and solution
Answer: (b) x − y2√3 √(1 + {2 − √3(x + y)x − y}2)
Method. length OT = √(a2 + H2) = H√(1 + cot2 θ)
OT = √(a2 + H2) = H√(1 + (aH)2) = H√(1 + cot2 θ).
With H = x − y2√3 and cot θ = 2 − √3(x + y)x − y:
OT = x − y2√3 √(1 + {2 − √3(x + y)x − y}2).
16. NDA 2022 (I)
The shadow of a tower becomes x metre longer, when the angle of elevation of sun changes from 60° to θ. If the height of the tower is √3 x metre, then which one of the following is correct?
Answer and solution
Answer: (b) 30° < θ < 45°
Method. find the new shadow length, then compare tan θ with tan 30° and tan 45°
Height h = √3 x. At 60° the shadow is h cot 60° = √3 x√3 = x.
The new shadow is x + x = 2x, so tan θ = √3 x2x = √32 ≈ 0.866.
tan 30° ≈ 0.577 < 0.866 < 1 = tan 45°, so 30° < θ < 45° (θ ≈ 40.9°).
17. NDA 2021 (I)
The shadow of a tower is found to be x metre longer, when the angle of elevation of the sun changes from 60° to 45°. If the height of the tower is 5(3 + √3) m, then what is x equal to?
Answer and solution
Answer: (b) 10 m
Method. shadow length = h cot(sun's elevation); take the difference
Shadow at 60°: h cot 60° = h√3; at 45°: h cot 45° = h.
x = h(1 − 1√3) = 5(3 + √3) × √3 − 1√3
(3 + √3)(√3 − 1) = 3√3 − 3 + 3 - √3 = 2√3
x = 5 × 2√3√3 = 10 m.
14.4 An observer above the ground: tops and bottoms of two heights
18. NDA 2026 (II)
For the following two (02) items that follow:
From the top of a building, the angles of depression of the top and bottom of a tower of height h are α and β respectively. Let H be the height of the building and D be the horizontal distance between the building and the tower.
From the top of a building, the angles of depression of the top and bottom of a tower of height h are α and β respectively. Let H be the height of the building and D be the horizontal distance between the building and the tower.
What is H2 + D2 equal to?
Answer and solution
Answer: (a) h2 cos2 αsin2(β − α)
Method. H2 + D2 is the square of the line from the building's top to the tower's foot; sine rule
Let A be the top of the building, B the foot and C the top of the tower. AB2 = H2 + D2.
In triangle ABC: ∠BAC = β − α; at C, between CB (vertically down) and CA (rising at α), ∠ACB = 90° + α; so ∠ABC = 90° - β.
Sine rule: ABsin(90° + α) = hsin(β − α) ⇒ AB = h cos αsin(β − α).
H2 + D2 = h2 cos2 αsin2(β − α).
19. NDA 2026 (II)
If α = 15° and β = 45°, then which one of the following is correct?
Answer and solution
Answer: (d) H = D
Method. the depression β of the tower's foot gives tan β = HD
The line from the building's top to the tower's foot drops H over the horizontal distance D, so tan β = HD.
With β = 45°: HD = 1 ⇒ H = D.
(α = 15° only fixes h = H − D tan 15° and does not change this.)
20. NDA 2018 (II)
The top of a hill observed from the top and bottom of a building of height h is at angles of elevation π6 and π3 respectively. What is the height of the hill?
Answer and solution
Answer: (b) 3h2
Method. two elevations from points vertically one above the other, sharing one horizontal distance
Let the hill be H high at horizontal distance d from the building.
From the bottom: tan 60° = Hd ⇒ d = H√3.
From the top: tan 30° = H − hd ⇒ H − h = d√3 = H3.
2H3 = h ⇒ H = 3h2.
21. NDA 2017 (II)
The angle of elevation of a stationary cloud from a point 25 m above a lake is 15° and the angle of depression of its image in the lake is 45°. The height of the cloud above the lake level is
Answer and solution
Answer: (b) 25√3 m
Method. cloud and its image: equal heights above and below the lake, one horizontal distance
Let the cloud be H m above the lake and d m horizontally from the observer O, who is 25 m above the lake. The image is H m below the lake level.
To the cloud: tan 15° = H − 25d. To the image: tan 45° = H + 25d, so d = H + 25.
H − 25 = (H + 25) tan 15° = (H + 25)(2 - √3)
H − 25 = 2H + 50 - √3H − 25√3 ⇒ H(√3 − 1) = 75 − 25√3 = 25√3(√3 − 1)
H = 25√3 m.
Shortcut. Standard result H = h(tan β + tan α)tan β − tan α = 25(1 + 2 - √3)1 − 2 + √3 = 25(3 - √3)√3 − 1 = 25√3.
22. NDA 2017 (II)
The angles of elevation of the top of a tower from the top and foot of a pole are respectively 30° and 45°. If hT is the height of the tower and hp is the height of the pole, then which of the following are correct?
(1) 2hphT3 + √3 = hP2
(2) hT - hP√3 + 1 = hP2
(3) 2(hP + hT)hP = 4 + √3
Select the correct answer using the code given below.
(1) 2hphT3 + √3 = hP2
(2) hT - hP√3 + 1 = hP2
(3) 2(hP + hT)hP = 4 + √3
Select the correct answer using the code given below.
Answer and solution
Answer: (c) 1 and 2 only
Method. express both heights through the common horizontal distance, then test each relation
Let the pole and tower be d apart. From the foot of the pole: tan 45° = hTd, so d = hT.
From the top of the pole: tan 30° = hT - hPd ⇒ hT - hP = hT√3 ⇒ hP = (√3 − 1)hT√3.
(1) 2hPhT3 + √3 = hP2 ⇔ hP = 2hT√3(√3 + 1) = 2(√3 − 1)hT2√3 = (√3 − 1)hT√3: true.
(2) hT - hP√3 + 1 = hT√3(√3 + 1) = (√3 − 1)hT2√3 = hP2: true.
(3) 2(hP + hT)hP = 2 + 2hThP = 2 + 2√3√3 − 1 = 2 + √3(√3 + 1) = 5 + √3 ≠ 4 + √3: false.
So 1 and 2 only.
14.5 Problems in three dimensions: directions, spheres and plots
23. NDA 2022 (II)
Consider the following for the next three (03) items that follow:
ABC is a triangular plot with AB = 16 m, BC = 10 m and CA = 10 m. A lamp post is situated at the middle point of the side AB. The lamp post subtends an angle 45° at the vertex B.
ABC is a triangular plot with AB = 16 m, BC = 10 m and CA = 10 m. A lamp post is situated at the middle point of the side AB. The lamp post subtends an angle 45° at the vertex B.
What is the height of the lamp post?
Answer and solution
Answer: (c) 8 m
Method. the post stands at the midpoint M of AB; its angle at B gives tan 45° = height/MB
M is the midpoint of AB, so MB = 162 = 8 m.
The post at M is vertical, so its top P, M and B form a right angle at M with ∠PBM = 45°.
PM = MB tan 45° = 8 m.
24. NDA 2022 (II)
What is ABsin C equal to?
Answer and solution
Answer: (b) 503 m
Method. sine rule: ABsin C = 2R, with R = abc4Δ
CA = CB = 10, so the median CM is the altitude: CM = √(102 - 82) = 6 m and Δ = 12 × 16 × 6 = 48 m2.
R = abc4Δ = 10 × 10 × 164 × 48 = 253.
ABsin C = 2R = 503 m.
Shortcut. sin C = 2 sin(C2) cos(C2) = 2 × 810 × 610 = 2425, so ABsin C = 16 × 2524 = 503.
25. NDA 2022 (II)
What is cos A + cos B + cos C equal to?
Answer and solution
Answer: (d) 3325
Method. isosceles triangle: base angles from the right triangle AMC, apex angle by the cosine rule
With CM = 6 the altitude: cos A = cos B = AMAC = 810 = 45.
cos C = 102 + 102 - 1622 × 10 × 10 = -56200 = -725.
cos A + cos B + cos C = 85 − 725 = 40 − 725 = 3325.
26. NDA 2019 (I)
The angle of elevation of a tower of height h from a point A due South of it is x and from a point B due East of A is y. If AB = z, then which one of the following is correct?
Answer and solution
Answer: (a) h2(cot2 y − cot2 x) = z2
Method. Pythagoras in the horizontal plane: south and east directions are perpendicular
Let O be the foot of the tower. Then OA = h cot x and OB = h cot y.
A is due south of O and B is due east of A, so ∠OAB = 90°.
OB2 = OA2 + AB2 ⇒ h2 cot2 y = h2 cot2 x + z2
h2(cot2 y − cot2 x) = z2.
27. NDA 2018 (I)
A spherical balloon of radius r subtends an angle α at the eye of an observer, while the angle of elevation of its centre is β. What is the height of the centre of the balloon (neglecting the height of the observer)?
Answer and solution
Answer: (a) r sin βsin(α2)
Method. the line of sight to the edge of the balloon is tangent, so the radius is perpendicular to it
Let O be the eye and C the centre. The tangents from O touch the balloon at points T with CT ⊥ OT, and each tangent makes angle α2 with OC.
In right triangle OTC: sin(α2) = rOC ⇒ OC = rsin(α2).
OC is inclined at β to the horizontal, so height of C = OC sin β = r sin βsin(α2).