Exercise 9.1 — Some Applications of Trigonometry

1. 12 marksNCERT Cl-10 Maths, Ex 9.1, Q1 (Fig. 9.11)

   A circus artist is climbing a 20 m long rope, tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the rope makes an angle of 30° with the ground.

   Hint. The rope is the sloping line, so it is the hypotenuse. You want the vertical side — that pairs hypotenuse with opposite, which is sine.

   Step 1 — Draw and label. The pole is vertical, the ground horizontal, and the rope runs from the top of the pole to a point on the ground, making 30° there. That gives a right triangle with the right angle at the foot of the pole.

   Step 2 — Decide which ratio. The rope (20 m) is the hypotenuse, and the pole height is the side opposite the 30° angle. Opposite over hypotenuse is sine.

   sin 30° = height / 20

   Step 3 — Substitute sin 30° = 1/2. 1/2 = height / 20

   Step 4 — Solve. height = 20 × 1/2 = 10

   ✦ Answer: the pole is 10 m high.

   A useful sanity check: the height must be less than the rope's length, and 10 < 20 ✓

   Where students slip. Using tan 30° and getting 20/√3 ≈ 11.55 m. That treats the 20 m as the horizontal distance, but the rope slopes — it is the hypotenuse, not a leg.

   Another way. A 30° right triangle is half an equilateral triangle, so the side opposite the 30° is always exactly half the hypotenuse. That gives 10 m with no formula at all.

2. 23 marksNCERT Cl-10 Maths, Ex 9.1, Q2

   A tree breaks in a storm and the broken part bends so that its top touches the ground, making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

   Hint. The tree is now in two pieces — the standing stump and the leaning broken part. The original height is the sum of both, so you need both lengths.

   Step 1 — Understand the figure. The stump stands vertically at the foot. The broken part leans from the top of the stump down to a point 8 m away, making 30° with the ground. So the broken part is the hypotenuse and the stump is the opposite side.

   The tree's original height = stump + broken part, because the broken piece was standing upright before the storm.

   Step 2 — Find the stump using tan. tan 30° = stump / 8 stump = 8 × 1/√3 = 8/√3

   Step 3 — Find the broken part using cos. cos 30° = 8 / broken broken = 8 / (√3/2) = 16/√3

   Step 4 — Add them. height = 8/√3 + 16/√3 = 24/√3

   Rationalise: 24/√3 = 24√3/3 = 8√3

   ✦ Answer: the tree was 8√3 m ≈ 13.86 m tall.

   Check: stump ≈ 4.62 m and broken part ≈ 9.24 m, and 4.62 + 9.24 ≈ 13.86 ✓

   Where students slip. Giving only the stump height, or only the broken length. The question asks for the height of the *tree*, which existed before it snapped — that is the two pieces added together.

   Another way. In a 30-60-90 triangle the sides are in the ratio 1 : √3 : 2. With the adjacent side 8, the opposite is 8/√3 and the hypotenuse is 16/√3, so the sum can be written down directly.

3. 33 marksNCERT Cl-10 Maths, Ex 9.1, Q3

   A contractor plans two slides for a park. For children under five the top is 1.5 m high and the slide is inclined at 30° to the ground; for older children the top is 3 m high and the slide is inclined at 60°. Find the length of each slide.

   Hint. A slide is the sloping surface — the hypotenuse. Height is opposite the given angle, so sine connects them.

   In both cases the slide is the hypotenuse and the given height is opposite the angle of inclination, so sin θ = height / length.

   Slide 1 — height 1.5 m, angle 30°. sin 30° = 1.5 / length 1/2 = 1.5 / length length = 1.5 × 2 = 3

   ✦ 3 m

   Slide 2 — height 3 m, angle 60°. sin 60° = 3 / length √3/2 = 3 / length length = 3 × 2/√3 = 6/√3 = 2√3

   ✦ 2√3 m ≈ 3.46 m

   ✦ Answer: the shorter slide is 3 m long and the steeper one is 2√3 m ≈ 3.46 m long.

   Worth noticing: the second slide is twice as tall but only about 15% longer, because it is far steeper. That is exactly the contractor's intention.

   Where students slip. Using tan and reporting 1.5√3 and √3 — those are the horizontal distances the slides cover, not the lengths of the slides themselves.

   Another way. For the 30° slide, the opposite side is half the hypotenuse, so the slide is 2 × 1.5 = 3 m immediately.

4. 42 marksNCERT Cl-10 Maths, Ex 9.1, Q4

   The angle of elevation of the top of a tower from a point on the ground 30 m from its foot is 30°. Find the height of the tower.

   Hint. You have the horizontal distance and want the vertical height. Two legs of the triangle means tangent.

   Step 1 — Set up. The tower is vertical, the 30 m is horizontal, and the angle of elevation at the observation point is 30°.

   Step 2 — Choose the ratio. Height is the side opposite the angle and 30 m is the side adjacent, so use tangent.

   tan 30° = height / 30

   Step 3 — Substitute tan 30° = 1/√3. 1/√3 = height / 30

   Step 4 — Solve and rationalise. height = 30/√3 = 30√3/3 = 10√3

   ✦ Answer: the tower is 10√3 m ≈ 17.32 m high.

   Since the angle is less than 45°, the height must come out smaller than the 30 m distance — and 17.32 < 30 ✓

   Where students slip. Multiplying by √3 instead of dividing, giving 30√3 ≈ 51.96 m. At an elevation of only 30° the tower cannot be taller than the distance you are standing away; the size of the answer flags the slip.

   Another way. Recognise the 30-60-90 ratios: with the adjacent side 30, the opposite side is 30/√3 = 10√3 directly.

5. 52 marksNCERT Cl-10 Maths, Ex 9.1, Q5

   A kite is flying at a height of 60 m above the ground. The string is tied to a point on the ground and makes an angle of 60° with it. Find the length of the string, assuming there is no slack.

   Hint. The taut string is a straight sloping line from the ground to the kite — the hypotenuse.

   Step 1 — Identify the parts. The height 60 m is vertical and opposite the 60° angle; the string is the hypotenuse. "No slack" is what allows us to treat the string as a straight line at all.

   Step 2 — Use sine. sin 60° = 60 / string

   Step 3 — Substitute sin 60° = √3/2. √3/2 = 60 / string string = 60 × 2/√3 = 120/√3

   Step 4 — Rationalise, because an answer should not be left with a surd underneath. 120/√3 = 120√3/3 = 40√3

   ✦ Answer: the string is 40√3 m ≈ 69.28 m long.

   The string has to be longer than the kite's height, and 69.28 > 60 ✓

   Where students slip. Using tan 60° and getting 20√3 ≈ 34.6 m — that is the horizontal distance from the tie-point to the spot below the kite, and it is shorter than the height, which no string could be.

   Another way. Work with the 30-60-90 ratios 1 : √3 : 2. The side opposite 60° is 60, so the unit is 60/√3, making the hypotenuse 2 × 60/√3 = 40√3.

6. 64 marksNCERT Cl-10 Maths, Ex 9.1, Q6

   A 1.5 m tall boy stands some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards it. Find the distance he walked.

   Hint. The boy's own height matters. The triangles sit at eye level, so work with the part of the building above 1.5 m.

   Step 1 — Reduce the height to eye level. The angles are measured from the boy's eyes, 1.5 m above the ground, so the relevant vertical rise is

   30 − 1.5 = 28.5 m

   This is the step most answers get wrong, and it changes the result.

   Step 2 — Distance at the first position, where the elevation is 30°. tan 30° = 28.5 / d₁ d₁ = 28.5 / (1/√3) = 28.5√3

   Step 3 — Distance at the second position, where the elevation is 60°. tan 60° = 28.5 / d₂ d₂ = 28.5 / √3 = 9.5√3

   (28.5/√3 = 28.5√3/3 = 9.5√3)

   Step 4 — The walk is the difference. d₁ − d₂ = 28.5√3 − 9.5√3 = 19√3

   ✦ Answer: the boy walked 19√3 m ≈ 32.91 m towards the building.

   The elevation grew, so he must have got closer — the first distance has to be the larger one, and 28.5√3 ≈ 49.4 > 16.5 ≈ 9.5√3 ✓

   Where students slip. Using 30 m instead of 28.5 m and getting 20√3 ≈ 34.64 m. The 1.5 m is given precisely because it must be subtracted; a question that did not want you to use it would not mention it.

   Another way. Work in units of 28.5: the horizontal distances are 28.5√3 and 28.5/√3, whose difference is 28.5(√3 − 1/√3) = 28.5 × 2/√3 = 19√3.

7. 74 marksNCERT Cl-10 Maths, Ex 9.1, Q7

   From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed on top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

   Hint. The bottom of the tower is the top of the building. Use the 45° to pin down the horizontal distance first.

   Step 1 — Use the 45° angle, which sees the top of the building at 20 m. tan 45° = 20 / d, and tan 45° = 1, so d = 20 m

   A 45° elevation always makes the horizontal distance equal to the height — that is why the question gives this angle for the known height.

   Step 2 — Use the 60° angle, which sees the very top of the tower. Let the tower's own height be h, so the total height above the ground is 20 + h.

   tan 60° = (20 + h) / 20 √3 = (20 + h)/20

   Step 3 — Solve. 20 + h = 20√3 h = 20√3 − 20 = 20(√3 − 1)

   Step 4 — Evaluate. 20(1.732 − 1) = 20 × 0.732 ≈ 14.64

   ✦ Answer: the tower is 20(√3 − 1) m ≈ 14.64 m tall.

   Check: the total height is 20 + 14.64 = 34.64 m, and 34.64/20 = 1.732 = tan 60° ✓

   Where students slip. Reporting 20√3 ≈ 34.64 m as the tower's height. That is the height of the building *and* tower together; the tower alone is what remains after subtracting the 20 m building.

   Another way. Compute the total height first (20√3, from the 60° triangle) and subtract the building at the end. Same two lines, arranged differently.

8. 84 marksNCERT Cl-10 Maths, Ex 9.1, Q8

   A statue 1.6 m tall stands on top of a pedestal. From a point on the ground the angle of elevation of the top of the statue is 60°, and of the top of the pedestal is 45°. Find the height of the pedestal.

   Hint. Both triangles share the same horizontal distance. Express it from each angle and set the two expressions equal.

   Step 1 — Put letters on the two things you do not know. Let the pedestal be p metres high and the observation point be d metres from its base. Since the statue sits on top, its own top is at p + 1.6 metres.

   Step 2 — Use the 45° angle for the top of the pedestal. tan 45° = p / d, and tan 45° = 1, so d = p … (1)

   The shared side d is now expressed in terms of p, which is what links the two triangles.

   Step 3 — Use the 60° angle for the top of the statue. tan 60° = (p + 1.6) / d √3 = (p + 1.6) / p (substituting d = p from (1))

   Step 4 — Solve for p. √3 p = p + 1.6 p(√3 − 1) = 1.6 p = 1.6/(√3 − 1)

   Step 5 — Rationalise by multiplying by (√3 + 1)/(√3 + 1), noting (√3 − 1)(√3 + 1) = 2. p = 1.6(√3 + 1)/2 = 0.8(√3 + 1)

   0.8 × 2.732 ≈ 2.19

   ✦ Answer: the pedestal is 0.8(√3 + 1) m ≈ 2.19 m high.

   Check: the statue's top is at 2.19 + 1.6 = 3.79 m, the distance is 2.19 m, and 3.79/2.19 ≈ 1.73 = tan 60° ✓

   Where students slip. Treating the two triangles as independent and trying to solve each alone. Neither has enough information on its own — it is the shared horizontal distance that makes the problem solvable.

   Another way. Because the elevation is 45°, the pedestal height and the distance are the same number. Naming that single quantity p from the start removes one variable before any algebra begins.

9. 94 marksNCERT Cl-10 Maths, Ex 9.1, Q9

   The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

   Hint. Both angles are measured across the same horizontal gap between the two feet. Find that gap from the tower first.

   Step 1 — Read the geometry carefully. The building and the tower stand a fixed distance d apart. One angle is measured standing at the tower's foot looking at the building's top; the other standing at the building's foot looking at the tower's top. Both use the same d.

   Step 2 — Use the tower, whose height is known. Looking from the building's foot at the tower's top, the elevation is 60°: tan 60° = 50 / d √3 = 50/d, so d = 50/√3

   Step 3 — Use the building, whose height b is wanted. Looking from the tower's foot at the building's top, the elevation is 30°: tan 30° = b / d b = d × tan 30° = (50/√3) × (1/√3)

   Step 4 — Simplify. The two √3 factors multiply to 3. b = 50/3

   ✦ Answer: the building is 50/3 m ≈ 16.67 m high.

   That the smaller angle belongs to the shorter structure is a good consistency check: from the same distance, a lower top subtends a smaller angle ✓

   Where students slip. Mixing up which angle belongs to which structure. Read the phrase 'from the foot of the tower' as naming where the observer stands — the object being viewed is the other one.

   Another way. Multiply the two tangent equations: (b/d)(50/d) is awkward, but dividing them gives b/50 = tan 30°/tan 60° = (1/√3)/√3 = 1/3, so b = 50/3 in a single step and d never has to be computed.

10. 105 marksNCERT Cl-10 Maths, Ex 9.1, Q10

    Two poles of equal height stand opposite each other on either side of an 80 m wide road. From a point between them on the road, the angles of elevation of the tops of the poles are 60° and 30°. Find the height of the poles and the distances of the point from each pole.

    Hint. Let the distance to the nearer pole be x, so the other is 80 − x. The poles are the same height, which gives you the equation.

    Step 1 — Set up with one unknown for the position. Let the point be x metres from the pole seen at 60°, so it is (80 − x) metres from the pole seen at 30°. Let each pole be h metres high.

    Step 2 — Write h from the 60° triangle. tan 60° = h / x, so h = x√3 … (1)

    Step 3 — Write h from the 30° triangle. tan 30° = h / (80 − x), so h = (80 − x)/√3 … (2)

    Step 4 — Equate, since the poles are of equal height — this is the fact the question hands you. x√3 = (80 − x)/√3

    Multiply both sides by √3: 3x = 80 − x 4x = 80, so x = 20

    Step 5 — Find the height from (1). h = 20√3 ≈ 34.64

    Step 6 — The two distances are x = 20 m and 80 − x = 60 m.

    Check with (2): (80 − 20)/√3 = 60/√3 = 20√3 ✓ — the same height, as required.

    ✦ Answer: each pole is 20√3 m ≈ 34.64 m high; the point is 20 m from one pole and 60 m from the other.

    The point is nearer the pole with the larger angle, which is what you would expect.

    Where students slip. Introducing two separate distance variables and then having too few equations. Writing the second distance as 80 − x uses the road's width immediately and keeps the problem to one unknown.

    Another way. Since tan 60° = 3 tan 30°, the nearer distance must be one third of the farther one. Splitting 80 in the ratio 1 : 3 gives 20 and 60 with no algebra.

11. 115 marksNCERT Cl-10 Maths, Ex 9.1, Q11 (Fig. 9.12)

    A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top is 60°. From a point 20 m away from this point, along the line joining it to the foot of the tower, the elevation is 30°. Find the height of the tower and the width of the canal.

    Hint. Let the canal width be d. The second point is a further 20 m back, so its distance to the tower is d + 20.

    Step 1 — Name the unknowns. Let the canal be d metres wide and the tower h metres high. The first point is d from the foot of the tower; the second is d + 20.

    Step 2 — From the near point, elevation 60°. tan 60° = h / d, so h = d√3 … (1)

    Step 3 — From the far point, elevation 30°. tan 30° = h / (d + 20), so h = (d + 20)/√3 … (2)

    Step 4 — Equate (1) and (2), since both express the same tower height. d√3 = (d + 20)/√3

    Multiplying through by √3: 3d = d + 20 2d = 20, so d = 10

    Step 5 — Substitute back into (1). h = 10√3 ≈ 17.32

    Check with (2): (10 + 20)/√3 = 30/√3 = 10√3 ✓

    ✦ Answer: the tower is 10√3 m ≈ 17.32 m high and the canal is 10 m wide.

    Where students slip. Writing the far distance as 20 rather than d + 20. The second point is 20 m *further back* from the first, not 20 m from the tower — misreading this makes the problem unsolvable.

    Another way. The ratio trick again: tan 60°/tan 30° = 3, so the far distance is three times the near one. Then d + 20 = 3d gives d = 10 at once.

12. 124 marksNCERT Cl-10 Maths, Ex 9.1, Q12

    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°. Find the height of the tower.

    Hint. This is the first question mixing elevation and depression. The depression angle gives the horizontal distance, because the building's height is known.

    Step 1 — Use the angle of depression to find the horizontal distance. Looking down from the roof to the foot of the tower, the depression is 45°. The vertical drop is the building's height, 7 m.

    The angle of depression from the roof equals the angle of elevation from the tower's foot back up to the roof, because they are alternate angles between two horizontals. So in that triangle:

    tan 45° = 7 / d, and tan 45° = 1, so d = 7 m

    Step 2 — Use the angle of elevation for the part of the tower above roof level. Let that upper part be x. From the roof, looking up at 60° across the same 7 m gap:

    tan 60° = x / 7 x = 7√3

    Step 3 — Add the building's height. The tower's foot is at ground level, so its full height is the roof height plus the part above it:

    height = 7 + 7√3 = 7(1 + √3)

    Step 4 — Evaluate. 7 × 2.732 ≈ 19.12

    ✦ Answer: the tower is 7(1 + √3) m ≈ 19.12 m tall.

    Where students slip. Reporting 7√3 ≈ 12.12 m. That is only the portion of the tower above the roof; the 7 m from the ground up to roof level is part of the tower's height too.

    Another way. Draw the horizontal from the roof across to the tower — it splits the tower into a lower piece equal to the building (7 m) and an upper piece found from the 60° elevation. Seeing the split makes the addition obvious.

13. 134 marksNCERT Cl-10 Maths, Ex 9.1, Q13

    From the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45°. One ship is exactly behind the other on the same side of the lighthouse. Find the distance between the two ships.

    Hint. Both triangles have the same vertical height of 75 m. Find each ship's distance from the lighthouse and subtract.

    Step 1 — Convert depressions to elevations. The angle of depression from the lighthouse down to a ship equals the angle of elevation from that ship up to the lighthouse top, so each triangle has a 75 m vertical side and the given angle at the ship.

    Step 2 — The nearer ship, at 45°. A larger angle means a closer ship, so 45° belongs to the near one. tan 45° = 75 / d₁, so d₁ = 75 m

    Step 3 — The farther ship, at 30°. tan 30° = 75 / d₂ d₂ = 75/(1/√3) = 75√3 ≈ 129.90 m

    Step 4 — Subtract, since one ship is directly behind the other on the same side. distance between = 75√3 − 75 = 75(√3 − 1)

    75 × 0.732 ≈ 54.90

    ✦ Answer: the ships are 75(√3 − 1) m ≈ 54.90 m apart.

    Where students slip. Adding the two distances instead of subtracting. Both ships are on the *same* side, so their distances are measured along the same line from the same foot — the gap between them is the difference.

    Another way. Work in units of 75: the distances are 75 × 1 and 75 × √3, so the gap is 75(√3 − 1) without evaluating either distance separately.

14. 144 marksNCERT Cl-10 Maths, Ex 9.1, Q14 (Fig. 9.13)

    A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m above the ground. The angle of elevation of the balloon from her eyes is 60°, and after some time it reduces to 30°. Find the distance travelled by the balloon during that interval.

    Hint. Subtract the girl's height first. The balloon stays at the same level, so both triangles share one vertical measurement.

    Step 1 — Work at eye level. The angles are taken from the girl's eyes at 1.2 m, so the balloon's height above her line of sight is

    88.2 − 1.2 = 87 m

    The balloon moves horizontally, so this 87 m is the same in both positions.

    Step 2 — Horizontal distance when the elevation is 60°. tan 60° = 87 / d₁ d₁ = 87/√3 = 29√3 (since 87/3 = 29)

    Step 3 — Horizontal distance when the elevation is 30°. tan 30° = 87 / d₂ d₂ = 87√3

    Step 4 — The balloon has drifted further away, so subtract. distance travelled = 87√3 − 29√3 = 58√3

    58 × 1.732 ≈ 100.46

    ✦ Answer: the balloon travelled 58√3 m ≈ 100.46 m.

    The angle dropped, so the balloon must have moved away — the second distance is the larger, as it should be ✓

    Where students slip. Using 88.2 m and getting 58.8√3 ≈ 101.84 m. The 1.2 m is stated so that it can be subtracted; the sightlines start at the girl's eyes, not at her feet.

    Another way. Factor from the start: the two distances are 87/√3 and 87√3, whose difference is 87(√3 − 1/√3) = 87 × 2/√3 = 58√3.

15. 155 marksNCERT Cl-10 Maths, Ex 9.1, Q15

    A straight highway leads to the foot of a tower. A man at the top of the tower observes a car at an angle of depression of 30°, approaching the foot of the tower at uniform speed. Six seconds later the angle of depression is 60°. Find the time the car takes to reach the foot of the tower from that point.

    Hint. The tower's height is never given and is never needed. Call it h, and it will cancel — the answer depends only on the ratio of the distances.

    Step 1 — Write both distances in terms of the tower height h. At 30° (the first sighting, further away): tan 30° = h / d₁, so d₁ = h√3

    At 60° (six seconds later, nearer): tan 60° = h / d₂, so d₂ = h/√3

    Step 2 — Find the distance covered in those six seconds. covered = d₁ − d₂ = h√3 − h/√3

    Putting both over √3: = (3h − h)/√3 = 2h/√3

    Step 3 — Convert to a speed. speed = (2h/√3) ÷ 6 = h/(3√3) per second

    Step 4 — Find the time for the remaining distance d₂ = h/√3. time = (h/√3) ÷ (h/(3√3)) = (h/√3) × (3√3/h)

    The h cancels and so does the √3, which is why the tower's height was never needed: time = 3

    ✦ Answer: the car reaches the foot of the tower 3 seconds later.

    A neat way to see it: the remaining distance h/√3 is exactly half of the 2h/√3 covered in six seconds, and at uniform speed half the distance takes half the time.

    Where students slip. Getting stuck because no height is given and assuming the question is incomplete. Naming the height h and carrying it through is the intended method — it cancels at the last step.

    Another way. Take the ratio directly: d₁ : d₂ = h√3 : h/√3 = 3 : 1. So of the three equal parts between the first sighting and the tower, the car covers two in six seconds, and the remaining one part takes 3 seconds.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter has a single exercise, 9.1, with 15 questions; the rationalised chapter runs 9.1 Heights and Distances followed by 9.2 Summary). Questions are referenced from the NCERT textbook for identification.