This MCQ module is based on: Heights and Distances – Advanced Problems and Exercises
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Heights and Distances – Advanced Problems and Exercises
🎓 Class 10
Mathematics
CBSE
Theory
Ch 9 — Some Applications of Trigonometry
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Heights and Distances – Advanced Problems and Exercises
Targeting Class 10 level in Trigonometry, with Intermediate difficulty.
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Multi-angle Problems (continued)
When two different angles of elevation or depression are given for the same object from two different points, we get two right-triangle equations that can be solved simultaneously. This technique allows us to find a height and a distance using only angular measurements.
Example 5 — Tower and building with two angles
The shadow of a tower standing on level ground is found to be 40 m longer when the sun's altitude is 30° than when it was 60°. Find the height of the tower. (Take \(\sqrt3 = 1.732\).)
Fig. 9.8 — Two shadow positions of a tower
Solution. Let CD = h (tower) and BC = x (shadow at 60°). Then AC = x + 40.
In △DCB: \(\tan 60°=h/x\Rightarrow h=x\sqrt3\) ... (1)
In △DCA: \(\tan 30°=h/(x+40)\Rightarrow 1/\sqrt3=h/(x+40)\Rightarrow h\sqrt3=x+40\) ... (2)
Substituting (1) into (2): \(x\sqrt3\cdot\sqrt3=x+40\Rightarrow 3x=x+40\Rightarrow x=20\). Hence \(h=20\sqrt3\approx 34.64\) m.
Example 6 — Two boats from a cliff
The angles of depression of two ships from the top of a 75 m cliff are 30° and 45°, with the second ship directly behind the first (on the same side). Find the distance between the ships.
Fig. 9.9 — Two ships viewed from top P of cliff
Solution. Let PC = 75 m be the cliff. Near ship at B: \(\tan 45°=75/CB\Rightarrow CB=75\) m. Far ship at A: \(\tan 30°=75/CA\Rightarrow 1/\sqrt3=75/CA\Rightarrow CA=75\sqrt3\) m.
Distance between the ships = CA − CB = \(75\sqrt3-75=75(\sqrt3-1)\approx 54.9\) m.
Example 7 — Bridge over a river
From a bridge 3 m above a river, the angles of depression of two banks are 30° and 45° (the banks are on opposite sides of the bridge). Find the width of the river.
Fig. 9.10 — Bridge 3 m above river, two banks
Solution. Let PD = 3 m be the height of the bridge above water D. At bank B (45° side): tan 45° = PD/DB ⇒ DB = 3 m. At bank A (30° side): tan 30° = PD/DA ⇒ DA = 3√3 m.
Width of the river = DA + DB = \(3\sqrt3 + 3 = 3(\sqrt3+1)\approx 8.196\) m.
Activity — Angle-pair puzzle
A tower is seen from two points at distances a and b from its base (a > b, both on the same side). The angle of elevation from the nearer point is β and from the farther point is α. Show that the height h of the tower is \(\dfrac{(a-b)\,\tan\alpha\,\tan\beta}{\tan\beta-\tan\alpha}\). Try deriving it!
h = b tan β = a tan α after shifting origin. But careful with setup: if tower height h and horizontal distances from tower base are d and d+k, tan α = h/(d+k) and tan β = h/d. Eliminate d: d = h cot β, d+k = h cot α, so k = h(cot α − cot β), giving h = k/(cot α − cot β) = k tan α tan β / (tan β − tan α). With k = a−b this matches.
Exercise 9.1
Q1. A circus artist is climbing a 20 m long rope which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground is 30°.
sin 30° = h/20 ⇒ h = 20 × (1/2) = 10 m.
Q2. A tree breaks due to storm and the broken part bends so that the top touches the ground making an angle 30° with it. Distance from the foot to where the top touches the ground is 8 m. Find the height of the tree.
Unbroken stump BC and bent part CA with A on ground, angle BAC = 30°, AB = 8. tan 30° = BC/AB ⇒ BC = 8/√3. cos 30° = AB/CA ⇒ CA = 8·2/√3 = 16/√3. Total height = BC + CA = 8/√3 + 16/√3 = 24/√3 = 8√3 m ≈ 13.86 m.
Q3. A contractor plans to install two slides for children. For children below 5, a slide at 1.5 m with inclination 30°; for elder ones, 3 m at 60°. Find the length of each slide.
Below 5: sin 30° = 1.5/L ⇒ L = 3 m. Elder: sin 60° = 3/L ⇒ L = 3·2/√3 = 2√3 ≈ 3.46 m.
Q4. The angle of elevation of the top of a tower from a point 30 m away from its foot is 30°. Find the height.
tan 30° = h/30 ⇒ h = 30/√3 = 10√3 m ≈ 17.32 m.
Q5. A kite is flying at 60 m above ground. The string attached makes 60° with the ground. Assuming string is taut, find the length of the string.
sin 60° = 60/L ⇒ L = 60·2/√3 = 40√3 m ≈ 69.28 m.
Q6. A 1.5 m tall boy stands away from a 30 m building. The angle of elevation from his eyes to the top changes from 30° to 60° as he walks towards the building. Find the distance walked.
Effective building height above eye = 30 − 1.5 = 28.5 m. At 30°: x₁ = 28.5/tan 30° = 28.5√3. At 60°: x₂ = 28.5/tan 60° = 28.5/√3 = 9.5√3. Walked = x₁ − x₂ = 28.5√3 − 9.5√3 = 19√3 m.
Q7. From a 20 m building, the angle of elevation of the top of a transmission tower on top is 60°, and the angle of elevation of the foot of the tower is 45°. Find the height of the tower.
Horizontal distance to building base: tan 45° = 20/x ⇒ x = 20. Now from same point, total height H (building + tower): tan 60° = H/20 ⇒ H = 20√3. Tower height = 20√3 − 20 = 20(√3 − 1) ≈ 14.64 m.
Q8. A statue 1.6 m stands on a pedestal. From a point on the ground, angle of elevation to the top of the statue is 60° and top of the pedestal 45°. Find the height of the pedestal.
Let pedestal height h, horizontal distance x. tan 45° = h/x ⇒ x = h. tan 60° = (h+1.6)/x = (h+1.6)/h ⇒ √3 h = h + 1.6 ⇒ h(√3 − 1) = 1.6 ⇒ h = 1.6/(√3 − 1) = 1.6(√3+1)/2 = 0.8(√3+1) m ≈ 2.19 m.
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 tower is 50 m high, find the height of the building.
Let distance between bases = d, tower height = 50, building height = h. tan 60° = 50/d ⇒ d = 50/√3. tan 30° = h/d ⇒ h = d/√3 = 50/3 ≈ 16.67 m.
Q10. Two poles of equal height stand on opposite sides of a 80 m wide road. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the heights of the poles and the distances of the point from them.
Let point be x from 60° pole and 80 − x from 30° pole, common height h. h = x·tan 60° = x√3. h = (80−x)tan 30° = (80−x)/√3. Equating: 3x = 80−x ⇒ x = 20. So height = 20√3 m ≈ 34.64 m, distances = 20 m and 60 m.
Q11. A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite, angle of elevation to the top is 60°. From another point 20 m further, it is 30°. Find height of the tower and width of the canal.
Let width = w, height = h. tan 60° = h/w ⇒ h = w√3. tan 30° = h/(w+20) ⇒ h√3 = w+20. Substituting: w·3 = w+20 ⇒ w = 10 m. h = 10√3 m ≈ 17.32 m; canal width = 10 m.
Q12. From the top of a 7 m building, angle of elevation to the top of a cable tower is 60° and angle of depression to its foot is 45°. Find the height of the tower.
From depression 45°: horizontal distance d = 7 m. From elevation 60°: let tower top be h above building's top ⇒ h = 7·tan 60° = 7√3. Total tower height = 7 + 7√3 = 7(1 + √3) m ≈ 19.12 m.
Q13. As observed from the top of a 75 m lighthouse above sea, angles of depression of two ships are 30° and 45°. If one ship is directly behind the other, find the distance between the ships.
Near ship at 45°: d₁ = 75. Far ship at 30°: d₂ = 75√3. Distance = 75√3 − 75 = 75(√3 − 1) m ≈ 54.9 m.
Q14. A 1.2 m girl spots a balloon moving at 88.2 m above ground. Its angle of elevation from her eyes is 60°, then after some time it becomes 30°. Find the distance travelled by the balloon.
Effective height of balloon above eye = 88.2 − 1.2 = 87 m. At 60°: x₁ = 87/√3 = 29√3. At 30°: x₂ = 87√3. Distance = x₂ − x₁ = 87√3 − 29√3 = 58√3 m ≈ 100.46 m.
Q15. A straight highway leads to the foot of a tower. A man driving a car observes the angle of depression of the tower from the top is 30°. Six seconds later, it changes to 60°. Find the time taken by the car to reach the foot of the tower from this new point.
Let tower height h, speed v. Distance from first point = h√3; from second = h/√3. Travelled in 6 s = h√3 − h/√3 = 2h/√3. Speed = h/(3√3). Time to reach from second point = (h/√3)/v = (h/√3) × (3√3/h) = 3 seconds.
9.2 Summary
Key Points
- The line of sight is the line drawn from the eye of the observer to the point in the object viewed.
- The angle of elevation is the angle between the line of sight and the horizontal when the object is above the horizontal level — we raise our head to view.
- The angle of depression is the angle between the line of sight and the horizontal when the object is below the horizontal level — we lower our head to view.
- The height or length of an object, or the distance between two distant objects, can be determined with the help of trigonometric ratios.
Looking Ahead
The next chapter takes us from angles to curves — Circles — where we study tangents, their properties, and prove theorems about the uniqueness of a tangent at a point and the equality of tangent lengths from an external point.Competency-Based Questions
Q1. A boy notices that the shadow of his 1 m scooter stand doubled from morning to noon. Apply the concept to find the two sun altitudes if shadows are 2 m and 1 m respectively.
L3 Applytan α = 1/2 ⇒ α ≈ 26.57° (morning). tan β = 1/1 = 1 ⇒ β = 45° (noon).
Q2. Analyse why, for equal angles of elevation from two sides of a building, the building appears at the mid-point on a straight road — even if the angles are 30° and not 45°.
L4 AnalyseEqual elevation means tan θ = h/d₁ = h/d₂ ⇒ d₁ = d₂. The two observers are equidistant from the foot, hence the building lies at the midpoint — independent of the specific angle (provided both are the same).
Q3. Evaluate whether a tower of height 100 m can be seen from a ship 500 m away at 10° elevation with the naked eye on a clear day (given tan 10° ≈ 0.176).
L5 EvaluateRequired tan = 100/500 = 0.2. Since 0.2 > 0.176, the actual elevation is slightly above 10°. So the tower subtends an angle of ≈ 11.3°, visible to the naked eye (angular threshold ≈ 1 arc-minute = 0.017°).
Q4. Create a trigonometry problem involving a drone flying at constant altitude, observed from two points on the ground 100 m apart, with angles of elevation differing by 15°. Give a worked solution.
L6 CreateProblem: "A drone at constant altitude is observed from two ground points 100 m apart on the same side. Elevations are 45° and 60°. Find altitude." Solution: Let altitude h. At 60°: d₁ = h/√3. At 45°: d₂ = h. d₂ − d₁ = 100 ⇒ h(1 − 1/√3) = 100 ⇒ h = 100√3/(√3 − 1) ≈ 236.6 m.
Assertion & Reason
A: If the sun's altitude increases, the shadow length decreases.
R: Shadow length = height / tan(altitude), and tan is an increasing function in (0°, 90°).
R: Shadow length = height / tan(altitude), and tan is an increasing function in (0°, 90°).
Answer: A. As altitude grows, denominator grows, so shadow shortens.
A: To compare two building heights using angles of elevation, the observer must stand at the same distance from each.
R: Only then does the tangent ratio give the height directly comparable to the other.
R: Only then does the tangent ratio give the height directly comparable to the other.
Answer: D. A is false — by measuring distances as well, you can compare heights from any two points. R is a valid shortcut but not necessary.
Frequently Asked Questions
What is the strategy for heights and distances word problems?
Draw a labelled diagram, mark the right angle, identify the known and unknown, write trigonometric ratios of the given angles, and solve algebraically.
Why do many problems use 30, 45 and 60 degrees?
These angles have simple exact trigonometric values (like tan 30 = 1/root3, tan 45 = 1, tan 60 = root3) making the algebra cleaner and answers exact.
What is a common mistake in these problems?
Confusing angle of elevation with angle of depression, mislabelling opposite and adjacent sides, or forgetting the height of the observer when it is given.
How do you find the height of a tower when the shadow length changes?
Write the tangent ratio for both elevation angles in terms of tower height and distance, then eliminate the distance and solve for the height.
Do heights and distances questions use sine rule or cosine rule?
No, NCERT Class 10 restricts these problems to right-triangle trigonometry only; sine and cosine rules appear in Class 11.
How important is this chapter for board exams?
Heights and distances regularly appears as a 3-4 mark word problem in CBSE Class 10 boards and is a high-scoring topic if the diagram is drawn correctly.
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