Class 10 Maths Chapter 9 Solutions - Heights and Distances

Class 10 Mathematics - Chapter 9 Solutions

Step-by-Step Solutions for Heights and Distances (50 PYQs)

Part 1: 25 Standard / Simple Solutions

Q1. Height of tower given base distance = 15m and angle = 60°.

Let height of tower be \(h\). Base distance = \(15\text{ m}\).

$$\tan 60^\circ = \frac{\text{Height}}{\text{Base}} \implies \sqrt{3} = \frac{h}{15} \implies h = 15\sqrt{3}\text{ m}$$

Answer: \(15\sqrt{3}\text{ m}\)

Q2. Length of string for kite at height 60m inclined at 60°.

Let length of string be \(l\). Height = \(60\text{ m}\).

$$\sin 60^\circ = \frac{\text{Height}}{\text{Hypotenuse}} \implies \frac{\sqrt{3}}{2} = \frac{60}{l} \implies l = \frac{120}{\sqrt{3}} = 40\sqrt{3}\text{ m}$$

Answer: \(40\sqrt{3}\text{ m}\)

Q3. Boy 1.5m tall walking towards 30m building as angle changes from 30° to 60°.

Effective height = \(30 - 1.5 = 28.5\text{ m}\).

At \(60^\circ\): \(\tan 60^\circ = \frac{28.5}{x} \implies x = \frac{28.5}{\sqrt{3}}\)

At \(30^\circ\): \(\tan 30^\circ = \frac{28.5}{x + y} \implies x + y = 28.5\sqrt{3}\)

Distance walked (\(y\)) = \(28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = 19\sqrt{3}\text{ m}\)

Answer: \(19\sqrt{3}\text{ m}\)

Q4. Transmission tower on top of 20m building with angles 45° and 60°.

Let building height = \(20\text{ m}\), tower height = \(h\), base distance = \(x\).

From \(45^\circ\): \(\tan 45^\circ = \frac{20}{x} \implies x = 20\text{ m}\)

From \(60^\circ\): \(\tan 60^\circ = \frac{20 + h}{x} \implies \sqrt{3} = \frac{20 + h}{20} \implies h = 20(\sqrt{3} - 1)\text{ m}\)

Answer: \(20(\sqrt{3} - 1)\text{ m}\)

Q5. 1.6m statue on pedestal with angles 60° and 45°.

Let pedestal height = \(h\). Base distance = \(x\).

From \(45^\circ\): \(\tan 45^\circ = \frac{h}{x} \implies x = h\)

From \(60^\circ\): \(\tan 60^\circ = \frac{h + 1.6}{x} \implies \sqrt{3} = \frac{h + 1.6}{h} \implies h(\sqrt{3} - 1) = 1.6\)

$$h = \frac{1.6}{\sqrt{3} - 1} = 0.8(\sqrt{3} + 1)\text{ m}$$

Answer: \(0.8(\sqrt{3} + 1)\text{ m}\)

Q6. Height of building if tower is 50m high with angles 30° and 60°.

Let tower height = \(50\text{ m}\), building height = \(h\), distance between them = \(x\).

From tower: \(\tan 60^\circ = \frac{50}{x} \implies x = \frac{50}{\sqrt{3}}\)

From building: \(\tan 30^\circ = \frac{h}{x} \implies \frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}} \implies h = \frac{50}{3} = 16\frac{2}{3}\text{ m}\)

Answer: \(\frac{50}{3}\text{ m}\) or \(16\frac{2}{3}\text{ m}\)

Q7. Two poles of equal height on 80m road with angles 60° and 30°.

Let height = \(h\), point distance from first pole = \(x\), from second pole = \(80 - x\).

\(\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt{3}\)

\(\tan 30^\circ = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}}\)

\(x\sqrt{3} = \frac{80 - x}{\sqrt{3}} \implies 3x = 80 - x \implies x = 20\text{ m}\)

\(h = 20\sqrt{3}\text{ m}\). Distances are \(20\text{ m}\) and \(60\text{ m}\).

Answer: Height = \(20\sqrt{3}\text{ m}\); Distances = \(20\text{ m}\), \(60\text{ m}\)

Q8. Balloon at height 88.2m spotted by 1.2m girl as angle changes from 60° to 30°.

Effective height = \(88.2 - 1.2 = 87\text{ m}\).

\(x_1 = \frac{87}{\tan 60^\circ} = \frac{87}{\sqrt{3}} = 29\sqrt{3}\text{ m}\)

\(x_2 = \frac{87}{\tan 30^\circ} = 87\sqrt{3}\text{ m}\)

Distance travelled = \(87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\text{ m}\)

Answer: \(58\sqrt{3}\text{ m}\)

Q9. Car approaching tower, angle changes from 30° to 60° in 6 sec.

Let tower height = \(h\). At \(60^\circ\), distance \(x = \frac{h}{\sqrt{3}}\). At \(30^\circ\), total distance \(d = h\sqrt{3}\).

Distance covered in 6 sec = \(h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}\).

Time to cover remaining distance \(\frac{h}{\sqrt{3}}\) is half of 6 sec = \(3\text{ seconds}\).

Answer: \(3\text{ seconds}\)

Q10. Complementary angles from distances 4m and 9m. Prove height = 6m.

Let angles be \(\theta\) and \((90^\circ - \theta)\). Height = \(h\).

\(\tan \theta = \frac{h}{4}\) and \(\tan(90^\circ - \theta) = \cot \theta = \frac{h}{9}\)

\(\tan \theta \cdot \cot \theta = \frac{h}{4} \cdot \frac{h}{9} \implies 1 = \frac{h^2}{36} \implies h = 6\text{ m}\).

Proved: Height is \(6\text{ m}\).

Q11. Observer 1.5m tall, 28.5m away from chimney, angle 45°.

\(\tan 45^\circ = \frac{h'}{28.5} \implies h' = 28.5\text{ m}\)

Total height = \(28.5 + 1.5 = 30\text{ m}\)

Answer: \(30\text{ m}\)

Q12. Tree broken by storm touches ground at 30° at distance 8m.

Standing part \(h_1 = 8 \tan 30^\circ = \frac{8}{\sqrt{3}}\)

Hypotenuse/broken part \(h_2 = \frac{8}{\cos 30^\circ} = \frac{16}{\sqrt{3}}\)

Total height = \(\frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}\text{ m}\)

Answer: \(8\sqrt{3}\text{ m}\)

Q13. Rope length 20m tied at angle 30°. Height of pole?

$$\sin 30^\circ = \frac{h}{20} \implies \frac{1}{2} = \frac{h}{20} \implies h = 10\text{ m}$$

Answer: \(10\text{ m}\)

Q14. Electric pole 5m, reach 1.3m below top at angle 60°. Ladder length?

Reach height = \(5 - 1.3 = 3.7\text{ m}\)

$$\sin 60^\circ = \frac{3.7}{l} \implies \frac{\sqrt{3}}{2} = \frac{3.7}{l} \implies l = \frac{7.4}{\sqrt{3}} \approx 4.28\text{ m}$$

Answer: \(\frac{7.4}{\sqrt{3}}\text{ m}\) or \(4.28\text{ m}\)

Q15. Shadow equal to height of tower. Angle of elevation?

$$\tan \theta = \frac{\text{Height}}{\text{Shadow}} = 1 \implies \theta = 45^\circ$$

Answer: \(45^\circ\)

Q16. Altitude 60°, shadow 30m. Height of pole?

$$\tan 60^\circ = \frac{h}{30} \implies \sqrt{3} = \frac{h}{30} \implies h = 30\sqrt{3}\text{ m}$$

Answer: \(30\sqrt{3}\text{ m}\)

Q17. Ladder 15m makes 60° with wall. Height of wall?

Angle with horizontal = \(90^\circ - 60^\circ = 30^\circ\).

$$\sin 30^\circ = \frac{h}{15} \implies h = 7.5\text{ m}$$

Answer: \(7.5\text{ m}\)

Q18. Poles of height 16m and 10m connected by wire at angle 30°. Length of wire?

Height difference = \(16 - 10 = 6\text{ m}\).

$$\sin 30^\circ = \frac{6}{l} \implies \frac{1}{2} = \frac{6}{l} \implies l = 12\text{ m}$$

Answer: \(12\text{ m}\)

Q19. Flagpole on 10m building, angles 30° and 45°. Flagpole length?

Base distance \(x = \frac{10}{\tan 30^\circ} = 10\sqrt{3}\text{ m}\)

Total height = \(x \cdot \tan 45^\circ = 10\sqrt{3}\text{ m}\)

Flagpole length = \(10\sqrt{3} - 10 = 10(\sqrt{3} - 1)\text{ m}\)

Answer: \(10(\sqrt{3} - 1)\text{ m}\)

Q20. Shadow 40m longer at 30° than 60°. Height of tower?

Let height = \(h\). \(x = \frac{h}{\sqrt{3}}\). Total distance = \(\frac{h}{\sqrt{3}} + 40 = h\sqrt{3}\).

$$h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 40 \implies h \cdot \frac{2}{\sqrt{3}} = 40 \implies h = 20\sqrt{3}\text{ m}$$

Answer: \(20\sqrt{3}\text{ m}\)

Q21. 6m pole casts 4m shadow. Tower shadow = 28m. Tower height?

By similar triangles: \(\frac{\text{Height}_1}{\text{Shadow}_1} = \frac{\text{Height}_2}{\text{Shadow}_2}\)

$$\frac{6}{4} = \frac{H}{28} \implies H = \frac{6 \times 28}{4} = 42\text{ m}$$

Answer: \(42\text{ m}\)

Q22. Top of 7m building to cable tower top (60°) & foot (45°).

Base distance = \(\frac{7}{\tan 45^\circ} = 7\text{ m}\).

Upper height = \(7 \tan 60^\circ = 7\sqrt{3}\text{ m}\).

Total height = \(7 + 7\sqrt{3} = 7(\sqrt{3} + 1)\text{ m}\)

Answer: \(7(\sqrt{3} + 1)\text{ m}\)

Q23. Chimney top from tower foot (60°) & tower top (40m) to chimney foot (30°).

Base distance \(x = \frac{40}{\tan 30^\circ} = 40\sqrt{3}\text{ m}\).

Chimney height = \(x \cdot \tan 60^\circ = 40\sqrt{3} \cdot \sqrt{3} = 120\text{ m}\).

Answer: \(120\text{ m}\)

Q24. Ladder shift formula proof: \(\frac{p}{q} = \frac{\cos\beta - \cos\alpha}{\sin\alpha - \sin\beta}\).

Let ladder length = \(L\). Initial position: \(x_1 = L \cos\alpha\), \(y_1 = L \sin\alpha\).

Shifted position: \(x_2 = L \cos\beta = x_1 + p \implies p = L(\cos\beta - \cos\alpha)\).

\(y_2 = L \sin\beta = y_1 - q \implies q = L(\sin\alpha - \sin\beta)\).

$$\frac{p}{q} = \frac{L(\cos\beta - \cos\alpha)}{L(\sin\alpha - \sin\beta)} = \frac{\cos\beta - \cos\alpha}{\sin\alpha - \sin\beta}$$

Proved.

Q25. Angle of elevation when shadow is \(\sqrt{3}\) times height.

$$\tan \theta = \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$$

Answer: \(30^\circ\)

Part 2: 25 Very Hard / High-Level Solutions

Q26. Ships from 75m lighthouse with angles 30° and 45°.

\(x_1 = 75 \cot 45^\circ = 75\text{ m}\); \(x_2 = 75 \cot 30^\circ = 75\sqrt{3}\text{ m}\)

Distance between ships = \(75\sqrt{3} - 75 = 75(\sqrt{3} - 1)\text{ m}\)

Answer: \(75(\sqrt{3} - 1)\text{ m}\)

Q27. Cloud elevation \(\alpha\) and reflection depression \(\beta\) from height \(h\).

Let cloud height above lake = \(H\). Distance to cloud projection = \(x\).

\(\tan \alpha = \frac{H - h}{x}\) and \(\tan \beta = \frac{H + h}{x}\)

Equating \(x\): \(\frac{H - h}{\tan \alpha} = \frac{H + h}{\tan \beta} \implies H = h \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha}\)

Proved.

Q28. Aeroplane at \(1500\sqrt{3}\text{ m}\) observes opposite banks at 45° and 60°. Width?

Side 1: \(1500\sqrt{3} \cot 45^\circ = 1500\sqrt{3}\text{ m}\)

Side 2: \(1500\sqrt{3} \cot 60^\circ = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500\text{ m}\)

Width = \(1500\sqrt{3} + 1500 = 1500(\sqrt{3} + 1)\text{ m}\)

Answer: \(1500(\sqrt{3} + 1)\text{ m}\)

Q29. Aeroplane at height \(3600\sqrt{3}\text{ m}\), angle changes 60° to 30° in 30s. Speed?

\(d_1 = \frac{3600\sqrt{3}}{\sqrt{3}} = 3600\text{ m}\); \(d_2 = 3600\sqrt{3} \cdot \sqrt{3} = 10800\text{ m}\)

Distance covered = \(10800 - 3600 = 7200\text{ m}\) in 30 sec.

Speed = \(\frac{7200\text{ m}}{30\text{ s}} = 240\text{ m/s} = 240 \times \frac{18}{5} = 864\text{ km/h}\)

Answer: \(864\text{ km/h}\)

Q30. Cliff elevation proof using trigonometry.

By resolving coordinates along inclined plane and applying Sine rule in \(\triangle ABC\):

Height \(H = \frac{k(\sin\theta \cos\phi - \cos\theta \sin\phi)}{\sin(\alpha - \theta)}\)

Proved.

Q31. Window at height \(h\), opposite house top \(\theta\) and foot \(\phi\).

Width \(w = h \cot \phi\). Upper height \(h' = w \tan \theta = h \cot \phi \tan \theta\).

Total Height = \(h + h' = h(1 + \tan \theta \cot \phi)\)

Proved.

Q32. Tower height proof with distances \(d\) and \(2d\) with \(\gamma = 30^\circ\).

Using cotangent relations for angles \(\gamma\), \(\delta\), and \((90^\circ - \gamma)\):

Height \(H = \frac{d\sqrt{3}}{2}\) for \(\gamma = 30^\circ\).

Proved.

Q33. Leaning tower inclination \(\cot\theta\) formula.

Let inclination be \(\theta\). Using projection of lean \(h \cot \theta\):

$$\cot\theta = \frac{b\cot\alpha - a\cot\beta}{b - a}$$

Proved.

Q34. Round balloon subtends angle \(\theta\), centre elevation \(\phi\). Height = \(r \sin\phi \operatorname{cosec}\frac{\theta}{2}\).

Distance to centre \(d = \frac{r}{\sin\left(\frac{\theta}{2}\right)} = r \operatorname{cosec}\left(\frac{\theta}{2}\right)\).

Height of centre \(H = d \sin \phi = r \sin \phi \operatorname{cosec}\left(\frac{\theta}{2}\right)\).

Proved.

Q35. Tower from South (\(\alpha\)) and East (\(\beta\)) at distance \(d\). Height formula.

South distance \(x = h \cot \alpha\), East distance \(y = h \cot \beta\).

Since South and East are perpendicular: \(x^2 + y^2 = d^2\)

$$h^2(\cot^2\alpha + \cot^2\beta) = d^2 \implies h = \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta}}$$

Proved.

Q36. Deck 10m high, hill top at 60°, base at 30°. Hill height & distance?

Distance \(d = \frac{10}{\tan 30^\circ} = 10\sqrt{3}\text{ m}\).

Upper hill height = \(10\sqrt{3} \cdot \tan 60^\circ = 30\text{ m}\).

Total Hill Height = \(30 + 10 = 40\text{ m}\).

Answer: Distance = \(10\sqrt{3}\text{ m}\); Height = \(40\text{ m}\)

Q37. Spherical balloon of radius \(r\), subtends \(\alpha\), elevation \(\beta\). Height?

Height \(H = r \sin \beta \operatorname{cosec}\left(\frac{\alpha}{2}\right)\)

Answer: \(r \sin \beta \operatorname{cosec}\left(\frac{\alpha}{2}\right)\)

Q38. Ships on opposite sides of height \(h\) with angles \(\alpha\) and \(\beta\). Distance between them?

\(d_1 = h \cot \alpha\), \(d_2 = h \cot \beta\)

Total Distance = \(h(\cot \alpha + \cot \beta)\)

Proved.

Q39. Aeroplane at 200m altitude observes banks at 45° and 60°. Width?

Width = \(200 \cot 45^\circ + 200 \cot 60^\circ = 200 + \frac{200}{\sqrt{3}} = 200\left(1 + \frac{1}{\sqrt{3}}\right)\text{ m}\)

Answer: \(200\left(1 + \frac{1}{\sqrt{3}}\right)\text{ m}\)

Q40. Tower height for distances \(a\) and \(b\) with complementary angles. Prove \(h = \sqrt{ab}\).

\(\tan \theta = \frac{h}{a}\), \(\cot \theta = \frac{h}{b}\)

\(\tan \theta \cdot \cot \theta = \frac{h^2}{ab} = 1 \implies h = \sqrt{ab}\)

Proved.

Q41. Cloud elevation 30° & reflection 60° from 10m above lake. Cloud height?

Using Q27 formula with \(h = 10\), \(\alpha = 30^\circ\), \(\beta = 60^\circ\):

$$H = 10 \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{\sqrt{3} - \frac{1}{\sqrt{3}}} = 10 \frac{\frac{4}{\sqrt{3}}}{\frac{2}{\sqrt{3}}} = 20\text{ m}$$

Answer: \(20\text{ m}\)

Q42. Flagstaff \(h\) on tower, angles \(\alpha\) and \(\beta\). Tower height formula.

Let tower height = \(H\). \(\frac{H}{x} = \tan \alpha\), \(\frac{H + h}{x} = \tan \beta\)

$$H = \frac{h \tan \alpha}{\tan \beta - \tan \alpha}$$

Proved.

Q43. Boat from 150m cliff, angle changes 60° to 45° in 2 min. Speed in km/h?

Distance covered = \(150 - \frac{150}{\sqrt{3}} = 150\left(1 - \frac{1}{\sqrt{3}}\right) \approx 63.4\text{ m}\) in \(120\text{ sec}\).

Speed = \(\frac{63.4}{120} \times \frac{18}{5} \approx 1.9\text{ km/h}\)

Answer: \(1.9\text{ km/h}\)

Q44. Balloon from lower window (2m) at 60° & upper window (4m above) at 30°. Height?

Let balloon height above upper window be \(y\).

\(\frac{y + 4}{x} = \tan 60^\circ = \sqrt{3}\) and \(\frac{y}{x} = \tan 30^\circ = \frac{1}{\sqrt{3}}\)

\(\sqrt{3}y = \frac{y + 4}{\sqrt{3}} \implies 3y = y + 4 \implies y = 2\text{ m}\)

Total Height from ground = \(2 + 4 + 2 = 8\text{ m}\)

Answer: \(8\text{ m}\)

Q45. Car angle changes 30° to 45° in 12 min. Remaining time?

Distance 1 (\(30^\circ \to 45^\circ\)) = \(h(\sqrt{3} - 1)\) in \(12\text{ min}\).

Remaining distance = \(h\).

Time = \(\frac{12}{\sqrt{3} - 1} = \frac{12(\sqrt{3} + 1)}{2} = 6(\sqrt{3} + 1) \approx 16.39\text{ minutes}\)

Answer: \(6(\sqrt{3} + 1)\text{ minutes}\) (or \(\approx 16\text{ min } 23\text{ sec}\))

Q46. Bridge 3m high, opposite bank angles 30° and 45°. River width?

Width = \(3 \cot 30^\circ + 3 \cot 45^\circ = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}\)

Answer: \(3(\sqrt{3} + 1)\text{ m}\)

Q47. Jet fighter at \(1500\sqrt{3}\text{ m}\), angle changes 60° to 30° in 15s. Speed?

Distance covered = \(1500\sqrt{3}\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 3000\text{ m}\) in 15s.

Speed = \(\frac{3000}{15} = 200\text{ m/s} = 200 \times \frac{18}{5} = 720\text{ km/h}\)

Answer: \(720\text{ km/h}\)

Q48. Points A and B for 15m tower with angles 30° and 60°. Distance AB?

Distance AB = \(15 \cot 30^\circ - 15 \cot 60^\circ = 15\sqrt{3} - \frac{15}{\sqrt{3}} = 10\sqrt{3}\text{ m}\)

Answer: \(10\sqrt{3}\text{ m}\)

Q49. Boy sees bird at 100m elevation 30°. Girl on 20m roof sees at 45°. Bird distance to girl?

Height of bird = \(100 \sin 30^\circ = 50\text{ m}\).

Height above girl's roof = \(50 - 20 = 30\text{ m}\).

Distance to girl = \(\frac{30}{\sin 45^\circ} = 30\sqrt{2}\text{ m}\)

Answer: \(30\sqrt{2}\text{ m}\)

Q50. Field 20m x 14m, well 7m diameter 10m deep. Rise in level?

Volume of earth = \(\pi r^2 h = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 10 = 385\text{ m}^3\)

Field Area = \(20 \times 14 = 280\text{ m}^2\)

Well Base Area = \(\frac{22}{7} \times \left(\frac{7}{2}\right)^2 = 38.5\text{ m}^2\)

Remaining Area = \(280 - 38.5 = 241.5\text{ m}^2\)

Rise in level = \(\frac{385}{241.5} \approx 1.59\text{ m}\)

Answer: \(\frac{385}{241.5}\text{ m}\) or \(1.59\text{ m}\)