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NCERT Solutions for Class 10 Maths Chapter 9: Some Applications of Trigonometry — Free PDF

Step-by-step solutions for heights and distances problems involving angles of elevation and depression.

CBSEClass 10
The SparkEd Authors (IITian & Googler)15 March 202638 min read
NCERT Solutions Class 10 Maths Chapter 9 Some Applications Of Trigonometry — SparkEd

Chapter 9 Overview: Some Applications of Trigonometry

Chapter 9 applies the trigonometric ratios you learnt in Chapter 8 to real-world problems involving heights and distances. You will learn to find the height of a tower, the width of a river, the distance of a ship, or the length of a bridge using angles of elevation and depression.

This is one of the most practical chapters in Class 10 Maths. Surveyors, architects, navigators, and astronomers have used these techniques for centuries. The good news for students is that the problems follow highly predictable patterns, and mastering the step-by-step approach guarantees full marks.

The chapter has 1 main exercise with around 16 word problems of varying difficulty. Every problem boils down to identifying right triangles and applying the appropriate trigonometric ratio. This chapter typically carries 4-6 marks in the CBSE board exam, often as a single long-answer question worth 4-5 marks.

Key Concepts and Definitions

Angle of Elevation: The angle above the horizontal line of sight when looking upward at an object. If you are standing on the ground and looking up at the top of a building, the angle between the horizontal and your line of sight is the angle of elevation.

Angle of Depression: The angle below the horizontal line of sight when looking downward at an object. If you are standing on top of a cliff and looking down at a boat, the angle between the horizontal and your line of sight is the angle of depression.

Line of Sight: The straight line from the observer's eye to the object being observed.

Horizontal Line: An imaginary horizontal line at the level of the observer's eye.

Key Relationship: The angle of depression from point AA to point BB equals the angle of elevation from point BB to point AA. This is because they are alternate interior angles formed by the line of sight crossing two parallel horizontal lines.

Standard Trigonometric Values Used:

Anglesin⁡\sincos⁡\costan⁡\tan
30∘30^\circ12\dfrac{1}{2}32\dfrac{\sqrt{3}}{2}13\dfrac{1}{\sqrt{3}}
45∘45^\circ12\dfrac{1}{\sqrt{2}}12\dfrac{1}{\sqrt{2}}11
60∘60^\circ32\dfrac{\sqrt{3}}{2}12\dfrac{1}{2}3\sqrt{3}

Key Problem-Solving Strategy

Every heights and distances problem follows the same systematic approach:

Step 1: Read the problem carefully and draw a clear diagram with all given information labelled. Mark the right angles, known lengths, and given angles.

Step 2: Identify right triangles in the figure. Most problems have one or two right triangles.

Step 3: Choose the appropriate trigonometric ratio based on which sides are known and which are needed:
- Know opposite and hypotenuse, or need them? Use sin⁡\sin.
- Know adjacent and hypotenuse, or need them? Use cos⁡\cos.
- Know opposite and adjacent, or need them? Use tan⁡\tan (most common in this chapter).

Step 4: Solve for the unknown using standard angle values (30∘,45∘,60∘30^\circ, 45^\circ, 60^\circ).

Step 5: Rationalise if the answer contains surds (e.g., write 13=33\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}).

Important: In about 80%80\% of the problems in this chapter, you will use tan⁡\tan because you typically know the vertical height (opposite) and need the horizontal distance (adjacent), or vice versa.

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Exercise 9.1 — Heights and Distances (Solved)

Here are the key solved problems from the exercise.

Problem 1: A tower is 30 m high. Find the distance from the foot if the angle of elevation is 30 degrees.

Solution:
Let the tower be AB=30AB = 30 m and the distance from the foot be BCBC.

In right △ABC\triangle ABC:

tan⁡30∘=ABBC=30BC\tan 30^\circ = \dfrac{AB}{BC} = \dfrac{30}{BC}

13=30BC\dfrac{1}{\sqrt{3}} = \dfrac{30}{BC}

BC=303 mBC = 30\sqrt{3} \text{ m}

Answer: The distance is 30330\sqrt{3} m ≈51.96\approx 51.96 m.

Problem 2: A kite's string is 100 m long at 60 degrees with the horizontal. Find the height.

Solution:
Let height =h= h, string =100= 100 m, angle with horizontal =60∘= 60^\circ.

sin⁡60∘=h100\sin 60^\circ = \dfrac{h}{100}

32=h100\dfrac{\sqrt{3}}{2} = \dfrac{h}{100}

h=503 m≈86.6 mh = 50\sqrt{3} \text{ m} \approx 86.6 \text{ m}

Answer: The height of the kite is 50350\sqrt{3} m.

Problem 3: From a 7 m building, the angles of elevation and depression of a tower's top and bottom are 60 degrees and 45 degrees.

Problem: Find the height of the tower.

Solution:
Let the building be AB=7AB = 7 m, tower be CDCD. Let EE be the point on the tower at the same height as AA.

So CE=CD−ED=CD−7CE = CD - ED = CD - 7 (since ED=AB=7ED = AB = 7 m).

From the angle of depression of 45∘45^\circ to the base of the tower:

tan⁡45∘=ABBD=7BD  ⟹  BD=7 m\tan 45^\circ = \dfrac{AB}{BD} = \dfrac{7}{BD} \implies BD = 7 \text{ m}

From the angle of elevation of 60∘60^\circ to the top of the tower:

tan⁡60∘=CEAE=CEBD=CE7\tan 60^\circ = \dfrac{CE}{AE} = \dfrac{CE}{BD} = \dfrac{CE}{7}

3=CE7  ⟹  CE=73\sqrt{3} = \dfrac{CE}{7} \implies CE = 7\sqrt{3}

CD=CE+ED=73+7=7(3+1) mCD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \text{ m}

Answer: Height of the tower =7(3+1)= 7(\sqrt{3} + 1) m ≈19.12\approx 19.12 m.

Problem 4: Two poles of heights 6 m and 11 m. Angle of elevation from a point between them is 60 degrees and 30 degrees.

Problem: Find the width of the road.

Solution:
Let point PP be on the road. Let the distance from PP to the 66 m pole be xx, and from PP to the 1111 m pole be yy.

tan⁡60∘=6x  ⟹  3=6x  ⟹  x=63=23\tan 60^\circ = \dfrac{6}{x} \implies \sqrt{3} = \dfrac{6}{x} \implies x = \dfrac{6}{\sqrt{3}} = 2\sqrt{3}

tan⁡30∘=11y  ⟹  13=11y  ⟹  y=113\tan 30^\circ = \dfrac{11}{y} \implies \dfrac{1}{\sqrt{3}} = \dfrac{11}{y} \implies y = 11\sqrt{3}

Width of road =x+y=23+113=133= x + y = 2\sqrt{3} + 11\sqrt{3} = 13\sqrt{3} m.

Answer: Width =133= 13\sqrt{3} m ≈22.52\approx 22.52 m.

Problem 5: From the top of a 60 m tower, angles of depression to two objects on the same side are 30 degrees and 60 degrees.

Problem: Find the distance between the objects.

Solution:
Let the tower be AB=60AB = 60 m. Let the two objects be at CC and DD with CC closer.

tan⁡60∘=60BC  ⟹  BC=603=203\tan 60^\circ = \dfrac{60}{BC} \implies BC = \dfrac{60}{\sqrt{3}} = 20\sqrt{3}

tan⁡30∘=60BD  ⟹  BD=603\tan 30^\circ = \dfrac{60}{BD} \implies BD = 60\sqrt{3}

CD=BD−BC=603−203=403 mCD = BD - BC = 60\sqrt{3} - 20\sqrt{3} = 40\sqrt{3} \text{ m}

Answer: Distance between objects =403= 40\sqrt{3} m ≈69.28\approx 69.28 m.

Additional Worked Examples

**Example 1: A statue 1.61.6 m tall stands on a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60∘60^\circ and the angle of elevation of the top of the pedestal is 45∘45^\circ. Find the height of the pedestal.**

Solution:
Let the pedestal height be hh m. Total height (pedestal + statue) =h+1.6= h + 1.6 m.

Let the distance from the point to the base be dd.

tan⁡45∘=hd  ⟹  d=h⋯(1)\tan 45^\circ = \dfrac{h}{d} \implies d = h \quad \cdots (1)

tan⁡60∘=h+1.6d  ⟹  3=h+1.6d⋯(2)\tan 60^\circ = \dfrac{h + 1.6}{d} \implies \sqrt{3} = \dfrac{h + 1.6}{d} \quad \cdots (2)

Substitute d=hd = h from (1) into (2):

3=h+1.6h\sqrt{3} = \dfrac{h + 1.6}{h}

3h=h+1.6\sqrt{3}h = h + 1.6

h(3−1)=1.6h(\sqrt{3} - 1) = 1.6

h=1.63−1=1.6(3+1)(3−1)(3+1)=1.6(3+1)2=0.8(3+1)≈2.19 mh = \dfrac{1.6}{\sqrt{3} - 1} = \dfrac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \approx 2.19 \text{ m}

---

**Example 2: From the top of a tower, the angle of depression of a car moving towards the tower is 30∘30^\circ. After 66 minutes, the angle of depression becomes 60∘60^\circ. How long will the car take to reach the tower?**

Solution:
Let the tower height be hh and the car's initial distance be d1d_1, final distance be d2d_2.

tan⁡30∘=hd1  ⟹  d1=h3\tan 30^\circ = \dfrac{h}{d_1} \implies d_1 = h\sqrt{3}

tan⁡60∘=hd2  ⟹  d2=h3=h33\tan 60^\circ = \dfrac{h}{d_2} \implies d_2 = \dfrac{h}{\sqrt{3}} = \dfrac{h\sqrt{3}}{3}

Distance covered in 66 minutes =d1−d2=h3−h33=2h33= d_1 - d_2 = h\sqrt{3} - \dfrac{h\sqrt{3}}{3} = \dfrac{2h\sqrt{3}}{3}.

Remaining distance =d2=h33= d_2 = \dfrac{h\sqrt{3}}{3}.

Since speed is constant: remaining distancedistance covered=h3/32h3/3=12\dfrac{\text{remaining distance}}{\text{distance covered}} = \dfrac{h\sqrt{3}/3}{2h\sqrt{3}/3} = \dfrac{1}{2}.

Time to cover remaining distance =12×6=3= \dfrac{1}{2} \times 6 = 3 minutes.

---

**Example 3: A vertical tower stands on a horizontal plane. From a point on the ground 4040 m away from the base, the angle of elevation of the top is 30∘30^\circ. Find the height of the tower.**

Solution:

tan⁡30∘=h40\tan 30^\circ = \dfrac{h}{40}

13=h40\dfrac{1}{\sqrt{3}} = \dfrac{h}{40}

h=403=4033≈23.09 mh = \dfrac{40}{\sqrt{3}} = \dfrac{40\sqrt{3}}{3} \approx 23.09 \text{ m}

Common Mistakes to Avoid

Mistake 1: Not drawing a diagram.
Many students try to solve the problem mentally and make errors. Always draw a labelled diagram first — it is worth marks and prevents confusion.

Mistake 2: Confusing angle of elevation with angle of depression.
Elevation is measured upward from the horizontal. Depression is measured downward. If you are on a building looking down, the angle is depression, not elevation.

Mistake 3: Using the wrong trigonometric ratio.
Most problems use tan⁡\tan (opposite/adjacent). Using sin⁡\sin or cos⁡\cos when tan⁡\tan is needed (or vice versa) gives wrong answers.

Mistake 4: Forgetting to rationalise the denominator.
303\dfrac{30}{\sqrt{3}} should be rationalised to 3033=103\dfrac{30\sqrt{3}}{3} = 10\sqrt{3}. CBSE expects rationalised answers.

**Mistake 5: Mixing up tan⁡30∘\tan 30^\circ and tan⁡60∘\tan 60^\circ.**
tan⁡30∘=13\tan 30^\circ = \dfrac{1}{\sqrt{3}} and tan⁡60∘=3\tan 60^\circ = \sqrt{3}. Swapping these is a very common error. Remember: 30∘30^\circ is a small angle, so tan⁡30∘\tan 30^\circ is the smaller value (<1< 1).

Tips for Scoring Full Marks

1. Always draw a diagram — it is worth marks and helps avoid errors. Label all known values and mark right angles.
2. The angle of depression from point AA to point BB equals the angle of elevation from BB to AA (alternate interior angles). Use this freely.
3. **Most problems use tan⁡\tan** because you typically know the vertical height and horizontal distance.
4. Rationalise surds: 13=33\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}. Always present answers in rationalised form.
5. When two angles are given (e.g., angles from the top and bottom of a tower), you need two right triangles and two equations.
6. For "moving object" problems (car approaching a tower), the two positions give two right triangles sharing the same height.
7. Memorise: tan⁡30∘=13\tan 30^\circ = \dfrac{1}{\sqrt{3}}, tan⁡45∘=1\tan 45^\circ = 1, tan⁡60∘=3\tan 60^\circ = \sqrt{3}. These three values cover almost every problem.
8. Write the final answer clearly with units (metres), and give both the exact (30330\sqrt{3} m) and approximate (≈51.96\approx 51.96 m) values.

Practice Questions with Answers

Q1. A tower is 5050 m high. The angle of elevation from a point on the ground is 45∘45^\circ. Find the distance from the base.

Answer: tan⁡45∘=50d  ⟹  d=50\tan 45^\circ = \dfrac{50}{d} \implies d = 50 m.

---

Q2. A tree breaks at a point and falls making a 30∘30^\circ angle with the ground. If the point where the top touches the ground is 2020 m from the base, find the total height.

Answer: Let the broken part be ll and the standing part be hh. tan⁡30∘=h20\tan 30^\circ = \dfrac{h}{20}, so h=203=2033h = \dfrac{20}{\sqrt{3}} = \dfrac{20\sqrt{3}}{3}. cos⁡30∘=20l\cos 30^\circ = \dfrac{20}{l}, so l=203/2=403=4033l = \dfrac{20}{\sqrt{3}/2} = \dfrac{40}{\sqrt{3}} = \dfrac{40\sqrt{3}}{3}. Total height =h+l=203+4033=6033=203≈34.64= h + l = \dfrac{20\sqrt{3} + 40\sqrt{3}}{3} = \dfrac{60\sqrt{3}}{3} = 20\sqrt{3} \approx 34.64 m.

---

Q3. From the top of a 120120 m building, the angle of depression of a car is 30∘30^\circ. Find the distance of the car from the base.

Answer: tan⁡30∘=120d\tan 30^\circ = \dfrac{120}{d}. d=1203≈207.84d = 120\sqrt{3} \approx 207.84 m.

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Q4. The shadow of a tower when the angle of elevation of the sun is 60∘60^\circ is 2525 m. Find the height of the tower.

Answer: tan⁡60∘=h25\tan 60^\circ = \dfrac{h}{25}. h=253≈43.3h = 25\sqrt{3} \approx 43.3 m.

---

Q5. Two observers on opposite sides of a 7575 m tower observe the top at angles of 30∘30^\circ and 60∘60^\circ. Find the distance between them.

Answer: d1=75tan⁡30∘=753d_1 = \dfrac{75}{\tan 30^\circ} = 75\sqrt{3}. d2=75tan⁡60∘=753=253d_2 = \dfrac{75}{\tan 60^\circ} = \dfrac{75}{\sqrt{3}} = 25\sqrt{3}. Total =753+253=1003≈173.2= 75\sqrt{3} + 25\sqrt{3} = 100\sqrt{3} \approx 173.2 m.

Key Takeaways

  • Heights and distances problems always involve right triangles and trigonometric ratios.
    - tan⁡\tan is the most-used ratio in this chapter (opposite/adjacent = height/distance).
    - Angle of depression from AA to BB == angle of elevation from BB to AA (alternate interior angles).
    - Always draw a clear, labelled diagram before solving.
    - Standard values to memorise: tan⁡30∘=1/3\tan 30^\circ = 1/\sqrt{3}, tan⁡45∘=1\tan 45^\circ = 1, tan⁡60∘=3\tan 60^\circ = \sqrt{3}.
    - Rationalise answers containing surds.
    - This chapter carries 44-66 marks in boards and the problems are highly predictable with practice.
    - The two-triangle technique (two angles, shared height or base) is the key to solving complex problems.

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