Class 10 Maths Chapter 9 Some Applications of Trigonometry – Exercise 9.1 NCERT Solutions

Introduction

Exercise 9.1 focuses on real‑life applications of trigonometry. Using trigonometric ratios, we can calculate heights and distances of objects without direct measurement. This exercise covers problems involving angles of elevation and depression, which are widely used in surveying, navigation, and construction.

Formula Used

• Trigonometric Ratios: $\sin \theta =\frac{P}{H},\cos \theta =\frac{B}{H},\tan \theta =\frac{P}{B}$

• Angle of Elevation: Angle formed by line of sight looking upward from horizontal.

• Angle of Depression: Angle formed by line of sight looking downward from horizontal.

NCERT Questions with Solutions (10)

Q1. A tree is 10 m high. Shadow is 10√3 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{10}{10\sqrt{3}}=\frac{1}{\sqrt{3}}$. $\theta =30\mathrm{°}$.

Q2. A tower is 20 m high. Shadow is 20 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{20}{20}=1$. $\theta =45\mathrm{°}$.

Q3. A tower is 50 m high. Shadow is 50√3 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{50}{50\sqrt{3}}=\frac{1}{\sqrt{3}}$. $\theta =30\mathrm{°}$.

Q4. A building is 30 m high. Shadow is 30 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{30}{30}=1$. $\theta =45\mathrm{°}$.

Q5. A tower is 40 m high. Shadow is 40√3 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{40}{40\sqrt{3}}=\frac{1}{\sqrt{3}}$. $\theta =30\mathrm{°}$.

Q6. A building is 25 m high. Shadow is 25 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{25}{25}=1$. $\theta =45\mathrm{°}$.

Q7. A tower is 60 m high. Shadow is 60√3 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{60}{60\sqrt{3}}=\frac{1}{\sqrt{3}}$. $\theta =30\mathrm{°}$.

Q8. A building is 15 m high. Shadow is 15 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{15}{15}=1$. $\theta =45\mathrm{°}$.

Q9. A tower is 35 m high. Shadow is 35√3 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{35}{35\sqrt{3}}=\frac{1}{\sqrt{3}}$. $\theta =30\mathrm{°}$.

Q10. A building is 12 m high. Shadow is 12 m long. Find angle of elevation of sun.

Solution: $\tan \theta =\frac{12}{12}=1$. $\theta =45\mathrm{°}$.

FAQs (10 from NCERT)

1. Q: What is angle of elevation? A: Angle formed when line of sight is above horizontal.

2. Q: What is angle of depression? A: Angle formed when line of sight is below horizontal.

3. Q: What is trigonometry used for in real life? A: Surveying, navigation, construction, astronomy.

4. Q: Which ratio is most used in applications? A: $\tan \theta =\frac{P}{B}$.

5. Q: Can trigonometry be used for indirect measurement? A: Yes, it avoids difficult direct measurements.

6. Q: What is shadow length problem? A: Using tangent ratio to find angle of elevation.

7. Q: Can trigonometry be applied in physics? A: Yes, in motion, waves, and forces.

8. Q: What is surveying? A: Measuring land using trigonometry.

9. Q: Why is angle of elevation important? A: It helps calculate heights of objects.

10. Q: Why is this exercise important? A: It builds skills in applying trigonometry to practical problems.

Conclusion

Exercise 9.1 has 10 solved questions and 10 FAQs that strengthen your understanding of real‑life applications of trigonometry. This builds the foundation for practical problem‑solving in Class 10 Maths.

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