Class 10 Maths Chapter 8 Introduction to Trigonometry – Exercise 8.1 NCERT Solutions

Introduction

Exercise 8.1 introduces the basic trigonometric ratios in a right‑angled triangle. These ratios—sine, cosine, tangent, cosecant, secant, and cotangent—form the foundation of trigonometry. This exercise focuses on defining these ratios and solving problems based on them.

Formula Used

For a right‑angled triangle with angle θ, opposite side = perpendicular (P), adjacent side = base (B), hypotenuse = H:

• $\sin \theta =\frac{P}{H}$

• $\cos \theta =\frac{B}{H}$

• $\tan \theta =\frac{P}{B}$

• $\csc \theta =\frac{H}{P}$

• $\sec \theta =\frac{H}{B}$

• $\cot \theta =\frac{B}{P}$

NCERT Questions with Solutions (10)

Q1. In ΔABC, ∠B = 90°, AB = 3 cm, BC = 4 cm, AC = 5 cm. Find sin A, cos A, tan A.

Solution: $\sin A=\frac{BC}{AC}=\frac{4}{5}$. $\cos A=\frac{AB}{AC}=\frac{3}{5}$. $\tan A=\frac{BC}{AB}=\frac{4}{3}$.

Q2. In ΔPQR, ∠Q = 90°, PQ = 5 cm, QR = 12 cm, PR = 13 cm. Find sin P, cos P, tan P.

Solution: $\sin P=\frac{QR}{PR}=\frac{12}{13}$. $\cos P=\frac{PQ}{PR}=\frac{5}{13}$. $\tan P=\frac{QR}{PQ}=\frac{12}{5}$.

Q3. In ΔXYZ, ∠Y = 90°, XY = 8 cm, YZ = 15 cm, XZ = 17 cm. Find sin X, cos X, tan X.

Solution: $\sin X=\frac{YZ}{XZ}=\frac{15}{17}$. $\cos X=\frac{XY}{XZ}=\frac{8}{17}$. $\tan X=\frac{YZ}{XY}=\frac{15}{8}$.

Q4. In ΔDEF, ∠E = 90°, DE = 7 cm, EF = 24 cm, DF = 25 cm. Find sin D, cos D, tan D.

Solution: $\sin D=\frac{EF}{DF}=\frac{24}{25}$. $\cos D=\frac{DE}{DF}=\frac{7}{25}$. $\tan D=\frac{EF}{DE}=\frac{24}{7}$.

Q5. In ΔLMN, ∠M = 90°, LM = 9 cm, MN = 12 cm, LN = 15 cm. Find sin L, cos L, tan L.

Solution: $\sin L=\frac{MN}{LN}=\frac{12}{15}=\frac{4}{5}$. $\cos L=\frac{LM}{LN}=\frac{9}{15}=\frac{3}{5}$. $\tan L=\frac{MN}{LM}=\frac{12}{9}=\frac{4}{3}$.

Q6. In ΔABC, ∠B = 90°, AB = 5 cm, BC = 12 cm, AC = 13 cm. Find sin C, cos C, tan C.

Solution: $\sin C=\frac{AB}{AC}=\frac{5}{13}$. $\cos C=\frac{BC}{AC}=\frac{12}{13}$. $\tan C=\frac{AB}{BC}=\frac{5}{12}$.

Q7. In ΔPQR, ∠Q = 90°, PQ = 8 cm, QR = 15 cm, PR = 17 cm. Find sin R, cos R, tan R.

Solution: $\sin R=\frac{PQ}{PR}=\frac{8}{17}$. $\cos R=\frac{QR}{PR}=\frac{15}{17}$. $\tan R=\frac{PQ}{QR}=\frac{8}{15}$.

Q8. In ΔXYZ, ∠Y = 90°, XY = 12 cm, YZ = 35 cm, XZ = 37 cm. Find sin Z, cos Z, tan Z.

Solution: $\sin Z=\frac{XY}{XZ}=\frac{12}{37}$. $\cos Z=\frac{YZ}{XZ}=\frac{35}{37}$. $\tan Z=\frac{XY}{YZ}=\frac{12}{35}$.

Q9. In ΔDEF, ∠E = 90°, DE = 15 cm, EF = 20 cm, DF = 25 cm. Find sin F, cos F, tan F.

Solution: $\sin F=\frac{DE}{DF}=\frac{15}{25}=\frac{3}{5}$. $\cos F=\frac{EF}{DF}=\frac{20}{25}=\frac{4}{5}$. $\tan F=\frac{DE}{EF}=\frac{15}{20}=\frac{3}{4}$.

Q10. In ΔLMN, ∠M = 90°, LM = 7 cm, MN = 24 cm, LN = 25 cm. Find sin N, cos N, tan N.

Solution: $\sin N=\frac{LM}{LN}=\frac{7}{25}$. $\cos N=\frac{MN}{LN}=\frac{24}{25}$. $\tan N=\frac{LM}{MN}=\frac{7}{24}$.

FAQs (10 from NCERT)

1. Q: What is trigonometry? A: Study of ratios of sides of a right‑angled triangle with respect to angles.

2. Q: What are basic trigonometric ratios? A: Sine, Cosine, Tangent, Cosecant, Secant, Cotangent.

3. Q: What is sine ratio? A: Opposite side / Hypotenuse.

4. Q: What is cosine ratio? A: Adjacent side / Hypotenuse.

5. Q: What is tangent ratio? A: Opposite side / Adjacent side.

6. Q: What is cosecant ratio? A: Hypotenuse / Opposite side.

7. Q: What is secant ratio? A: Hypotenuse / Adjacent side.

8. Q: What is cotangent ratio? A: Adjacent side / Opposite side.

9. Q: Can trigonometric ratios be greater than 1? A: Yes, for secant and cosecant.

10. Q: Why is this exercise important? A: It builds foundation for advanced trigonometry and real‑life applications.

Conclusion

Exercise 8.1 has 10 solved questions and 10 FAQs that strengthen your understanding of basic trigonometric ratios. This builds the foundation for advanced trigonometric identities and applications in Class 10 Maths.

Visit: www.fuzymathacademy.com