Some Applications of Trigonometry l Exercise 9.1 l Class 10 l NCERT Solutions

Class 10 Mathematics

Chapter 9: Some Applications of Trigonometry

Exercise 9.1 – Solutions (Q8 to Q15)

8. Statue on Pedestal

Given:
Height of statue = 1.6 m
Angle to top of statue = 60°
Angle to top of pedestal = 45°

Let height of pedestal = h

tan 45° = h / x ⇒ 1 = h/x ⇒ x = h

tan 60° = (h + 1.6) / x

√3 = (h + 1.6)/h

√3 h = h + 1.6

h(√3 − 1) = 1.6

h = 1.6 / (√3 − 1)

Multiply by (√3 + 1)/(√3 + 1)

h = 1.6(√3 + 1)/(3 − 1)

h = 0.8(√3 + 1)

Height of pedestal = 0.8(√3 + 1) m

9. Tower and Building

Given:
Height of tower = 50 m
Angle from foot of tower to building top = 30°
Angle from foot of building to tower top = 60°

Let distance between them = x

tan 60° = 50 / x

√3 = 50/x ⇒ x = 50/√3

tan 30° = Height of building / x

1/√3 = h / (50/√3)

h = 50/3

Height of building = 50/3 m

10. Two Poles

Let distance from point to first pole = x
Distance to second pole = 80 − x
Height of poles = h

tan 60° = h/x ⇒ √3 = h/x ⇒ h = √3x

tan 30° = h/(80 − x) ⇒ 1/√3 = h/(80 − x)

Substitute h:

1/√3 = √3x/(80 − x)

80 − x = 3x

80 = 4x

x = 20 m

So other distance = 60 m

Height = √3 × 20 = 20√3 m

11. TV Tower and Canal

Let width of canal = x
Height of tower = h

From point C:

tan 60° = h/x ⇒ √3 = h/x ⇒ h = √3x

From point D (20 m farther):

tan 30° = h/(x + 20)

1/√3 = √3x/(x + 20)

x + 20 = 3x

20 = 2x

x = 10 m

h = √3 × 10 = 10√3 m

Width of canal = 10 m
Height of tower = 10√3 m

12. Building and Cable Tower

Given:
Building height = 7 m
Angle of depression to foot = 45°
Angle of elevation to top = 60°

Let distance = x

tan 45° = 7/x ⇒ x = 7 m

tan 60° = (H − 7)/7

√3 = (H − 7)/7

H − 7 = 7√3

H = 7 + 7√3

Height of tower = 7(1 + √3) m

13. Lighthouse and Ships

Given:
Height = 75 m
Angles = 30° and 45°

Distance of nearer ship:

tan 45° = 75/x₁ ⇒ x₁ = 75

Distance of farther ship:

tan 30° = 75/x₂

1/√3 = 75/x₂ ⇒ x₂ = 75√3

Distance between ships = x₂ − x₁

= 75√3 − 75

= 75(√3 − 1)

Distance between ships = 75(√3 − 1) m

14. Balloon Problem

Given:
Balloon height = 88.2 m
Girl height = 1.2 m

Effective height = 88.2 − 1.2 = 87 m

At 60°:

tan 60° = 87/x₁ ⇒ √3 = 87/x₁

x₁ = 87/√3

At 30°:

tan 30° = 87/x₂ ⇒ 1/√3 = 87/x₂

x₂ = 87√3

Distance travelled = x₂ − x₁

= 87√3 − 87/√3

= 87( (3 − 1)/√3 )

= 174/√3

= 58√3 m

Distance travelled = 58√3 m

15. Man on Tower and Car

Let height of tower = h
Initial distance = x

tan 30° = h/x

1/√3 = h/x ⇒ x = √3h

After 6 seconds:

tan 60° = h/y

√3 = h/y ⇒ y = h/√3

Distance covered in 6 seconds:

= x − y

= √3h − h/√3

= (3h − h)/√3

= 2h/√3

Speed = (2h/√3)/6 = h/(3√3)

Remaining distance = y = h/√3

Time to reach = (h/√3) / (h/(3√3))

= 3 seconds

Time required = 3 seconds

Final Answers Summary

• 8) 0.8(√3 + 1) m
• 9) 50/3 m
• 10) Height = 20√3 m; distances = 20 m and 60 m
• 11) Height = 10√3 m; width = 10 m
• 12) 7(1 + √3) m
• 13) 75(√3 − 1) m
• 14) 58√3 m
• 15) 3 seconds