Class 10 Mathematics
Chapter 9: Some Applications of Trigonometry
Exercise 9.1 – Stepwise Solutions
1. Circus Artist Problem
Given:
Length of rope (hypotenuse) = 20 m
Angle with ground = 30°
To Find: Height of pole
Using sin θ = Opposite / Hypotenuse
sin 30° = Height / 20
1/2 = Height / 20
Height = 20 × 1/2 = 10 m
2. Broken Tree Problem
Given:
Angle with ground = 30°
Distance from foot to touching point = 8 m
Let total height = h
In right triangle:
tan 30° = Height of broken part / 8
1/√3 = Broken height / 8
Broken height = 8/√3
Length of broken part = 8 / cos 30°
= 8 / (√3/2) = 16/√3
Total height = Broken height + Remaining height
= 8/√3 + 16/√3 = 24/√3
= 8√3 m
Height of tree = 8√3 m
3. Slides in Park
(i) For small children:
Height = 1.5 m
Angle = 30°
sin 30° = 1.5 / Length
1/2 = 1.5 / Length
Length = 1.5 × 2 = 3 m
(ii) For elder children:
Height = 3 m
Angle = 60°
sin 60° = 3 / Length
√3/2 = 3 / Length
Length = (3 × 2)/√3 = 6/√3 = 2√3 m
Slide length = 2√3 m
4. Height of Tower
Given:
Distance from foot = 30 m
Angle of elevation = 30°
tan 30° = Height / 30
1/√3 = Height / 30
Height = 30/√3 = 10√3 m
Height of tower = 10√3 m
5. Kite Problem
Given:
Height = 60 m
Angle = 60°
sin 60° = 60 / String length
√3/2 = 60 / L
L = (60 × 2)/√3 = 120/√3 = 40√3 m
Length of string = 40√3 m
6. Boy and Building Problem
Given:
Building height = 30 m
Boy height = 1.5 m
Effective height = 30 − 1.5 = 28.5 m
Let initial distance = x
tan 30° = 28.5 / x
1/√3 = 28.5 / x
x = 28.5√3
Let final distance = y
tan 60° = 28.5 / y
√3 = 28.5 / y
y = 28.5/√3
Distance walked = x − y
= 28.5√3 − 28.5/√3
= 28.5( (3 − 1)/√3 )
= 28.5 × 2/√3
= 57/√3 = 19√3 m
Distance walked = 19√3 m
7. Transmission Tower Problem
Given:
Height of building = 20 m
Angle to bottom of tower = 45°
Angle to top of tower = 60°
Let distance from building = x
tan 45° = 20 / x
1 = 20 / x
x = 20 m
Let tower height = h
tan 60° = (20 + h) / 20
√3 = (20 + h)/20
20√3 = 20 + h
h = 20√3 − 20
Height of tower = 20(√3 − 1) m
Final Answers Summary
- 1) 10 m
- 2) 8√3 m
- 3) 3 m and 2√3 m
- 4) 10√3 m
- 5) 40√3 m
- 6) 19√3 m
- 7) 20(√3 − 1) m
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