Home
Class 12
MATHS
A tower subtends an angle alpha at a poi...

A tower subtends an angle `alpha` at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b ft just above A is `beta`. Then the height of the tower is

A

`b tan alpha.tanbeta`

B

`b(cot beta)/(cot alpha)`

C

`b(tan beta)/(tan alpha).sinalpha`

D

`bcot alpha.cotbeta`

Text Solution

AI Generated Solution

The correct Answer is:
To find the height of the tower given the angles α and β, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Geometry**: - Let the height of the tower be \( h \). - Let point \( A \) be at the base of the tower, and point \( B \) be a point \( b \) feet above point \( A \). - The angle \( \alpha \) is the angle subtended by the tower at point \( A \). - The angle \( \beta \) is the angle of depression from point \( B \) to point \( A \). 2. **Identify the Right Triangles**: - In triangle \( ABC \) (where \( C \) is the top of the tower), we can use the tangent function: \[ \tan(\alpha) = \frac{h}{AB} \] - Here, \( AB \) is the horizontal distance from point \( A \) to the base of the tower. 3. **Express \( AB \) in terms of \( h \)**: - Rearranging the equation gives: \[ AB = \frac{h}{\tan(\alpha)} \] 4. **Analyze Triangle \( ABD \)**: - In triangle \( ABD \) (where \( D \) is directly below point \( B \)), we have: \[ \tan(\beta) = \frac{h}{AB + b} \] - Here, \( AB + b \) is the distance from point \( B \) to the foot of the tower. 5. **Express \( AB + b \) in terms of \( h \)**: - Rearranging gives: \[ AB + b = \frac{h}{\tan(\beta)} \] 6. **Substitute \( AB \) into the equation**: - Substitute \( AB = \frac{h}{\tan(\alpha)} \) into the equation: \[ \frac{h}{\tan(\alpha)} + b = \frac{h}{\tan(\beta)} \] 7. **Clear the fractions**: - Multiply through by \( \tan(\alpha) \tan(\beta) \): \[ h \tan(\beta) + b \tan(\alpha) \tan(\beta) = h \tan(\alpha) \] 8. **Rearrange to isolate \( h \)**: - Rearranging gives: \[ h \tan(\beta) - h \tan(\alpha) = -b \tan(\alpha) \tan(\beta) \] - Factor out \( h \): \[ h (\tan(\beta) - \tan(\alpha)) = -b \tan(\alpha) \tan(\beta) \] 9. **Solve for \( h \)**: - Finally, we can express \( h \) as: \[ h = \frac{b \tan(\alpha) \tan(\beta)}{\tan(\beta) - \tan(\alpha)} \] ### Final Result: The height of the tower \( h \) is given by: \[ h = \frac{b \tan(\alpha) \tan(\beta)}{\tan(\beta) - \tan(\alpha)} \]
Promotional Banner

Topper's Solved these Questions

  • HEIGHTS & DISTANCE

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (OBJECTIVE) LEVEL-I|20 Videos
  • FUNCTION

    FIITJEE|Exercise NUMERICAL BASED|3 Videos
  • HYPERBOLA

    FIITJEE|Exercise NUMERICAL BASED|4 Videos

Similar Questions

Explore conceptually related problems

A tower subtends an angle alpha at a point in the plane of its base and the angle of depression of the foot of the tower at a point b ft. just above A is beta . Then , height of the tower is

A tower subtends an angle alpha at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b metres just above A is beta .Prove that the height of the tower is b tan alpha cot beta .

A tower subtends and angle 60^(@) at a point A in the plane of its base and the angle of depression of the foot of the tower at a point 10 meters just above A is 30^(@). The height of the tower is

A tower subtends an angle of 30o at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60o . The height of the tower is h/2m (b) sqrt(3)h m (c) h/3m (d) h/(sqrt(3))m

A tower subtends an angle of 30^(@) at a point on the same level as its foot.At a second point h metres above the first,the depression of the foot of the tower is 60^(@). The height of the tower is:

FIITJEE-HEIGHTS & DISTANCE -ASSIGNMENT PROBLEMS (OBJECTIVE) LEVEL-II
  1. The height of a house subtends a right angle at the opposite street li...

    Text Solution

    |

  2. The angle of elevation of the top of a tower at a point A due south of...

    Text Solution

    |

  3. The angle of elevation of a tower PQ at a point A due north of it is 3...

    Text Solution

    |

  4. An electric pole stands at the vertex A of the equilateral triangular ...

    Text Solution

    |

  5. A pole on the ground lenses 60^(@) to the vertical. At a point a meter...

    Text Solution

    |

  6. A 10 meters high tower is standing at the centre of an equilateral tri...

    Text Solution

    |

  7. A tower subtends an angle alpha at a point A in the plane of its base ...

    Text Solution

    |

  8. A tower subtends angle theta, 2theta, 3theta at the three points A,B a...

    Text Solution

    |

  9. A spherical balloon of radius 50 cm, subtends an angle of 60^(@) at a ...

    Text Solution

    |

  10. PQ is a post of height a, AB is a tower of height h at a distance x fr...

    Text Solution

    |

  11. The angle of elevation of the top of a hill from a point is alpha. Aft...

    Text Solution

    |

  12. The base of a cliff is circular. From the extremities of a diameter of...

    Text Solution

    |

  13. A round balloon of radius r subtends an angle alpha at the eye of the ...

    Text Solution

    |

  14. .A house of height 100 m substends a right angle at the window of an o...

    Text Solution

    |

  15. A tower subtends an angle alpha at a point A in the plane of its...

    Text Solution

    |

  16. ABCD is a rectangular field. A vertical lamp post of height 12 m stand...

    Text Solution

    |

  17. A tower of x metres height has flag staff at its top. The tower and th...

    Text Solution

    |

  18. An aeroplane flying horizontally , 1km above the ground , is observed...

    Text Solution

    |

  19. A man in a boat rowed away from a cliff 150 m high takes 2 min, to cha...

    Text Solution

    |

  20. The angle of elevation of the cloud at a point 2500 m high from the la...

    Text Solution

    |