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The Angle of Elevation of the Top of a Tower Standing on a Horizontal Plane from a Point a is α. After Walking a Distance D Towards the Foot of the Tower the Angle of Elevation is Found T - Mathematics

MCQ

The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance d towards the foot of the tower the angle of elevation is found to be β. The height of the tower is

Options

  • \[\frac{d}{cot \alpha + cot \beta}\] 

  • \[\frac{d}{cot \alpha + cot \beta}\]

  • \[\frac{d}{\tan \beta - \tan \alpha}\]

  • \[\frac{d}{\tan \beta - \tan \alpha}\]

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Solution

The given information can be represented with the help of a diagram as below.

Here, CD = h is the height of the tower. Length of BC is taken as x.

In`Δ ACD`

`tan A=(CD)/(AC)`

`tan∝=h/(d+x)` 

`h=(d+x)tan∝ `............(1)

In ΔBCD. 

`tan ß = CD/BC` 

`tan ß=h/x`

`x=h cot ß`          ...............(2) 

From (1) and (2) 

`h=(d+h cot ß)tan ∝`

`h=d tan ∝+h cot ß tan ∝` 

`h(1-cot ß tan ∝ )= d tan ∝` 

`h=d tan ∝/((1-cot ß tan ∝ ))=d/(cot ∝-cot ß)` 

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 12 Trigonometry
Q 7 | Page 42
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