The Angle of Elevation of the Top of a Tower from Ta Point on the Same Level as the Foot of the Tower is 30° ,Show that the Height of the Tower is 129.9 Metres. - Mathematics

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The angle of elevation of the top of a tower from ta point on the same level as the foot of the tower is 30° . On advancing 150 m towards foot of the tower, the angle of elevation becomes 60° Show that the height of the tower is 129.9 metres.

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Solution

Let AB be the tower
We have:
CD =150m,∠ACB = 30° and ∠ADB = 60°
Let:
AB = hm and BD = xm

In the right ΔABD,we have:

`(AB)/(AD) = tan 60° = sqrt(3)

`⇒h/x = sqrt(3)`

`⇒ x = h/sqrt(3)`

Now, in the right  ΔACB,we have:

`(AB)/(AC) = tan 30° = 1/ sqrt(3)`

`⇒ h/(x +150) = 1/ sqrt(3)`

`sqrt(3) h =x +150`

On putting `x= h/ sqrt(3) ` in the above equation, we get:

`sqrt(3)h = h/ sqrt(3) + 150`

`⇒ 3h = h + 150 sqrt(3)`

`⇒ 2h = 150sqrt(3)`

`⇒ h = (150 sqrt(3)) /2 =75 sqrt(3) = 75 xx1.732 = 129.9m`

Hence, the height of the tower is 129.9 m

 

 

Concept: Heights and Distances
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Chapter 14: Height and Distance - Exercises

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RS Aggarwal Secondary School Class 10 Maths
Chapter 14 Height and Distance
Exercises | Q 26
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