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If the Angles of Elevation of a Tower from Two Points Distant a and B (A>B) from Its Foot and in the Same Straight Line from It Are 30° and 60°, Then the Height of the Tower is - Mathematics


If the angles of elevation of a tower from two points distant a and b (a>b) from its foot and in the same straight line from it are 30° and 60°, then the height  of the tower is


  • \[\sqrt{a + b}\]

  • \[\sqrt{ab}\]

  • \[\sqrt{a - b}\]

  • \[\sqrt{\frac{a}{b}}\]

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Let h be the height of tower AB 

Given that: angle of elevation are`∠C=60°` and`∠D=60°` .`∠D=30°`

Distance `BC=b`  and`BD=a` 

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle `ABC`, 

`⇒ tan C=( AB)/(BC)`  

`⇒ tan 60°=(AB)/(BC)` 

`⇒ tan 60°= h/b` 

Again in a triangle ABD

`⇒ tan D=(AB)/(BD)` 

`⇒ tan 30°=h/a`

Again in a triangle ABD,

`⇒ tan D=(AB)/(BD)` 

`⇒ tan 30°=h/a`

`⇒ tan (90°-60°)=h/a` 

`⇒ cot 60°=h/a` 

`⇒ 1/tan 60°=h/a` 

`⇒ b/h=h/a`             ` "put tan "60°=h/b` 

`⇒ h^2=ab` 




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RD Sharma Class 10 Maths
Chapter 12 Trigonometry
Q 4 | Page 41
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