Question

on a point of land there is a tower whose angle of elevation is such that its tangent is5/12. If on moving 192m distance on this line tangent becomes 3/4 ,then find the height of tower.​

Answer 1

${ \bold { \underline{\large\purple{Given : - }}}} \:$

★ There is a tower whose angle of elevation is such that its tangent is 5/12. If on moving 192m distance on this line tangent becomes 3/4

${ \bold { \underline{\large\pink{Find: - }}}} \:$

★ Find the height of tower.

☆ Answer ☆

Let AB be the tower andet the angle of elevation be of its top at C be " a ".Let " d " be a point at distance 192 metres from C such that the angle of elevation of the top of the tower at D be " b ".

Let " h " be the height of the tower and AD = " x "

Now, from given :-

TanA = 5/12 and TanB = 3/4

★ In ∆ CAB, we have

TanA = AB/AC

5/12 = h/( x + 192 ) ---------------(1)

★ In ∆ DAB, we have

TanB = AB/AD

TanB = h/x

3/4 = h / x -------(2)

We have to find " h " . This means we have to eliminate x from the equation (1) and (2).

From equation (2) , we have

3x = 4h

= x = 4h /3

Substituting the value of " x " in the equation (1) , we get

5/12 = h/ 192 + 4h/3

By cross - multiplication

➹ 5 (192 + 4h/3) = 12 h

➹ 5( 576 + 4h ) = 36 h

➹ 2880 + 20h = 36 h

➹ 2880 = 36h - 20 h

➹ 16h = 2880

➹ h = 2880/16

$\longmapsto\tt{h=\cancel\dfrac{2880}{16}}$

h = 180 metres.

Therefore,

• Height of tower is 180 metres.

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