Question

guys plzz... solve......​

Answer 1

Answer :

Here's yours answer,

Solutions :-

Given that , AB is a building with height 60m.

Angle of depression from top of building to top of lamp is 30° and to bottom is 60° .

Now , AB = AO +OB

Let us consider that AO is X m.

So ,

AB = AO +OB

60 = X + OB

OB = 60 - X..............(i)

Also , OB = CD........(ii)

OC= BD.....(iii)

Now ,

In ∆AOC ,

$tan30 ^{o} = \frac{ao}{co}$

$\frac{1}{ \sqrt{3} } = \frac{x}{co}$

$co = \sqrt{3} x.........(iv)$

Similarly , In ∆ ABD ,

$tan {60}^{o} = \frac{ab}{bd}$

$\sqrt{3} = \frac{60}{bd}$

$bd = \frac{60}{ \sqrt{3} }$

On Multiplying numerator and denominator by √3 , we get ;

$bd = \frac{60}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$bd = \frac{60 \sqrt{3} }{3} = 20 \sqrt{3} .......(v)$

From equation (iii) , we get ;

OC = BD ,

So ,

$\sqrt{3} x = 20 \sqrt{3}$

$x = \frac{20 \sqrt{3} }{ \sqrt{3} } = 20$

Thus ,

AO = 20m

And ,

BO=CD = (60-x) = (60-20 ) = 40m

So , height of vertical lamp is 40m

(ii)Putting value of X in equation (iii) , we get ;

OC = √3x

OC= √3×20

OC= BD= 20√3m

(I)So , Distance between them is 20√3m

(iii) Distance between height of building and lamp is = (60-40) = 20m

Hope it helps you :)

Answer 2

Answer:

hope it helps you

Step-by-step explanation:

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