# From the foot of a hill the angle of elevation of the top of a tower is found to be 45o . After walking 2 km upwards along the slope of the hill which is inclined at30o , the same is found to be 60o . Find the height of the tower.

Let ,
We have height of the tower = h

From the image =

tan 45$°$

1 =                                            ( As we know tan 45$°$ =  1 )

h  + h '  =  b1 + b2                                                      ------------------ ( 1 )

And

cos 30$°$  = $\frac{{b}_{1}}{2}$

$\frac{\sqrt{3}}{2}$ = $\frac{{b}_{1}}{2}$                                                                           ( As we know cos30$°$  = $\frac{\sqrt{3}}{2}$ )

b1 = $\sqrt{3}$                                                     ------------------ ( 2 )

And
sin 30$°$  = $\frac{h\text{'}}{2}$

$\frac{1}{2}$  =$\frac{h\text{'}}{2}$                                                                                                     ( As we know sin 30$°$  =  $\frac{1}{2}$  )

SO,

h'  = 1                                                    ------------------ ( 3 )

And
tan 60$°$  = $\frac{h}{{b}_{2}}$

$\sqrt{3}$    = $\frac{h}{{b}_{2}}$

h  =  $\sqrt{3}$  b2                                                     ------------------ ( 4 )

from equation 1 , 2 , 3  and 4 , we get

h  = $\sqrt{3}$  + $\frac{h}{\sqrt{3}}$ -  1

$\frac{h}{\sqrt{3}}$= $\sqrt{3}$  -  1

= $\sqrt{3}$  -  1

$\sqrt{3}$ h - h  =  3 - $\sqrt{3}$

$\sqrt{3}$ h - h  =  $\sqrt{3}$ ( $\sqrt{3}$ - 1 )

h ( $\sqrt{3}$  -  1 ) = $\sqrt{3}$  ( $\sqrt{3}$  - 1 )

h  =  $\sqrt{3}$    = 1.732 km                                                                              ( Ans )

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