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Question

The angle of elevation of the top of a tower $$ 30\,m $$ high from the foot of another tower in the same plane is $$ 60^{\circ} $$ , and the angle of elevation of the top of second  tower from the foot of the first tower is $$ 30^{\circ} $$ . Find the distance between the two towers and also the height of the other tower.


Solution

Two vertical towers $$ TW = 30 \,m $$ and $$ ER = x \,m $$ (let) are standing on a horizontal plane $$ RW = y $$ (let) . The angle of elevation from R to top $$ 90\,m $$ high tower is $$ 60^{\circ} $$ and the angle of elevation of second tower from $$ W $$ is $$ 30^{\circ} $$ . 
In $$ \Delta \, ERW$$ ,
$$ tan \, 30^{\circ} = \dfrac{x}{y} $$ 
$$ \Rightarrow \, \dfrac{1}{\sqrt{3}} = \dfrac{x}{y} $$ 
$$ \Rightarrow \, y = \sqrt{3} x \, (i) $$ 
Now , In $$ \Delta \, TWR $$ , 
$$ tan \, 60^{\circ} = \dfrac{30^{\circ}}{y} \, (from\ i) $$ 
$$ \Rightarrow \, \sqrt{3} = \dfrac{30}{\sqrt{3}x} $$  [From $$(i)$$] 
$$ \Rightarrow \, 3x = 30 $$ 
$$ \Rightarrow \, x = 10\,m $$  
Now , $$ y = \sqrt{3}x $$ 
$$ \Rightarrow \, y = \sqrt{3}(10) $$ 
$$ \Rightarrow \, y = 1.732 \times 10 $$ 
$$ \Rightarrow \, y = 17.32 \, m $$ 
Hence , the distance between the two towers is $$ 17.32 \,m $$ and the height of the second tower is $$ 10\,m $$. 
1765151_1796671_ans_bb1d9b9d415e456d95bde8db1d4b816a.png

Mathematics

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