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  • 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° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and

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° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.

Answers (1)

Solution.         According to the question
                        
Here 30 m is the length of tower AB.
Let h is the height of tower DC
Let the distance between them is x
In \bigtriangleupABC
\tan 60^{\circ}= \frac{30}{x}     \left [ \tan \theta = \frac{Perpendicular}{Base} \right ]
\sqrt{3}= \frac{30}{x}
x= \frac{30}{\sqrt{3}}\cdots \left ( 1 \right )
In \bigtriangleupBDC
\tan 30^{\circ}= \frac{h}{x}
\frac{1}{\sqrt{3}}= \frac{h}{30}\times \sqrt{3}    (using (1))
\frac{30}{\sqrt{3}}=h\sqrt{3}
\frac{30}{\sqrt{3}\times\sqrt{3} }= h
\frac{30}{3}= h
h = 10m
Hence the height of the second tower is 10
Distance between them

=\frac{30}{\sqrt{3}}m .

Posted by

infoexpert27

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