Question

please answer this question it's urgent.

Answer 1

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Here is your answer in the attachment.

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Answer 2

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$<b><i><font face=Copper black size=4 color=red><h3>$
Here is your answer.
Let the distance between foot of tower and building is x m
and Height of the building is h m

Now
In I case
tan 60° = Perpendicular/Base
$\sqrt{3} = \frac{50}{x} \\ x = \frac{50}{ \sqrt{3} }$
In II Case

tan 30° = Perpendicular/Base
$\frac{1}{ \sqrt{3} } = \frac{h}{x} \\ x = \sqrt{3} \times h$
As x is same so From above two case we get

$\sqrt{3} \times h = \frac{50}{ \sqrt{3} } \\ \\ h = \frac{50}{ \sqrt{3} \times \sqrt{3} } \\ \\ h = \frac{50}{3} \\ \\ h = 16.66 \: m$
Therefore
Height of the building is 16.66 m

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