A tower stands vertically on the ground. From a point on the ground which is 25 in away in the foot of the tower, the angle of the elevation of the top of the tower is found to be $45^{o}$. Then the height $(in\ meter)$ of the tower is:
$( A) \ 25\sqrt{2}$
$( B) \ 25\sqrt{3}$
$( C) \ 25$
$( D) \ 12.5$
Given: Here a tower is given, from a point its distance to the foot of the tower $25 \ m$, and from the same point the angle of the elevation of the top of the tower $45^{o}$.
To do: To find the height of the tower.
Solution: Let us draw a figure
And let the tower is BC. And from point A distance of the foot of the tower $AB =25 \ m$
As given angle of elevation of the top of the tower from point A, $\angle BAC=45^{o}$
In $\vartriangle ABC$,
$tan45^{o} =\frac{BC}{AB} =\frac{BC}{25}$
$\frac{BC}{25} =1$ $( on\ substituting\ tan45^{o} =1)$
Or
$BC=25\ m$
$\therefore$ Option $( C)$ is correct.
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