A flag-staff 5 m high stands on a building o | vedclass

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(D) Let the observer be at point $O$ at a height of $30 \ m$. The building has a height of $25 \ m$ and the flag-staff has a height of $5 \ m$,so the

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none of these

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(D) Let the observer be at point $O$ at a height of $30 \ m$. The building has a height of $25 \ m$ and the flag-staff has a height of $5 \ m$,so the total height of the building plus the flag-staff is $30 \ m$. Let the horizontal distance between the observer and the building be $x$.Let the angle subtended by the flag-staff at the observer be $\alpha$ and the angle subtended by the building at the observer be $\alpha$.From the geometry,the angle subtended by the building is $\alpha$ and the angle subtended by the flag-staff is also $\alpha$. Thus,the total angle subtended by the building and the flag-staff together is $2\alpha$.In the right-angled triangle formed by the observer and the top of the building,$	an \alpha = \frac{5}{x}$ (since the flag-staff is $5 \ m$ high).In the right-angled triangle formed by the observer and the base of the building,$	an 2\alpha = \frac{30}{x}$.Using $	an 2\alpha = \frac{2	an \alpha}{1 - 	an^2 \alpha}$,we have $\frac{30}{x} = \frac{2(5/x)}{1 - (5/x)^2} = \frac{10/x}{(x^2 - 25)/x^2} = \frac{10x}{x^2 - 25}$.So,$30 = \frac{10x^2}{x^2 - 25} \implies 3(x^2 - 25) = x^2 \implies 3x^2 - 75 = x^2 \implies 2x^2 = 75 \implies x^2 = 37.5$.$x = \sqrt{37.5} = \sqrt{\frac{75}{2}} = 5\sqrt{\frac{3}{2}}$.The distance of the observer from the top of the flag-staff is the hypotenuse of the triangle with base $x$ and height $5$,which is $\sqrt{x^2 + 5^2} = \sqrt{37.5 + 25} = \sqrt{62.5} = \sqrt{\frac{125}{2}} = 5\sqrt{\frac{5}{2}}$.Since this value is not among the options,the correct answer is $(D)$.

$A$ flag-staff $5 \ m$ high stands on a building of height $25 \ m$. An observer at a height of $30 \ m$ observes that the flag-staff and the building subtend equal angles. The distance of the observer from the top of the flag-staff is:

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