The angles of elevation of the top of a tower | vedclass

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(C) Let the height of the tower be $H = a + x$,where $x$ is the height of the top of the tower above the building level.In $\Delta ABC$,$	an{30^\circ

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$\frac{a(3+\sqrt{3})}{2}$

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(C) Let the height of the tower be $H = a + x$,where $x$ is the height of the top of the tower above the building level.In $\Delta ABC$,$	an{30^\circ} = \frac{AC}{BC} = \frac{x}{BC}$. Since $	an{30^\circ} = \frac{1}{\sqrt{3}}$,we have $BC = x\sqrt{3}$.In $\Delta ADE$,$	an{45^\circ} = \frac{AE}{DE} = \frac{a+x}{DE}$. Since $	an{45^\circ} = 1$ and $DE = BC = x\sqrt{3}$,we have $1 = \frac{a+x}{x\sqrt{3}}$.Thus,$x\sqrt{3} = a + x$,which implies $x(\sqrt{3}-1) = a$,so $x = \frac{a}{\sqrt{3}-1}$.The total height of the tower is $H = a + x = a + \frac{a}{\sqrt{3}-1} = a \left( 1 + \frac{1}{\sqrt{3}-1} ight) = a \left( \frac{\sqrt{3}-1+1}{\sqrt{3}-1} ight) = \frac{a\sqrt{3}}{\sqrt{3}-1}$.Rationalizing the denominator: $\frac{a\sqrt{3}}{\sqrt{3}-1} 	imes \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{a(3+\sqrt{3})}{3-1} = \frac{a(3+\sqrt{3})}{2}$.

The angles of elevation of the top of a tower $A$ from the top $B$ and bottom $D$ of a building of height $a$ are $30^\circ$ and $45^\circ$ respectively. If the tower and the building stand at the same level,then the height of the tower is

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