Question : The angle of elevation of the top of a tower from the top of a building whose height is 680 m is $45^{\circ}$ and the angle of elevation of the top of the same tower from the foot of the same building is $60^{\circ}$. What is the height (in m) of the tower?
Option 1: $340(3 + \sqrt3)$
Option 2: $310(3 - \sqrt3)$
Option 3: $310(3 + \sqrt3)$
Option 4: $340(3 - \sqrt3)$
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Correct Answer: $340(3 + \sqrt3)$
Solution :
Let AB be the building and CD be the tower.
AB = DE = 680 m
$\tan 45^{\circ}$ = $\frac{CE}{AE}$
⇒ 1 = $\frac{CE}{BD}$
⇒ CE = BD ------------(i)
$\tan 60^{\circ}$ = $\frac{CD}{BD}$
⇒ $\sqrt3$ = $\frac{CE+DE}{CE}$
⇒ $\sqrt3$ × CE = CE + DE
⇒ CE ($\sqrt3-1$) = 680
⇒ CE = $\frac{680}{\sqrt3-1}$
⇒ CE = $\frac{680(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$
⇒ CE = $340(\sqrt3+1)$
CD = DE + CE
= $680 + 340(\sqrt3+1)$
= $340(\sqrt3+3)$
Hence, the correct answer is $340(\sqrt3+3)$.
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