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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