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An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

Options:
A .  21.6 m
B .  23.2 m
C .  24.72 m
D .  None of these
Answer: Option A

Let AB be the observer and CD be the tower


An Observer 1.6 M Tall Is 203 away From A Tower. The Angle ...


\(Draw BE\bot CD.\)


Then, CE = AB = 1.6 m,


      BE = AC = 203 m.


\(\frac{DE}{BE}=\tan30^{0}=\frac{1}{3}\)


\(\Rightarrow DE = \frac{203}{3}m=20m.\)


Therefore CD = CE + DE = (1.6 + 20) m = 21.6 m.


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