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Home  >>  EAMCET  >>  Mathematics
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A person observes the top of a tower from a point A on the ground. The elevation of the tower from this point is $60^{\circ}$. He moves $60\;m$ in the direction perpendicular to the line joining A and base of the tower. The angle of elevation of the tower from this point is $45^{\circ}.$ Then the height of the tower (in meters) is

\[(a)\;60 \sqrt {\frac{3}{2}} \quad (b)\;60 \sqrt{2} \quad (c)\;60 \sqrt{3}\quad (d)\;60 \sqrt {\frac{2}{3}}\]

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1 Answer

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$(a)\;60 \sqrt {\frac{3}{2}}$
answered Nov 7, 2013 by pady_1
 

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