A) \[20\text{ }m\]
B) \[30\text{ }m\]
C) \[40\text{ }m\]
D) \[50\text{ }m\]
Correct Answer: C
Solution :
Let the height of tower \[AD=a\text{ }m\] From ACED, \[\tan \,30{}^\circ =\frac{60-a}{x}\] \[x=\sqrt{3}\,\,(60-a)\] ...(i) and from \[\Delta ABC\,\,\tan 60{}^\circ =\frac{60}{x}\] \[x=\frac{60}{\sqrt{3}}\] ..?(ii) Equating equation (i) and (ii), we get \[\frac{60\sqrt{3}}{\sqrt{3}}=\sqrt{3\,\,(60-a)}\] \[60=\sqrt{3}\times \sqrt{3}\,(60-a)=180-3a\] \[3a=180-60=120=\]You need to login to perform this action.
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