A) 120
B) \[60\sqrt{3}\]
C) \[120\sqrt{3}\]
D) \[60\]
Correct Answer: B
Solution :
Let h be the height of the object. In \[\Delta CAD\] \[\tan {{30}^{o}}=\frac{CD}{AC}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{h}{120+x}\] \[\Rightarrow \] \[\sqrt{3}h=120+x\] ?(i) and in \[\Delta CBD\] \[\tan {{60}^{o}}=\frac{CD}{BC}\] \[\Rightarrow \] \[\sqrt{3}=\frac{h}{x}\] \[\Rightarrow \] \[h=\sqrt{3}x\] ?(ii) From Eqs (i) and (ii),we get \[3x=120+x\] \[\Rightarrow \] \[x=60\,m\] On putting \[x=60\] in Eq. (i), we get Height of the object \[=60\sqrt{3}m\]You need to login to perform this action.
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