A) \[\frac{10}{\sqrt{3}}\]m
B) \[10\sqrt{3}\] m
C) \[\frac{20}{\sqrt{3}}\] m
D) 30 m
Correct Answer: D
Solution :
Let RQ be the height of building, then RQ = 10 m. S be the position of helicopter. Then In \[\Delta \] PQR, \[\frac{RQ}{PQ}=\tan 30{}^\circ \] \[\Rightarrow \,\,\,PQ=\frac{RQ}{\tan \,30{}^\circ }=10\sqrt{3}\] \[\therefore \] In \[\Delta \,\,\,SPQ,\,\,\tan \,60{}^\circ =\frac{SQ}{PQ}\] \[\Rightarrow \,\,\frac{SQ}{PQ}=\sqrt{3}\] \[\Rightarrow \,\,SQ=PQ\times \sqrt{3}\] \[\Rightarrow 10\sqrt{3}\times \sqrt{3}=30\] mYou need to login to perform this action.
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