A) \[\frac{20}{\sqrt{3}}m\]
B) \[30\sqrt{3}m\]
C) \[20\sqrt{3}m\]
D) \[\frac{30}{\sqrt{3}}m\]
Correct Answer: C
Solution :
(c) In \[\Delta ABF,\,\,\]\[\tan 60{}^\circ =\frac{30}{x}\] \[\therefore \]\[x=\frac{30}{\tan 60{}^\circ }=\frac{30}{\sqrt{3}}\] In \[\Delta ABE,\]\[\tan 30{}^\circ =\frac{30}{x+y}\] \[\therefore \] \[x+y=\frac{30}{\tan 30{}^\circ }\] \[\Rightarrow \] \[x+y=30\sqrt{3}\] \[\Rightarrow \] \[\frac{30}{\sqrt{3}}+y=30\sqrt{3}\] \[\Rightarrow \] \[y=30\sqrt{3}-\frac{30}{\sqrt{3}}=\frac{60}{\sqrt{3}}=20\sqrt{3}\]You need to login to perform this action.
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