A) \[40\sqrt{2}\,m\]
B) \[50\sqrt{3}\,m\]
C) \[42\sqrt{3}\,m\]
D) \[41\sqrt{2}\,m\]
Correct Answer: A
Solution :
Let the length of second string be x m. In \[\Delta \,ABC,\] \[\sin {{30}^{o}}=\frac{AC}{AB}\] or \[\frac{1}{2}=\frac{AC}{100}\]\[\Rightarrow \] \[AC=50\,m\] In \[\Delta \,AEF,\] \[\sin {{45}^{o}}=\frac{AF}{AE}\Rightarrow \frac{1}{\sqrt{2}}=\frac{AC-FC}{x}\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}=\frac{50-10}{x}\] \[[\because \,\,AC=50m,\,\,FC=ED=10\,m]\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}=\frac{40}{x}\Rightarrow x=40\sqrt{2}m\] So, the length of string that the second boy must have so that the two kites meet \[=40\sqrt{2}\,m\]You need to login to perform this action.
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