A) \[100\sqrt{3}m\]
B) \[\frac{100}{\sqrt{3}}m\]
C) \[50\,m\]
D) \[150(\sqrt{3}+1)m\]
Correct Answer: A
Solution :
Given \[B{{P}_{2}}=300m\] In \[\Delta AB{{P}_{2}}\] \[\frac{B{{P}_{2}}}{AB}=\tan {{60}^{o}}=\sqrt{3}\] \[\therefore \]\[AB=\frac{300}{\sqrt{3}}=100\sqrt{3}m\] Also in\[\Delta ABP1\] \[\frac{{{P}_{1}}B}{AB}=\tan {{45}^{o}}=1\] \[\therefore \]\[PB=100\sqrt{3}m\]You need to login to perform this action.
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