A) \[\frac{40}{\sqrt{3}}m\]
B) \[20\sqrt{3}m\]
C) \[40\sqrt{3}m\]
D) 30m
Correct Answer: B
Solution :
Let MN be the telegraph post and it gets broken at A in such a way that its top M touches the ground and thus makes an angle of 30° with the ground at P. \[\Rightarrow \] MA = AP In \[\Delta ANP\] \[\tan 30{}^\circ =\frac{AN}{NP}\,\,;\frac{1}{\sqrt{3}}=\frac{AN}{20}\] \[\Rightarrow AN=\frac{20}{\sqrt{3}}m\] Similarly, \[cos\text{ }30{}^\circ =\frac{NP}{AP}\] \[\frac{\sqrt{3}}{2}=\frac{20}{AP}\Rightarrow =\frac{40}{\sqrt{3}}m=MA\] Height of the pole = AN + MA \[\frac{20}{\sqrt{3}}+\frac{40}{\sqrt{3}}=\]You need to login to perform this action.
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