A) \[k/2\]
B) \[\frac{k}{2\sqrt{2}}\]
C) \[k\]
D) None of these
Correct Answer: B
Solution :
Let \[PQ\] and \[RS\] be the two posts, such that\[PQ=2RS\]. If \[M\] is the midpoint of\[RP\] \[RM=PM=\frac{k}{2}\] \[\because \] \[\angle RMS=\theta \] and \[\angle QMP={{90}^{o}}-\theta \] Let\[RS=h\], then\[PQ=2h\] Now, in\[\Delta PMQ\], \[\tan ({{90}^{o}}-\theta )=\frac{PQ}{MP}\] \[\frac{PQ}{MP}=\cot \theta \Rightarrow \frac{2h}{k/2}=\cot \theta \] or \[\cot \theta =\frac{4h}{k}\] ... (i) In\[\Delta SRM\], \[\tan \theta =\frac{SR}{RM}\] \[\Rightarrow \] \[\frac{h}{k/2}=\tan \theta \] \[\Rightarrow \] \[\frac{2h}{k}=\tan \theta \] ... (ii) Multiplying Eq. (i) by Eq. (11), we get \[\frac{4h}{k}\times \frac{2h}{k}=1\] \[\Rightarrow \] \[8{{h}^{2}}={{k}^{2}}\] \[\Rightarrow \] \[{{h}^{2}}=\frac{{{k}^{2}}}{8}\] \[\Rightarrow \] \[h=\frac{k}{2\sqrt{2}}m\]You need to login to perform this action.
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