question_answer

Two posts are \[k\]metre apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in metre) of the shorter post is

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             \[\frac{4h}{k}\times \frac{2h}{k}=1\] \[\Rightarrow \]   \[8{{h}^{2}}={{k}^{2}}\] \[\Rightarrow \]   \[{{h}^{2}}=\frac{{{k}^{2}}}{8}\] \[\Rightarrow \]   \[h=\frac{k}{2\sqrt{2}}m\]