question_answer

A man standing on a level plane observes the elevation of the top of a pole to be \[\theta \]. He then walks a distance equal to double the height of the pole and then finds that the elevation is now \[2\theta \]. Then \[\cot \theta \]is equal to

A) \[\sqrt{2}+1\]              

B)        \[2-\sqrt{3}\]               

C)  \[\sqrt{2}-1\]               

D)         \[2+\sqrt{3}\]

Correct Answer: B

Solution :

[b] In \[\Delta ABC,\] \[x=h\,\,\cot 2\theta \]                          ...(i) In \[\Delta ABD,\] \[h=(2h+x)\,\tan \theta \]                       ...(ii) Putting value of x from (i) into (ii), we get \[1=\left( 2+\frac{1}{\tan 2\theta } \right)\tan \theta \] \[\Rightarrow \,\,\,\,\,\,\,\,\,1=\left( 2+\frac{1-{{\tan }^{2}}\theta }{2\tan \theta } \right)\tan \theta \] \[\therefore \,\,\,\,\,{{\tan }^{2}}\theta -4\tan \theta +1=0\] \[\therefore \,\,\,\,\,\tan \theta =2-\sqrt{3}\] (as \[\tan \theta =2+\sqrt{3}\] not possible)