question_answer

If the angle of elevation of an object from a point 200 meters above the lake is found to be \[30{}^\circ \] and the angle of depression of its image in the lake is \[45{}^\circ \], then the height of the object above the lake is

A) \[\frac{200(\sqrt{3}-1)}{(\sqrt{3}+1)}\,\,meters\]

B) \[\frac{200(\sqrt{3}-1)}{\sqrt{3}}\,\,meters\]

C) \[\frac{200(\sqrt{3}+1)}{\sqrt{3}}\,\,meters\]

D)   \[\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)}\,\,meters\]  

Correct Answer: D

Solution :

 Let \[OL=h,\] therefore \[LO'=H\] From \[\Delta \,PQO',\,\,PQ=QO'=QL+LO'\]                 \[=200+h\,\,\,[OL=LO'=h]\] From \[\Delta \,PQO,\,\frac{OQ}{PQ}=\tan {{30}^{o}}\] or         \[OQ=PQ\text{ }tan\text{ }{{30}^{o}}\] or            \[H-200=(200+h)\frac{1}{\sqrt{3}}\] \[\therefore \]  \[h=\frac{200\left( \sqrt{3}+1 \right)}{\sqrt{3}-1}\]