question_answer

A man observes the elevation of a tower to be \[30{}^\circ \]. After advancing 11 cm towards it, he finds that the elevation is \[45{}^\circ \]. The height of the tower to the nearest metre is

A)  10                         

B)         15  

C)  20                         

D)         22

Correct Answer: B

Solution :

 Since     \[\angle OQT={{45}^{o}}\] therefore   \[OT=OQ=h\,(say)\] From  \[\Delta OPT,\,\frac{OP}{OT}=\cot \,{{30}^{o}}\] or            \[\frac{h+11}{h}=\sqrt{3}\] or            \[h=\frac{11}{\sqrt{3}-1}=\frac{11}{2}(\sqrt{3}+1)\] or             \[h=1.366\times 11\]                 \[=15.026=15\]