question_answer

A person observes the angle of depression of a building as \[{{30}^{o}}\]. The person proceeds towards the building with a speed of \[25(\sqrt{3}-1)m/h\]. After two hours, he observes the angle of elevation as \[{{45}^{o}}\]. The height of the building (in \[m)\] is:

A) \[50(\sqrt{3}-1)\]                            

B) \[50(\sqrt{3}+1)\]

C) \[50\]                                   

D) \[100\]

Correct Answer: C

Solution :

The distance covered by a person from \[A\] to\[B=25(\sqrt{3}-1)\times 2\]                 \[=50(\sqrt{3}-1)\] In\[\Delta DBC\],\[\tan {{45}^{o}}=\frac{DC}{BC}\] \[\Rightarrow \]               \[BC=h\] In\[\Delta DAC\],\[\tan {{30}^{o}}=\frac{h}{AB+BC}\] \[\Rightarrow \]               \[\frac{1}{\sqrt{3}}=\frac{h}{50(\sqrt{3}-1)+h}\] \[\Rightarrow \]               \[50(\sqrt{3}-1)+h=h\sqrt{3}\] \[\Rightarrow \]               \[h=50\,\,m\]