question_answer

From the top of a house 32 meters high, if the angle of elevation of the top of a tower is \[45{}^\circ \] and the angle of depression of the foot of the tower is \[30{}^\circ \], then the height of the tower is

A) \[\frac{32}{\sqrt{3}}(\sqrt{3}+1)\,\,meters\]

B) \[32(\sqrt{3}+1)\,\,meters\]

C) \[32\sqrt{3}\,\,meters\]

D) \[\frac{32}{3}(\sqrt{3}+1)\,\,meters\]

Correct Answer: B

Solution :

 From \[\Delta \,OBH,\] \[\frac{BH}{OB}=\cot \,{{30}^{o}}=\sqrt{3}\]                 or            \[BH=32\sqrt{3}\]                                 From \[\Delta \,BHT\]                 \[BT=BH=32\sqrt{3}\] \[\therefore \] Height of Tower                                 \[=OT=OB+BT=32(\sqrt{3}+1)\]