question_answer

    The angle of elevation of the top of a tower at the top and the foot of a pole 10 m high are\[30{}^\circ \]and\[60{}^\circ ,\]respectively. The height of the tower is

A)  15m                                     

B)  20m

C)  \[10\sqrt{3}m\]              

D)  \[25\sqrt{3}m\]

Correct Answer: A

Solution :

                Let\[OP=h\]be the height of the tower, A, B be the foot and the top of the pole AB of height 10 m. Then\[\angle PAO=60{}^\circ \]and\[\angle PBL=30{}^\circ \]. So, that \[BL=AO=OP\cot 60{}^\circ =\frac{h}{\sqrt{3}}\] Also,      \[BL=PL\cot 30{}^\circ =(h-10)\sqrt{3}\] \[\Rightarrow \]               \[\frac{h}{\sqrt{3}}=(h-10)\sqrt{3}\] \[\therefore \]  \[h=15\]