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, \[BL=PL\cot 30{}^\circ =(h-10)\sqrt{3}\] \[\Rightarrow \] \[\frac{h}{\sqrt{3}}=(h-10)\sqrt{3}\] \[\therefore \] \[h=15\]You need to login to perform this action.
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