A) \[24\sqrt{3}\text{ }m\]
B) \[24\,m\]
C) \[24\sqrt{2}\,m\]
D) \[96\,m\]
Correct Answer: B
Solution :
[b] Suppose height of the tower\[=h\,m\] In \[\Delta PBC,\] \[\tan 30{}^\circ =\frac{h}{x}\] \[\Rightarrow \] \[1\sqrt{3}=\frac{h}{x}\]\[\Rightarrow \]\[x=\sqrt{3}\,h\] (i) In \[\Delta PAC,\] \[\tan 15{}^\circ =\frac{h}{x+48}\] \[(\because tan15{}^\circ =2-\sqrt{3})\] \[\Rightarrow \] \[2-\sqrt{3}=\frac{h}{\sqrt{3}h+48}\] [from Eq.(i)] \[\Rightarrow \] \[h=2\sqrt{3}h+96-3h-48\sqrt{3}\] \[\Rightarrow \] \[4h-2\sqrt{3}h-96-48\sqrt{3}\] \[\Rightarrow \] \[2h-\sqrt{3}h=48-24\sqrt{3}\] \[\Rightarrow \] \[h\,(2-\sqrt{3})=48-24\sqrt{3}\] \[\Rightarrow \] \[h=\frac{96+48\sqrt{3}\times (2+\sqrt{3})}{(2-\sqrt{3})\times (2+\sqrt{3})}\] \[\Rightarrow \] \[h=\frac{96+48\sqrt{3}-48\sqrt{3}-72}{4-3}\] \[=96-72=24\,m\] |
You need to login to perform this action.
You will be redirected in
3 sec