question_answer

The angles of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are \[15{}^\circ \] and \[30{}^\circ \] respectively. If A and B are on the same side of the tower and AB = 48 m, then the height of the tower is

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)] \[\Rightarrow \]   \[h=2\sqrt{3}h+96-3h-48\sqrt{3}\] \[\Rightarrow \]   \[4h-2\sqrt{3}h-96-48\sqrt{3}\] \[\Rightarrow \]   \[2h-\sqrt{3}h=48-24\sqrt{3}\] \[\Rightarrow \]   \[h\,(2-\sqrt{3})=48-24\sqrt{3}\] \[\Rightarrow \]   \[h=\frac{96+48\sqrt{3}\times (2+\sqrt{3})}{(2-\sqrt{3})\times (2+\sqrt{3})}\] \[\Rightarrow \]   \[h=\frac{96+48\sqrt{3}-48\sqrt{3}-72}{4-3}\] \[=96-72=24\,m\]