question_answer

The angles of elevation of the top of a tower  from the top  and bottom  at a building of height a are\[{{30}^{o}}\]and\[{{45}^{o}}\]respectively. If the tower and the building stand at the same level, then the height of the tower is     [Karnataka CET 2000]

A) \[a\sqrt{3}\]

B) \[\frac{a\sqrt{3}}{\sqrt{3}-1}\]

C) \[\frac{a\,(3+\sqrt{3})}{2}\]

D) \[a\,(\sqrt{3}-1)\]

Correct Answer: C

Solution :

In \[\Delta ABC,\,\,\tan {{30}^{o}}=\frac{AC}{BC}\,\text{or }\frac{1}{\sqrt{3}}=\frac{x}{BC},\]where \[AC=x\,\] or \[\text{ }BC=x\sqrt{3}\] and in \[\Delta ADE\], \[\tan {{45}^{o}}=\frac{a+x}{DE}\] or \[1=\frac{a+x}{x\sqrt{3}}\] or \[x\sqrt{3}=a+x\],\[x(\sqrt{3}-1)=a\]or \[x=\frac{a}{\sqrt{3}-1}.\]  Therefore height of the tower, \[a+x=a+\frac{a}{\sqrt{3}-1}\] \[=a\left[ \frac{\sqrt{3}-1+1}{\sqrt{3}-1} \right]\] \[=\frac{a\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\]\[=\frac{a(3+\sqrt{3)}}{2}.\]