question_answer

There is a small island in the middle of a \[100\,\,m\]wide river. There is a tall tree on the island. Points P and Q are points directly opposite to each other on the two banks and in line with the tree. If the angles of elevation of the top of the tree at P and Q are \[{{30}^{o}}\]and\[{{45}^{o}}\], then the height of tree is:

A) \[50(\sqrt{3}-1)m\]                       

B) \[50\sqrt{3}m\]

C) \[50(\sqrt{3}+1)m\]       

D)        \[\frac{100}{\sqrt{3}-1}m\]

Correct Answer: A

Solution :

 Here height of tree\[=AB\] In\[\Delta APB\]             \[\tan {{30}^{o}}=\frac{AB}{BP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{x}\] or         \[x=\sqrt{3}AB\]                                    ? (i) In         \[\Delta AQB,\,\,\tan {{45}^{o}}=\frac{AB}{BQ}\] \[\Rightarrow \]   \[\frac{AB}{100-x}=1\] \[\Rightarrow \]   \[x=100-AB\]                             ? (ii) So from (i) and (ii)             \[\sqrt{3}AB=100-AB\] \[\Rightarrow \]   \[AB\left( \sqrt{3}+1 \right)=100\] \[\Rightarrow \]   \[AB=\frac{100}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\]             \[=50(\sqrt{3}-1)\] \[\therefore \]Height of tree             \[=50(\sqrt{3}-1)\]metre.