question_answer

There is a small island in the middle of a 100 m wide river. There is 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 \[30{}^\circ \]and \[45{}^\circ .\]Then, the height of tree is

A)  \[50\,\,(\sqrt{3}-1)\,\,m\]

B)  \[50\sqrt{3}\,\,\text{m}\]

C)  \[50\,\,(\sqrt{3}+1)\,\,\text{m}\]

D)  \[=(18\times 15\times 12)c{{m}^{3}}\]

Correct Answer: A

Solution :

Here, height of tree = AB

In \[\Delta APB,\]

\[\tan 30{}^\circ =\frac{AB}{BP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{x}\]   ?(i)

In \[\Delta AQB,\]\[\tan 45{}^\circ =\frac{AB}{BQ}\]

\[\frac{AB}{100-x}=1\]

\[\Rightarrow \]               \[x=100-AB\]                 ?(ii)

So, from Eqs. (i) and (ii)

\[\sqrt{3}AB=100-AB\]

\[AB=\frac{100}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=50(\sqrt{3}-1)\]

\[\therefore \] Height of tree \[=50\,\,(\sqrt{3}-1)\,\,\text{m}\]