question_answer

The angle of elevation of a tower at a point is \[45{}^\circ .\] After going 40 m towards the foot of the tower, the angle of elevation of the tower becomes \[60{}^\circ .\] Then, the height of the tower is

A)  \[20(\sqrt{3}+1)\,\,m\]

B)         \[\frac{20\sqrt{3}}{\sqrt{3}+1}\,\,m\]

C)  \[\frac{40\sqrt{3}}{\sqrt{3}+1}\,\,m\]                

D)  \[\frac{40\sqrt{3}}{\sqrt{3}-1}\,\,m\]

Correct Answer: D

Solution :

Let PQ (= h) be the height of the tower. Let S and R be the points where the angles subtended are \[45{}^\circ \]and \[60{}^\circ .\]

In \[\Delta PQR,\]

\[\tan 60{}^\circ =\frac{PQ}{RQ}\]

\[\Rightarrow \]   \[\sqrt{3}=\frac{h}{x}\]

\[\Rightarrow \]   \[x=\frac{h}{\sqrt{3}}\]                         ?(i)

In \[\Delta PQS,\]

\[\tan 45{}^\circ =\frac{PO}{SQ}=\frac{h}{x+40}\]

\[\Rightarrow \]   \[h=40+x\]

\[\Rightarrow \]   \[h=40+\frac{h}{\sqrt{3}}\]                   [from Eq. (i)]

\[h\,\,\left( 1-\frac{1}{\sqrt{3}} \right)=40\]

\[\Rightarrow \]   \[h\,\,\left( \frac{\sqrt{3}-1}{\sqrt{3}}=40 \right)\]\[\Rightarrow \]\[h=\frac{40\sqrt{3}}{\sqrt{3}-1}\,\,m\]