question_answer

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is \[60{}^\circ \]. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is \[45{}^\circ \]. Find the height of the tower PQ and the distance PX (Use \[\sqrt{3}=1.73\])

Answer:

We have, PQ as a vertical tower

In \[\Delta \,YZQ\]                       \[tan\text{ }45{}^\circ =\frac{QZ}{YZ}\]

\[\frac{QZ}{YZ}=1\]

\[QZ=YZ\]                           ?(i)

And, in \[\Delta \text{ }XPQ\]              \[tan60{}^\circ =\frac{QP}{XP}\]

\[\sqrt{3}=\frac{QZ+40}{XP}\]

\[\sqrt{3}=\frac{QZ+40}{YZ}\]                    \[(\because \,XP=YZ)\]

\[\sqrt{3}\,QZ=QZ+40\]              [Using (i)]

\[\sqrt{3}\,QZ-QZ=40\]

\[QZ\left( \sqrt{3}-1 \right)=40\]

\[QZ=\frac{40}{\sqrt{3}-1}=\frac{40}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\]

\[=20\left( \sqrt{3}+1 \right)\]

\[=20(2.73)\]

\[=54.60\,m\]

\[\therefore PX=54.6\,m\]

And \[PQ=(54.6+40)m=94.6\,m\]