question_answer

From a point on the ground, the angle of elevation of the top of a tower is observed to be \[60{}^\circ \]. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is \[30{}^\circ \]. Find the height of the tower and its horizontal distance from the point of observation.

Answer:

We have, PQ as a vertical tower.

Now, in \[\Delta \,YZQ\]

\[\tan \,30{}^\circ =\frac{QZ}{YZ}\]

\[\frac{1}{\sqrt{3}}=\frac{QZ}{YZ}\]

\[YZ=QZ\sqrt{3}\]                                        ?(i)

And, in \[\Delta \,XPQ\]

\[\tan \,60{}^\circ =\frac{QP}{XP}\]

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

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

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

\[3QZ=QZ+40\]

\[2QZ=40\]

\[QZ=20\]

\[\therefore \] Height of tower \[=(40+20)m=60\,m\]

and Horizontal, distance \[=QZ\sqrt{3}=20\sqrt{3}\,m\]