• # 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 $\sqrt{3}=1.73$)

 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 $PQ=(54.6+40)m=94.6\,m$