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