A) 20 m
B) 22 m
C) 25 m
D) 30 m
Correct Answer: D
Solution :
Let AB be the tower of height h and CD be the observer of height 1.5 metre at a distance of 28.5 metre from the tower AB. In \[\Delta \]AED, \[\tan 45{}^\circ =\frac{AE}{DE}\] \[\Rightarrow \] \[1=\frac{AE}{DE}\] \[\Rightarrow \] \[AE=DE=28.5\] metre \[\therefore \]h=AE+BE=AE+DC \[=28.5+1.5=30\]metreYou need to login to perform this action.
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