question_answer

An observer 1.5 metre tall is 28.5 metre away from a tower. The angle of elevation of the top of the tower from her eyes is \[45{}^\circ \]. What is the height of the tower?

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 \[=28.5+1.5=30\]metre