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Answer :
Answer:
Approximately [tex]2.8\; \text{m}[/tex].
Step-by-step explanation:
Refer to the diagram attached. Let [tex]{\sf E}[/tex] denote the eyes of the observer, and let [tex]{\sf L}[/tex] denote the point where the building is at the same height as [tex]{\sf E}[/tex]. [tex]m \angle {\sf ELA} = m \angle {\sf ELB} = 90^{\circ}[/tex].
It is given that [tex]{\sf EL} = 20\; \text{m}[/tex], [tex]m\angle {\sf AEL} = 47^{\circ}[/tex], and [tex]m\angle {\sf BEL} = 43^{\circ}[/tex].
In right triangle [tex]\triangle {\sf AEL}[/tex], the angle [tex]m\angle {\sf AEL} = 47^{\circ}[/tex] is opposite to the leg [tex]{\sf AL}[/tex], and adjacent to the leg [tex]{\sf EL} = 20\; \text{m}[/tex]. Hence, the length of the [tex]{\sf AL}[/tex] can be found using [tex]\tan \angle {\sf AEL}[/tex]:
[tex]\displaystyle \tan \angle {\sf AEL} = \frac{(\text{opposite})}{(\text{adjacent})} = \frac{(\textsf{AL})}{(\textsf{EL})}[/tex].
[tex](\textsf{AL}) = (\textsf{EL})\, \tan \angle \textsf{AEL} = (20\; \text{m})\, \tan(47^{\circ})[/tex].
Similarly, in right triangle [tex]\triangle \textsf{BEL}[/tex]:
[tex](\textsf{BL}) = (\textsf{EL})\, \tan \angle \textsf{BEL} = (20\; \text{m})\, \tan(43^{\circ})[/tex].
The height of the antenna, [tex]\textsf{AB}[/tex], would be:
[tex](\textsf{AB}) = (\textsf{AL}) - (\textsf{BL}) = (20\; \text{m})\, \left(\tan(47^{\circ}) - \tan(43^{\circ})\right) \approx 2.8\; \text{m}[/tex].
Note that the height of the eyes does not matter. The reason is that both the height of the rooftop and the height of the antenna are relative to the height of the eyes. When subtracting, the height of the eyes is eliminated from the equation.
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