Answer:
Approximately [tex]105\; \rm ft[/tex] (assuming that the street is level.)
Step-by-step explanation:
Refer to the diagram attached. Consider the height of this building as the sum of two parts: the part below the window of the observer, and the part above the window of the observer.
The question states that the window of the observer is [tex]33\; \rm ft[/tex] above the street. Assume that the street is level. The height of the part of that building below the window of this observer would also be [tex]33\; \rm ft\![/tex].
Make use of the angle [tex]18^{\circ}[/tex] to calculate the horizontal distance between that building and the window of the observer.
Refer to the diagram attached. There are two right triangles in this diagram. Consider the one on the bottom (the one with the angle that measures [tex]18^{\circ}[/tex].) The two sides of this right triangle adjacent to the right angle correspond to:
The angle that measures [tex]18^{\circ}[/tex] is adjacent to the side that corresponds to the horizontal distance between the two buildings. The tangent of this angle would relate the lengths of the two sides:
[tex]\displaystyle \tan(18^{\circ}) = \frac{33\; \rm ft}{\text{horizontal distance}} \quad \genfrac{}{}{0}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}[/tex].
Hence, the horizontal distance between the two buildings would be:
[tex]\begin{aligned}\text{horizontal distance} = \frac{33\; \rm ft}{\tan(18^{\circ})}\end{aligned}[/tex].
In the other right triangle in this diagram (the one with the angle that measures [tex]45^{\circ}[/tex], the two sides adjacent to the right angle correspond to:
Again, the tangent of the [tex]45^{\circ}[/tex] angle would correspond to:
[tex]\displaystyle \tan(45^{\circ}) = \frac{\text{height of opposite building above the window}}{\text{horizontal distance}} \quad \genfrac{}{}{0}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}[/tex].
Rearrange to find the height of the building above the window of the observer:
[tex]\begin{aligned}& \text{height of opposite building above the window} \\ =\; & (\tan(45^{\circ})) \, (\text{horizontal distance}) \\ =\; & \tan(45^{\circ}) \cdot \frac{33\; \rm ft}{\tan(18^{\circ})} \\ \approx\; &102\; \rm ft\end{aligned}[/tex].
The height of the opposite building would be the sum of the two parts, which is approximately:
[tex](102 + 33) \; \rm ft = 135\; \rm ft[/tex].
