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19. What is the angle of elevation, tothe nearest tenth of a degree, tothe top of a 45-foot building from85 feet away?

User Manung
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Given data:

The given height of the building is H=45 ft.

The given distance is D=85 ft.

The expression for the angle of elevation is,


\begin{gathered} \tan \theta=(H)/(D) \\ \theta=\tan ^(-1)((45)/(85)) \\ =27.9^(\circ) \end{gathered}

Thus, the angle of elevation is 27.9 degrees.

User Javierfdr
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