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pav-90 [236]
3 years ago
10

An observer in a hot air balloon sights a building that is 50 m from the balloon's launch point. The balloon has risen 165 m. Wh

at is the angle of depression from the balloon to the building? Round to the nearest degree

Mathematics
1 answer:
Kaylis [27]3 years ago
4 0
Notice the picture,
recall your SOH, CAH, TOA
\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}

you have,
opposite side, 165
adjacent side, 50
and the angle

that means, we'll need Mrs. tangent
thus 
\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50}
\\ \quad \\
 tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right]
\\ \quad \\
\theta=tan^{-1}\left[ \cfrac{165}{50}\right]
\\ \uparrow  \\
 \textit{angle of elevation}\iff\textit{angle of depression}

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