Question

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

Answer 1

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}`