Trigonometry Solutions
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A balloon is rising at the rate of 10 ft a second and is being carried horizontally by a wind which has a velocity of 15 miles an hour. Find the actual velocity and the angle that its path makes with the vertical.
\(15~mi/hr\) can be converted as \(22 ~ft /s\)
The actual velocity ,\(v\) of the wind is the hypotenuse of the triangle.
Thus, \(v^2 = 10^2+22^2\)
\(v = \sqrt{10^2+22^2}\)
\(v =24.17~ft/sec\)
The value of \(\theta\) can be calculated using \(\tan\).
Thus:
\(\tan \theta = \frac{22}{10} \)
\( \theta = \arctan {\frac{22}{10}} \)
\( \theta = 65.55^o \)