## Algebra 2 (1st Edition)

$\theta =\tan^{-1} (\dfrac{44t}{100})$ or, $\tan^{-1} (\dfrac{11t}{25})$
Remember: $30 mph= 44 ft/sec$ Now, $\tan \theta= 44t \times \dfrac{1}{100}=\dfrac{44t}{100}$ Re-write as: $\theta = \tan^{-1} (\dfrac{44t}{100})$ This implies that $\theta =\tan^{-1} (\dfrac{44t}{100})$ or, $\tan^{-1} (\dfrac{11t}{25})$