## Calculus: Early Transcendentals 8th Edition

Based in the figure for problem 5 in part d: $a= 24 sin(θ)$ $b = 24 cos(θ)$
Considering: $cos(θ)= \frac{Adjacent leg}{Hypotenuse}$ $sin(θ)= \frac{Opposite leg}{Hypotenuse}$ then 24= hypotenuse a= Opposite Leg b= Adjacent Leg