Question

1. a small plane takes off from an airport and begins to climb at a 14.5° angle of elevation at 725ft/min. after 20.0 min, how far along the ground will the plane have flown? 2 marks

Explanation:

Step1: Calculate total vertical climb

The plane climbs at a rate of 725 ft/min for 20.0 min. So the total vertical climb (opposite - side of the right - triangle formed by the plane's path) is the product of the rate and time.
$725\times20 = 14500$ ft

Step2: Use tangent function

We know that in a right - triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Here, $\theta = 14.5^{\circ}$ and the opposite side (vertical climb) is 14500 ft, and we want to find the adjacent side (distance along the ground).
Let $x$ be the distance along the ground. Then $\tan(14.5^{\circ})=\frac{14500}{x}$.
We can solve for $x$: $x=\frac{14500}{\tan(14.5^{\circ})}$
Since $\tan(14.5^{\circ})\approx0.2586$, $x=\frac{14500}{0.2586}\approx56069$ ft

Answer:

56069 ft