15.07.2022 - 21:54

# Mid-Air Refueling. This exercise illustrates a mid-air refueling scenario that military aircraft often enact. Assume the elevation angle that the hose makes with the plane being fueled is\theta = 36^circ.

Question:

Mid-Air Refueling. This exercise illustrates a mid-air refueling scenario that military aircraft often enact. Assume the elevation angle that the hose makes with the plane being fueled is {eq}theta = 36^circ {/eq}. If the hose is 150 feet long, what should the altitude difference, a, be between the two planes?

a {eq}approx {/eq} _____ feet

• We can use trigonometry to solve this problem. Let's draw a diagram to represent the situation: In the diagram, we have two planes flying at different altitudes. Plane A is the plane being fueled, and plane B is the refueling plane. The hose connecting the two planes is shown at an angle of {eq}theta = 36^circ {/eq} with respect to plane A. We are given that the hose is 150 feet long. Let's call the altitude of plane A (the plane being fueled) h and the altitude of plane B (the refueling plane) H. We want to find the altitude difference, a = H - h. To do this, we can use the tangent function: {eq}tan(theta) = \frac{a}{150} {/eq} Rearranging this equation, we get: {eq}a = 150 tan(theta) {/eq} Plugging in {eq}theta = 36^circ {/eq}, we get: {eq}a = 150 tan(36^circ) \approx 109 {/eq} Rounding to the nearest foot, we get: {eq}a \approx 109 \text{ feet} {/eq} Therefore, the altitude difference between the two planes should be \approx imately 109 feet.