As a plane takes off it ascends at a 20 degree angle of elevation. if the plane has been traveling at an average rate
of 270 ft/s and continues to ascend at the same angle, then how high is the plane after 10 seconds (the plane has
traveled 2700 ft).
2700 ft
923.5 feet
2537.2 feet
2873.3 feet
7894.3 feet

The plane was 923.5 feet high after 10 seconds.

Step-by-step explanation:

Angle made by plane with ground during take off = x = 20°

Average speed of plane = 270 ft/s

Let the height of Plane after 10 seconds from the takeoff = H

Distance covered by plane in 10 seconds = 270 ft/s × 10 sn= 2700 ft

According to trigonometric ratios:

Here , perpendicular = H

Hypotenuse = 2700 ft

The plane was 923.5 feet high after 10 seconds.

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west