an airplane is flying at a constant altitude of 28000 meters. when first seen to the east of the observer, the angle to the airplane is 71.5 degrees. after 73 seconds, the angle of elevation is 51.6 degrees. find the speed of the airplane.

Answer 1

draw two triangles, one that has an angle of 71.5º and another with an angle of 51.6º

lets find the base in each one, call the first one x, the other one y

tan71.5 = 28000/x
x = 28000/tan71.5 = 9368.67

tan51.6 = 2800/y
y = 22192.53

So the horizontal distance travelled in 73 seconds is y-x = 12823.86

speed = distance/time = 12823.86/73
= 128.34 m/second

BTW, that airplane must be a spacecraft to fly at a height of 28 km, lol

Answer 2

I need help with Fraction vocabulary worksheet

Answer 3

If that is the case, then this should be a new question and thus a new topic.