An airplane is traveling at a speed of 200 miles/hour with a bearing of 240°. The wind velocity is 60 miles/hour at a bearing of 25°. What are the plane's actual speed and direction angle?

vx=-200cos60+60cos25

vy=-200sin60+60sin25

v=(vx^2+vy^2)^(1/2)

v=154.73mph

tana=vy/vx

a=arctan(vy/vx)

a≈72.85 and this is in the third quadrant so:

The bearing is 180+72.85=252.85

So:

The plane is traveling at a speed of 154.73mph with a bearing of 252.85°

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RiverShinesDragon RiverShinesDragon · Answer 2 · 2019-05-21T21:59:10+02:00

Answer:

<154.72, 197.15 degrees> (D)

Step-by-step explanation:

First we turn bearings into workable angles:

Bearing to angle = 90 - θ

Airplane bearing to angle = 90 - 240 = -150 = 210 (add 360 to -150)

Wind bearing to angle = 90 - 25 = 65

Now let's work on the vectors of the plane and the wind

Airplane vector: <200cos(210), 200sin(210)> = <-173.2,-100>

Wind vector: <60cos(65), 60sin(65)> = <25.36, 54.38>

Add the two to get the component form of the actual speed vector and we get:

<-147.84, -45.62>

Trig form: <sqrt(h^2+v^2), arctan(v/h)>

<sqrt(147.84^2+45.62^2), arctan(v/h)> = <154.72, 197.15 degrees>

*the trig form shows that u = <speed, angle>

Let me know if I missed anything; was kind of in a rush.