I NEED HELP PLEASE THANKS!
An airplane is taking off headed due north with an air speed of 173 miles per hour at an angle of 18° relative to the horizontal. The wind is blowing with a velocity of 42 miles per hour at an angle of S47°E. Find a vector that represents the resultant velocity of the plane relative to the point of takeoff. Let i point east, j point north, and k point up.
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The ' horizontal ' may act as the x - axis in this case, the airplane taking off at an angle of 173 cos 18 respective to this x - axis. Respectively it travels restricted to an angle of 173 sin 18 from the y - axis. The following shows this angle at vector( s ) j and k relative to the air -

j - ( 173 cos 18 ),

k - ( 173 sin 18 )

Thus, one can assume such -

[tex]0i + ( 173 cos 18 )j + ( 173 sin 18 )k[/tex]

Knowing that, this second bit here should be similar to the first bit above, given that the wind is now blowing with a velocity of 42 miles per hour at an angle of 47 degrees. Therefore, j = 42 cos 47, i = 42 sin 47 -

[tex]( 42 sin 47 )i + ( 42 cos 47 )j + 0k[/tex]

Adding the two we should get the following -

[tex]30.7i +135.9j + 53.4k[/tex]

sqdancefan sqdancefan · Answer 2 · 2020-07-10T22:09:09+02:00

Answer:

30.72i+ 135.89j +53.46k

Step-by-step explanation:

If we measure angle φ up from the horizontal and angle θ CCW from east, the direction vector of the airplane at take-off is ...

(ρ, θ, φ) = (173 mph, 90°, 18°)

The rectangular expression of this vector will be ...

(ρ·cos(θ)·cos(φ), ρ·sin(θ)·cos(φ), ρ·sin(φ)) = (0, 164.53, 53.46) . . . mph

__

The wind vector is ...

(ρ, θ, φ) = (42, -43°, 0°) ⇒ (30.72, -28.64, 0) . . . mph

And the rectangular coordinate sum of these vectors is ...

(0, 164.53, 53.46) +(30.72, -28.64, 0) = (30.72, 135.89, 53.46)

The resultant velocity vector of the airplane is ...

30.72i+ 135.89j +53.46k