What are the new speed and direction of the plane relative to the grou

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SUMMARY

The problem involves calculating the new speed and direction of a plane moving at 300 mi/h east when it encounters a wind blowing at 100 mi/h at an angle of 30 degrees north of east. The solution requires vector addition, where the plane's velocity vector and the wind's velocity vector are combined. The resultant vector is determined using trigonometric methods, specifically the Pythagorean theorem for magnitude and the tangent function for direction. The final answer provides both the speed and the angle of the plane relative to the ground.

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Homework Statement


a plane moving initially at 300mi/h to the east, suddenly enters region where the wind is blowing at 100 mi/h toward the direction 30 degree's north of east. What are the new speed and direction of the plane relative to the ground.


Homework Equations



Magnitude= sqrt(A^2 + B^2 - 2ABcos ([STRIKE]0[/STRIKE]))
Angle= [STRIKE]0[/STRIKE] = tan-1(B/A)

The Attempt at a Solution



This is suppose to be a pretty straight forward question but I am not getting it. I first tried the first equation A=300 B=100. I am not coming up with the answer.
 
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The soln is the sum of the two vectors (300 m/h E + 100m/h 30 deg N of E). Draw the first vector (arrow), then the second beginning with its tail at the tip of the first. Complete the triangle, then use triangle trigonometry to find the magnitude and direction of the third side. Alternatively, write both vectors in terms of their rectangular components and find the resultant that way.
 


Show your calculations.
 


OK got it. Thanks!
 

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