**An airplane is traveling at a fixed altitude with an outside wind factor. The airplane is headed N 40o W at a speed of 600 miles per hour. As a airplane comes to a certain point, it comes across a wind in the direction N 45o E with a velocity of 80 miles per hour. What are the resultant speed and direction of the airplane? Round your Answers to the nearest hundredth.**

**Answer:**

Split the two vectors in both x and y directions.

Assume that E=east is the positive direction, and W=west as the negative direction.

X-Direction:

(600 mi/hr)cos(-40)=-460

(80 mi/hr)cos(45)=57

Result: (-460) + (57) = -403 or 403 West

Y-Direction:

(600 mi/hr)sin(40)=386

(80 mi/hr)sin(45)=57

Result: (386) + (57) = +443 or 443 North

Use Pythagorean Theorem to solve the resultant speed.

Resultant speed:

c^2=a^2+b^2

c^2=(-403)^2+(443)^2

c^2=358658

c=599 mi/hr <—Resultant Speed

Use Tangent in Trig to find direction.

tanθ=y/x

tanθ=(443)/(-403)

θ= -47.7\textdegree

or 47.7\textdegree West <—Direction

Answer:

599mi/hr at 47.7\textdegree West