A pilot flies on a bearing of 160° for 30 miles. The pilot then executes a quick turn and flies another 15 miles at a bearing of 205°. What bearing should the pilot use (to the nearest second) and how far must she travel(to the nearest of a mile) to make it back to her starting position?

34.65 miles at an angle of 325.39°

Step-by-step explanation:

Ok, the initial position is (0,0)

Then she flies 30 miles at an angle of 160°, if we count the angle counterclokwise from the x-axis, the new position will be:

p = (30*cos(120°), 30*sin(120°))

Then she travels another 15 miles at an angle of 205°, the new position is:

p = (30*cos(120°) + 15*cos(205°), 30*sin(120°) + 15*sin(205°))

p = (-28.59 , 19.64)

If she now travles X miles at an angle Y, we must have that the final position is the point (0,0)

this means that:

X*cos(Y) = -(-28.59) = 28.59

X*sin(Y) = -19.64

Now, we can find the quotient between those two equations and use that tan(x) = sin(x)/cos(x)

X*(sin(Y))/(X*cos(Y)) = -19.64/28.59

Tg(Y) = -0.69

Y = ATg(-0.69) = -34.61°

If we use only positive angles, this angle is equivalent to:

360° - 34.61° = 325.39°

now lets find the distance:

Xcos(325.39°) = 28.59

X = 28.59/cos(325.39°) = 34.65 miles.

only 10/60 isn't equivalent

hope this : )