Trig 11, please help!!?

Question

Please bear with me, I just have a few more questions to ask!

I'm so grateful for any help you can give me =)

Ok here goes:

An airplane leaves an airport and travels for 100 mi in a direction of 300 degrees. How far north of the airport is the plane then? How far west?

Thank you!!

Please explain as much as possible =)

Answer 1

300 degrees is the same as N60W, so one angle is 60.
As you are taking N as one bearing the other angle is 90 so the third must be 30.
You have a right angled triangle with the angles 60 and 30 if you draw the diagram.
Let x be the dist N and y be dist W.
x/100 = cos 60
x = 100 x cos 60
x = 50 mls.(N)
y/100 = sin60
y = 100 x sin 60
y = 86.6 mls. (W).

Answer 2

You never mentioned 300 degrees with reference to what direction. I suppose the airplane travels at 300 degrees NORTH of the airport.

If you sketch the situation, you are actually asking for the base and the perpendicular of the right triangle the path of the airplane makes with the vertical and horizontal axis.

By simple geometric manipulations work out the remaining angles of the triangle, one being the right angle.

The angles come out to be 60 degrees (with the vertical) and 30 degrees (with the horizontal).

Now, apply the Sin and Cosine to find the two distances asked for.

Let, the distance to the North of airport = N
And, the distance to the West of airport = W

N = 100 * sin (30) = 100 * 0.5 = 50mi
W = 100 * cos (30) = 100 * 0.866 = 86.6mi

(Sketch the airplane's path to have a better understanding.)

Answer 3

starting from the North and going 300 deg clockwise, you come to 60 deg North -West.
Now you have to find the vertical and horizontal components of the distance travelled.
Vertical component = 100 x cos 60
V.C = 50 mi, North of the airport
Horizontal component = 100 x sin60
H.C = 86.8 mi , West of the airport

Answer 4

North is 360°, so 300° is 60° W of N.
100cos60° ≈ 86.6 mi. N
100sin60° ≈ 50 mi. W

Answer 5

isnt 300 degrees south-south east?