Trigniometry help please....?

Question

An airplane flying 450 miles per hour has a bearing of N19degree W.After flying 2 hours,how far north has the plane traveled from its point of departure.

I did this:450 times 2=900 miles for distance.I don't know what to do after this.What is the problem looking for?an angle or distance?Thanks.

Answer 1

soh-cah toa

Make a triangle the longest side the hypotonose is 900miles long.

The adjacent is how far north its gone....

Cos angle x hypotonose = adjacent

You then can use Cos 19 x 900miles = 850.966718 miles

A diagram really helps here......

Answer 2

Distance north= 900 x Sin.71deg.

And if you have to write it up, it's spelt hypotenuse. And you don't need to give an answer like this to more than one decimal place, if that.

Answer 3

Yes, trigonometry provides the solution.

What you have to solve is the sides of a right triangle.  What you have solved thus far is the hypotenuses.  I don't understand your N19degrees W, unless the problem says you are flying from the west on 19 degrees.  North would be 0 and East would be 90 degrees, so the angle at the bottom left would be 71 degrees and the angle at the top right would be 19 degrees.  Remembering that the total of the angles must equal 180.  Do they?  19+71+90=180.

So, you are solving for the length of the side that goes the same direction as North.

Answer 4

You are looking at right triangle on its point. The base of the triangle is north or 0 degrees, the  19 degree to the right is the hypotenuse which is 900 miles (450mph X 2 hours) . Back from the end of the flight or 71 degrees to the base gives you 90 degrees at the base. So you have right triangle with the other two angles being 19 and 71 which give us 180 degrees any right triangle should have. Now try your math and go from there.

Best Wishes.

Answer 5

900mile cos19º

Answer 6

Huh?