Trig, I need some major help?

Question

I made a picture of the problem, just click the link and click on the picture to make it bigger. Please help, thanks!
http://i34.photobucket.com/albums/d141/darealbigtigga/trig.jpg
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Answer 1

first leg = 140 mph *1.3 hrs = 182 miles
second leg = 140 mph*2.3 hrs = 322 miles
since the triangle is a right triangle (120-30=90)
the distance = sqrt(182^2+322^2)=369.9 miles

Answer 2

ok, given the diagram supplied by the link, let's first find the lengths the plane travelled. Since it travelled at 140mph along both paths, we can multiply this by the times it flew in those directions to find the lengths, which are also the lengths for the triangle.

@30deg: (1.3h)*(140mph)=182m =a
@120deg: (2.3h)*(140mph)=322m =b

Now that we have the lengths we need to find c. If you notice the plane turned 90degrees on its second leg of the trip relative to its inital path. This means that the triangle is indeed a right triangle, which you can verify using supplementary angles in from geometry.

So since we have a right triangle, the Pythagorean Theorem holds and we have:

c^2=a^2 +b^2
c^2=(182)^2 +(322)^2 or c=369.876m

Hope this helped.

Answer 3

The angle at the lower left is 90º - 30º = 60º. The angle at the top is 180º - 120º + 30º = 90º. This is therefore a right triangle with the distance you want as the hypotenuse. One leg is 1.3hrs x 140mph, and the other is 2.3hrs x 140mph. The distance is then the square root of the sum of the squares of the legs.

Answer 4

the projection of the first part of the flight= 140*1.3 cos60*=91
the projection of the secondpart=140*2.3*sin30*=161
so the distance from the starting point=91+161
=252 miles

Answer 5

D = 1.3*140*sen 30º +2.3*140*cos 30º

D = 369,86 miles

₢

Answer 6

well, I could tell you in hours, but since the problem doesn't give miles, it's hard to tell you.

if it helps, it's (the square root of 6.98) hours away...