Trig Bearing Problems

• Oct 6th 2009, 06:50 PM
Thatoneguy12345
Trig Bearing Problems
Any help would be appreciated!

I just have no clue where to start on these problems.

Well basically for this first one it says to find the bearing of an airplane located at that point. Express the Bearing using both methods.

(-4,0)

Is this one just North 90 degrees West? I don't know where to draw the angle at.

And if someone can help me through this problem step to step, I would be glad :D.

A plane flies at 1.3hr at 110mph on a bearing of 38 degrees. It then turns and flies 1.5 hr at the same speed on a bearing of 128 degrees. How far is the plane from the starting point?

I have no idea where to start!
• Oct 6th 2009, 07:45 PM
TKHunny
Bearing One Way:

North: 0º
East: 90º
South: 180º
West: 270º
• Oct 6th 2009, 08:48 PM
pacman
"A plane flies at 1.3hr at 110mph on a bearing of 38 degrees. It then turns and flies 1.5 hr at the same speed on a bearing of 128 degrees. How far is the plane from the starting point?"

Solution:

Mark the points where the plane starts as A, then where it landed 1.3 hr later as B and 1.5 hr later as C. You now constructed a triangle ABC.

Mark all indicated angles, then angle ABC = 38 + (180 - 128) = 38 + 52 = 90 degrees.

Heyy! its a right triangle, hmnnn . . . .

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distance AB = distance x speed = (1.3 hr)(110 mi/hr) = 143 miles,

distance BC = distance x speed = (1.5 hr)(110 mi/hr) = 165 miles,

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By Pythagorean Theorem:

AC^2 = AB^2 + BC^2,

distance travelled after 2.8 hours = AC = sqrt (AB^2 + BC^2) = sqrt (143^2 + 165^2) = (11) sqrt 394 = 218.34 miles
• Oct 6th 2009, 10:43 PM
Thatoneguy12345
Thanks a lot!