If a plane takes off bearing Upper N 23 Superscript ring operator Baseline Upper W and flies 2 miles and then makes a right turn left-parenthesis 90 Superscript ring operator Baseline right-parenthesis and flies 9 miles farther, what bearing will the traffic controller use to locate the plane?

Guest Aug 7, 2015

#1**+5 **

If we let the origin be the starting point, the segment drawn from the origin to the final destination point will will form a right triangle with the other two segments of travel with the hypotenuse given by: √[2^2 + 9^2] = √85

And, using the Law of Cosines, the angle Θ formed between the two sides having the origin as a vertex is given by:

9^2 = 2^2 + 85 - 2 (2)(√85) cosΘ =

81 = 89 - 4√85 cosΘ

-8 = -4√85 cosΘ

cosΘ = 2/√85 aad using the cos inverse, we have

cos^{-1} (2/√85) = about 77.47°

Then, the "bearing" will be given by N (77.47 - 23)°E = N (54.47)° E

BTW......"heading" is usually used with respect to aviation, rather than "bearing".......in this case the heading is just 54.47°

CPhill Aug 7, 2015

#1**+5 **

Best Answer

If we let the origin be the starting point, the segment drawn from the origin to the final destination point will will form a right triangle with the other two segments of travel with the hypotenuse given by: √[2^2 + 9^2] = √85

And, using the Law of Cosines, the angle Θ formed between the two sides having the origin as a vertex is given by:

9^2 = 2^2 + 85 - 2 (2)(√85) cosΘ =

81 = 89 - 4√85 cosΘ

-8 = -4√85 cosΘ

cosΘ = 2/√85 aad using the cos inverse, we have

cos^{-1} (2/√85) = about 77.47°

Then, the "bearing" will be given by N (77.47 - 23)°E = N (54.47)° E

BTW......"heading" is usually used with respect to aviation, rather than "bearing".......in this case the heading is just 54.47°

CPhill Aug 7, 2015