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?
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°
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°