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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?

 Aug 7, 2015

Best Answer 

 #1
avatar+111396 
+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°

 

 

 Aug 7, 2015
 #1
avatar+111396 
+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

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