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Adam must fly home to city A from a business meeting in city B. One flight option flies directly to city A from​ B, a distance of about 467.3 miles. A second flight option flies first to city C and then connects to A. The bearing from B to C is N28.7​E, and the bearing from B to A is N60.7E. The bearing from A to B is S60.7​W, and the bearing from A to C is N79.1W. How many more frequent flyer miles will Adam receive if he takes the connecting flight rather than the direct​ flight?

User Lei
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1 Answer

2 votes

Answer:

The value is
k =109.6 \ miles

Explanation:

The diagram illustrating the question is shown on the first uploaded image

From the question we are told that

The distance from city A to B is AB = 467.3 miles

The bearing from B to C is
\theta_(BC) =  N 28.7E

The bearing from B to A is
\theta_(BA) =  N 60.7E

The bearing from A to B is
\theta_(AB) =  S60.7W

The bearing from A to C is
\theta_(AC) =  S79.1W

Generally from the diagram


\theta_A  =  180 - 60.7 -79.1

=>
\theta_A  =  40.2 ^o

Also


\theta_B  =  32^o

and


\theta_C  =  180 - (\theta_A  +\theta_B )

=>
\theta_C  =  180 - (40.2  + 32 )

=>
\theta_C  =  107.8 ^o

Generally according to Sine Rule


(BC)/(sin (\theta_A))  = (CA)/(sin (\theta_B)) =(AB)/(sin (\theta_C))

=>
(BC)/(sin (40.2))  = (CA)/(sin (32)) =(467.3 )/(sin (107.8))

So


(BC)/(sin (40.2))  = (467.3 )/(sin (107.8))

=>
BC = 316.8 \ miles

Also


(CA)/(sin (32)) =  (467.3 )/(sin (107.8))


CA = 260 .1 \ miles

Generally the additional flyer miles that Adam will receive if he takes the connecting flight rather than the direct​ flight is mathematically represented as


k = [CA +BC]  - AB

=>
k = [260 .1 +316.8]- 467.3

=>
k =109.6 \ miles

Adam must fly home to city A from a business meeting in city B. One flight option-example-1
User AlbertK
by
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