Chapter 4, Problem 144CR

To determine

To Compute : The distance between city A to city C and the bearing from city A to city C.

Answer to Problem 144CR

The distance between city A to city C is 1221 miles and the bearing from city A to city C is $85.6°$ .

Explanation of Solution

Given information : A plane flies 650 miles from city A to city B at a bearing of $48°$ , and flies 810 miles from city B to city C at a bearing of $115°$ . As shown below.

  

  $\angle B=48°+65°=113°$ , using alternate interior angle and supplementary angle properties.

Using above diagram,

  $\sin 48°=\frac{AF}{650}$

Implies,

  $AF=650\times \sin 48°=483$

  $\sin 65°=\frac{DC}{810}$

Implies,

  $DC=810\times \sin 65°=734=EF$

Therefore,

  $AE=AF+FE$

  $AE=483+734=1217$

  $\cos 48°=\frac{BF}{650}$

Implies,

  $BF=650\times \cos 48°=435$

  $\cos 65°=\frac{BD}{810}$

Implies,

  $BD=810\times \cos 65°=342$

Therefore,

CE = FD = BF - BD = 435 - 342 = 93

Using Pythagoras theorem,

AC²= AE²+ CE²

AC²= 1217² + 93²

AC = 1221, That is the distance between city A and city C is 1221 miles.

Considering,

  $\tan \angle CAE=\frac{CE}{AE}=\frac{93}{1217}$

Implies,

  $\angle CAE={\tan }^{-1}\frac{93}{1217}=4.4°$

Therefore the bearing from city A to city C is,

  $90-\angle CAE=85.6°$

Thus,

The distance between city A to city C is 1221 miles and the bearing from city A to city C is $85.6°$ .