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A plane flew 150 miles on a course of 220 degrees and then 130 miles on a course of 130 degrees. Then the plane returned to its starting point via the shortest route possible. Find that shortest distance. 

I need your help!

Sagot :

Wilviedelacruzidn
as you draw your equation into a triangle ;the <A=220°, a small is 150 miles which we place to the opposite of the angle .THE <C=130°,c small is 130 miles which we place into the opposite of the angle. and the <B is unknown and its small one.

as we solve it:
to get the <B follow the formula of A+B+C=180° so we just only need the <B coz we the <A <C. <B=180°-220°-130° =<B is -170° it cannot be because no negative dimension.as the question needed just the shortest distance.

solve:
sin A/ a=sin B/ b=sin C/c
sin 220° /150 miles= sin 130°/ 130 miles= sin -170° /c
sin 220° /150 miles=sin -170° /c; make it cross multiply.
c=150 miles(sin -170°)/(sin 220°); do it in a calculator.
c=40.52 miles; round to nearest tens.
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