Maligayang pagdating sa Imhr.ca, kung saan ang iyong mga tanong ay masasagot ng mga eksperto at may karanasang miyembro. Tuklasin ang malalim na mga sagot sa iyong mga tanong mula sa isang malawak na network ng mga eksperto sa aming madaling gamitin na Q&A platform. Sumali sa aming platform upang kumonekta sa mga eksperto na handang magbigay ng eksaktong sagot sa iyong mga tanong sa iba't ibang larangan.

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 :

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.