Question 925918
Here is a sketch that illustrates the situation. (The angles and proportions are not right on purpose).
{{{drawing(500,200,-15,35,-5,15,
green(line(10,10,15,15)),
red(circle(-12,0,0.5)),
red(circle(32,0,0.5)),locate(18.5,6,x),
line(-12,0,32,0),line(0,0,10,10),line(10,10,32,0),
locate(-15,-1,start),locate(26,-1,destination),
red(arc(0,0,10,10,-45,0)),red(arc(10,10,10,10,-45,24.4)),
locate(1.5,2.5,red(22^o)),locate(11,12,red(30^o)),
locate(-7,1.5,80),locate(15,1.5,130),locate(9,-3,210),
arrow(8.5,-3.5,-12,-3.5),arrow(11.5,-3.5,32,-3.5)
)}}} The plane was {{{210-80=130}}} miles from its destination when it made the first {{{(22^o)}}} turn.
It was {{{x}}} miles from its destination when it made the second {{{(30^o)}}} turn.
We can use law of sines on the triangle you can see in the sketch.
The angle at the top of that triangle is the supplementary angle to the {{{30^o}}} angle of the second turn.
It measures {{{180^o-30^o=150^o}}} , and like all supplementary angles, has the same sine:
{{{sin(150^o)=sin(30^o)=0.5}}} (an exact {{{0.5=1/2}}} number).
Law of sines says that
{{{130/sin(150^o)=x/sin(22^o)}}}
We use {{{sin(22^o)=0.3746}}} (rounded) to get a very accurate answer.
{{{130/sin(150^o)=x/sin(22^o)}}}
{{{130/0.5=x/0.3746}}}--->{{{130*0.3746/0.5=x}}}--->{{{highlight(x=97.4)}}} (rounded).
The plane was {{{highlight(97.4)}}} miles from its destination when it made the {{{30^o}}} turn.