First, you should know what the Law of Sines is. It is simply this:
#" "a/sinA=b/sinB=c/sinC#
Now let's look at our given. We have:
#A=57°#, #a=11#, and #b=10#
We need to look for #B#, #C#, and #c#.
Step 1 - Solving for #B#
Law of Sines
#[1]" "a/sinA=b/sinB#
Plug in the values of #A#, #a#, and #b#.
#[2]" "11/sin(57°)=10/sinB#
Multiply both sides by #(sinB)[sin(57°)]#.
#[3]" "11/cancelsin(57°)(sinB)cancel[sin(57°)]=10/cancelsinBcancel(sinB)[sin(57°)]#
#[4]" "11sinB=10sin(57°)#
Divide both sides by #11#.
#[5]" "sinB=(10sin(57°))/11#
Apply #arcsin# on both sides.
#[6]" "arcsin(sinB)=arcsin((10sin(57°))/11)#
#[7]" "color(blue)(B=arcsin((10sin(57°))/11)~~49.678696410324°)#
Step 2 - Solving for #C#
The sum of the interior angles of all triangles is #180°#.
#[1]" "A+B+C=180°#
Isolate #C#.
#[2]" "C=180°-A-B#
Plug in the values of #A# and #B#.
#[3]" "C=180°-57°-arcsin((10sin(57°))/11)#
#[4]" "color(blue)(C=123°-arcsin((10sin(57°))/11)~~73.321303589676°)#
Step 3 - Solving for #c#
Law of Sines
#[1]" "c/sinC=a/sinA#
Multiply both sides by #sinC#
#[2]" "c/cancelsinCcancel(sinC)=a/sinA(sinC)#
#[3]" "c=(asinC)/sinA#
Plug in the values of #A#, #a#, and #C#.
#[4]" "color(blue)(c=(11sin(123°-arcsin((10sin(57°))/11)))/sin(57°)~~12.564196743012485)#
