First, convert the degrees with minutes to just degrees:
#A=83^@20'#
#color(white)A=83^@+20'#
#color(white)A=83^@+20'color(blue)(*1^@/(60'))#
#color(white)A=83^@+20color(red)cancelcolor(black)'color(blue)(*1^@/(60color(red)cancelcolor(blue)'))#
#color(white)A=83^@+20^@/60#
#color(white)A=83^@+1^@/3#
#color(white)A=color(black)(83 1/3)^@#
Now, we can construct a triangle based on what we know so far:
We can see that we can use the law of sines to compare between #angleC#, #c#, #angleA#, and #a# (you should probably use a calculator for the last step):
#color(white)=>sinA/a=sinC/c#
#=>sin(color(black)(83 1/3)^@)/a=sin(54.6^@)/18.1#
#color(white)=>a/sin(color(black)(83 1/3)^@)=18.1/sin(54.6^@)#
#color(white)=>a=sin(color(black)(83 1/3)^@)*18.1/sin(54.6^@)#
#color(white)(=>a)~~22.05#
Now, since we know that the angles in a triangle must add up to #180^@#, we can do some math to compute the measure of #angleB#:
#mangleA+mangleB+mangleC=180^@#
#color(black)(83 1/3)^@+mangleB+54.6^@=180^@#
#mangleB=180^@-color(black)(83 1/3)^@-54.6^@#
#qquadqquad qquadqquad qquad vdots quad# After some annoying work...
#color(white)(mangleB)=color(black)(42 1/15)^@#
We can use the law of sines once again to solve for #b#:
#color(white)=>sinB/b=sinC/c#
#=>sin(color(black)(42 1/15)^@)/b=sin(54.6^@)/18.1#
#color(white)=>b/sin(color(black)(42 1/15)^@)=18.1/sin(54.6^@)#
#color(white)=>b=sin(color(black)(42 1/15)^@)*18.1/sin(54.6^@)#
#color(white)(=>b)~~14.88#
We have solved all the parts of the triangle. Hope this helped!
