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:
https://www.geogebra.org/geometry
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!