We have triangle with #AB=6,hat(A)=pi/3 and hat(B)=pi/4#.How to find #AC and BC#?

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Answer 1 · 2017-04-22T00:20:50

See below.

Using the so called sinus law.

#(sin hat C)/[AB] = (sin hat A)/[BC] = (sin hat B)/[AC]#

but

#hat A + hat B + hat C = pi# and

#hat A = pi/3#
#hat B = pi/4#

so

#hat C = pi -pi/3-pi/4 = (5pi)/12#

and now

#BC= AB( (sin hat A)/(sin hat C)) = 6(sin(pi/3)/sin((5pi)/12))=#
#9sqrt2-3sqrt6#

and also

#AC = 6 (sin(pi/4)/sin((5pi)/12))=#
#6sqrt3-6#