We have triangle #ABC# with #BC=13,AC=14,hat(C)=arccos(5/13)#.How to find the other two angles?

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Answer 1 · 2017-04-21T20:20:27

See below.

Using the so called sinus law

#(sin hatA)/[BC] = (sin hat B)/[AC]#

and

#hatA+hatB+hatC=pi# but #hatC=arccos(5/13)#

and

#sin hatB = sin(pi-hat A-hat C)=sin(hatA+hat C)# and

#sin(hatA+hat C)=sin hat A cos hat C+cos hat A sin hat C#

and

#sin hat C=12/31# and #cos hat C = 5/13#

so

#14 sin hat A = 13(5/13 sin hat A+12/13cos hat A)# or

#9 sin hat A = 12 cos hat A# and

#tan hat A = 12/9 = 4/3#

and #hat A = arctan(4/3)#

and now

#hat B = pi - (hat A + hat C)#

This is left to the reader.