Given triangle ABC with AB 5 and BC 332 The measure of ang
Solution
AB=c=5 and BC=a=5sqrt3 /3
a/sinA=c/sinC
(5sqrt3 /3)/(sin300)=5/sin(c)
(2*5sqrt3 /3)=5/sin(c)
(2/sqrt3)=1/sin(c)
c =pi/3, 2pi/3
2choices

AB=c=5 and BC=a=5sqrt3 /3
a/sinA=c/sinC
(5sqrt3 /3)/(sin300)=5/sin(c)
(2*5sqrt3 /3)=5/sin(c)
(2/sqrt3)=1/sin(c)
c =pi/3, 2pi/3
2choices
