Question 247756
How do I calculate side BC of an Isosceles triangle when the two equal sides AB and AC are 12 and angle A [BAC] is 130°?? 
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You can use the Law of Sines:
a/sin(A) = b/sin(B)
Angles B and C are 25 degs (since it's an isosceles)
a = 12*sin(130)/sin(25)
a =~ 21.75
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You can also drop an altitude from A.  Then a/2 = 12*sin(65)
a = 24*sin(65)
a =~ 21.75 (again, it's monotonous)