Question

三角形ABC中，三个内角ABC的对边分别为a b c ,

若a^2+c^2=b^2+ac，且a:c=（根号3 + 1 ）：2，求角C的大小

Answer 1

a²+c²=b²+ac
即a²+c²-b²=ac
因为a²+c²-b²=2ac
cosB
则cosB=1/2
B=π/3
即A+C=2π/3
a/c
=
sinA/sinC
=
sin(2π/3
-
C)/sinC
=1/2
+
√3/2
cotC
=
(1
+
√3)/2
故cotC
=
1
C=π/4