Question

已知三角形的周长为根号2+1.且 sinA+sinB=根号2sinc,三角形面积为1/6sinC,求角C的度数

Answer 1

sinA+sinB=√2sinC
sinA/sinC+sinB/sinC=√2
a/c+b/c=√2
(a+b)/c=√2
a+b=√2c
周长为根号2+1
a+b+c=√2+1
√2c+c=√2+1
c=1，即AB=1

a+b=√2+1-c=√2
三角形ABC的面积为1/6sinC
1/2absinC=1/6sinC
ab=1/3
cosC=(a^2+b^2-c^2)/2ab
={(a+b)^2-2ab-c^2}/(2ab)
={(√2)^2-2*1/3-1^2}/(2*1/3)
={2-2/3-1}/(2/3)
=1/2
C=60°

Answer 2

a/sinA=b/sinB=c/sinC
sinA+sinB=根号2sinc 则a+b=√2c

a+b+c=√2+1
所以c=1
所以a+b=√2
a²+b²+2ab=2

S=1/2absinC=1/6sinC
所以ab=1/3
则a²+b²=4/3
cosC=(a²+b²-c²)/2ab=1/2
C=60度