已知三角形ABC的周长为根号2+1，且sinA+sinB=根号2乘以sinC

Question

（1）求边AB的长 （2）若三角形ABC的面积为六分之一sinC，求角C的度数

详细解答过程啦！谢谢

Answer 1

sinA + sinB = √2sinC
a/sinA = b/sinB = c/sinC
有：
(a+b+c)/(sinA+sinB+sinC) =(a+b)/(sinA+sinB)=c/sinC
所以有：
(√2 + 1 - c)/(√2sinC) = c/sinC
那么：
√2 + 1 - c = √2c
所以，AB = c = 1

S = 1/2 * a * b * sinC = 1/6 sinC
所以，a * b = 1/3。
又 a + b = (√2 + 1) - c = √2
因此，可以得到：
(√2 - a) * a = √2*a - a^2 = 1/3
化简，得到：
3a^2 - 3√2 * a + 1 = 0
a = (3√2 - √6)/6 = √2/2 - √6/6，b = √2/2 + √6/6
或者，a = (3√2 + √6)/6 = √2/2 + √6/6, b = √2/2 - √6/6
又因为：cosC = (c^2 - a^2 - b^2)/(2ab)
= [1 - (1/2 + √12/6 + 1/6) - (1/2 - √12/6 + 1/6)]/[2 * (1/2 - 1/6)]
= [-1/3]/(2/3)
= -1/2
所以，C = 120°

Answer 2

(1)由sinA+sinB=√2sinC两边乘以外接圆直径2R即得
a+b=√2c
代入a+b+c=1+√2得到
(1+√2)c=1+√2
∴AB=c=1
(2)∵S=(1/2)absinC=(1/6)sinC
∴ab=1/3
又a+b=√2
∴a^2+b^2=(a+b)^2-2ab=2-2/3
由余弦定理得cosC=(a^2+b^2-c^2)/(2ab)=1/2
故∠C=60°