高一解三角形三题。

a=4，b=3，C=60

c^2=a^2+b^2-2abcosc=4^2+3^2-2*4*3*1/2 = 13

c=√13

sinA = asinC/c = 4 *√3/2/√13 = 2√39/13

b^2=ac

b^2=a^2+c^2-2accosB

ac=a^2+c^2-2accosB

cosb =(a^2+c^2-ac)/(2ac)=(a^2+c^2)/(2ac)-1/2

∵(a-c)^2≥0,a^2+c^2≥2ac,(a^2+c^2)/(2ac)≥1

∴cosb =(a^2+c^2)/(2ac)-1/2≥1/2

∴0

(b+c):(c+a):(a+b)= 12:8:10

(b+c)/(c+a)=12/8，(b/c+1)/(1+a/c)=3/2，2b/c-3a/c=1...(1)

(c+a)/(a+b)=8/10，(1+a/c)/(a/c+b/c)=4/5，4b/c-a/c=5...(2)

(2)-(1)*2: 5a/c=3，a/c = 3/5。

4b/c=5+a/c=5+3/5=28/5，b/c=7/5

a^2+c^2-b^2)/(2ac

= {(a/c)^2+1-(b/c)^2 }/(2a/c)

= { (3/5)^2 + 1 - (7/5)^2 } / (2*3/5)

= ( 3^2 + 5^2 - 7^2 ) / (2*3*5)

= -15/30

= -1/2

B=120