高中数学题！求解
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(1)
a/sinA =b/sin2B
a/sinA =b/(2sinB.cosB)
1/(2cosB) = 1
cosB =1/2
B=π/3
(2)
b=√13
a-c=2
(a-c)^2 =4
a^2+c^2 -2ac =4
[a^2+c^2 -2ac.cos(π/3)] -2ac.cos(π/3)= 4
b^2 -2ac.cos(π/3)=4
13- ac =4
ac = 9
SΔABC
=(1/2)ac.sinB
=(1/2)(9)(√3/2)
=9√3/4
a/sinA =b/sin2B
a/sinA =b/(2sinB.cosB)
1/(2cosB) = 1
cosB =1/2
B=π/3
(2)
b=√13
a-c=2
(a-c)^2 =4
a^2+c^2 -2ac =4
[a^2+c^2 -2ac.cos(π/3)] -2ac.cos(π/3)= 4
b^2 -2ac.cos(π/3)=4
13- ac =4
ac = 9
SΔABC
=(1/2)ac.sinB
=(1/2)(9)(√3/2)
=9√3/4
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