Question

在△ABC中,已知sinA+sinC=2sinB, 且∠B=π/6,若△ABC的面积为1/2,则∠B的对边b?急！谢谢！

Answer 1

sinB=1/2, cosB=√3/2
S△ABC=1/2*acsinB=ac/4=1/2∴ac=2
b^2=a^2+c^2-2accosB=(a+c)^2-2ac-2accosB=4b^2-4-2√3
∴3b^2=4+2√3=(√3+1)^2
b=(√3+1)/√3=1+√3/3

Answer 2

sinB = 1/ 2, cosB = sqrt(3) / 2
By law of sine, sinA / a = sinC / c = sinB / b = k (k != 0) => sinA = ka, sinC = kc, sinB = kb, so
sinA + sinC = 2sinB => ka + kc = 2kb => a + c = 2b, and S△ABC = (1/2) * ac * sinB = 1/2.
so {a + c = 2b, ac = 2}

By law of cosine, b^2 = a^2 +c^2 - 2ac * cosB = (a + c) ^ 2 - 2ac - 2ac * cosB = 4b^2 - 4 - 2 * sqrt(3), b = sqrt({4 + 2 * sqrt(3)} / 3)