这是题目以及第二小题选择条件1的答案，请生成讲解视频---**Question Stem:**
12. (2022 北京平谷高三一模) 在锐角△ABC中，角A, B, C的对边分别为a, b, c, 且√3c - 2bsin C = 0.
(1) 求角B的大小;
(2) 在下面两个条件中任选一个作为已知，求△ABC的面积。
条件①: b = 3√3, a = 2;
条件②: a = 2, A = π/4.
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**Answer:**
见解析
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**Analysis/Solution:**
**Part (1): Finding angle B**
Given the condition: √3c - 2bsin C = 0.
By the Sine Rule, we have c/sinC = b/sinB, which implies c = (b/sinB)sinC.
Substitute c into the given condition:
√3 * (b/sinB)sinC - 2bsinC = 0.
Since b is a side length and C is an angle in a triangle, b > 0 and sinC ≠ 0. We can divide the entire equation by bsinC:
√3 / sinB - 2 = 0.
√3 / sinB = 2.
sinB = √3 / 2.
Since △ABC is an acute triangle, all angles A, B, C must be acute.
Therefore, B must be π/3 (or 60°).
(Handwritten notes also show: `0° < C < 180°`, `sinB = √3/2`, `0° < B < 180°`, `B = π/3`).
**Part (2): Finding the area of △ABC (Choosing condition ①)**
We choose condition ①: b = 3√3, a = 2.
From Part (1), we know B = π/3.
By the Sine Rule: a/sinA = b/sinB.
Substitute the known values: 2/sinA = (3√3) / (√3/2).
2/sinA = 6.
sinA = 2/6 = 1/3.
Since △ABC is an acute triangle, A must be acute, so cosA > 0.
cosA = √(1 - sin²A) = √(1 - (1/3)²) = √(1 - 1/9) = √(8/9) = 2√2 / 3.
(Handwritten notes confirm `cosA = √(1-1/9) = 2√2/3`).
Now, we find sinC. Since A + B + C = π in a triangle, C = π - (A + B).
So, sinC = sin(π - (A + B)) = sin(A + B).
Using the sine addition formula: sin(A + B) = sinAcosB + cosAsinB.
Substitute the values:
sinC = (1/3) * cos(π/3) + (2√2 / 3) * sin(π/3).
sinC = (1/3) * (1/2) + (2√2 / 3) * (√3 / 2).
sinC = 1/6 + (2√6) / 6.
sinC = (1 + 2√6) / 6.
(Handwritten notes confirm this calculation: `sinC = (1/3)*(1/2) + (2√2/3)*(√3/2) = (1+2√6)/6`).
Finally, calculate the area of △ABC using the formula S_△ABC = (1/2)absinC.
S_△ABC = (1/2) * 2 * (3√3) * ((1 + 2√6) / 6).
S_△ABC = (3√3) * ((1 + 2√6) / 6).
S_△ABC = (√3 / 2) * (1 + 2√6).
S_△ABC = (√3 + 2√18) / 2.
S_△ABC = (√3 + 2 * 3√2) / 2.
S_△ABC = (√3 + 6√2) / 2.
S_△ABC = √3/2 + 3√2.
(Handwritten notes confirm this calculation: `S_△ABC = (1/2) * 2 * 3√3 * (1+2√6)/6 = (√3+6√2)/2 = √3/2 + 3√2`).
The student's solution is correct.
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**Supplementary Instruction:**
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