三角形面积计算

질문

在三角形ABC中，已知AB=1，AC=2，B-C=2π/3，求三角形的面积

사고

Analyzing angle relationships

I’m working through the angles in triangle ABC, noting that the maximum difference between B and C is π, leading to π/3. This helps pinpoint the needed angle B and angle C for calculating the area.

Relating angles to sides

First, I express angle B. Then, I apply the law of sines to link angle C to side B, using trigonometric identities to solve for sin(C + 120°), involving cos 120° and sin 120°.

Establishing associations

OK, let's cross-multiply and use trigonometric identities to link angle C to tan C. From there, angle B and angle A are derived, completing the angle relationships.

Crunching the calculations

I’m piecing together the formula for the area, which involves calculating sin(angle A) using sin and cos of angle C. This requires determining tan C and then finding the values of sin C and cos C.

Breaking down the components

I start by pulling together sin(angle A) using the sine of a difference formula, leveraging the values for sin(pi/3), cos(pi/3), cos(2C), and sin(2C).

Thought for 7 seconds

답변

在三角形 $ABC$ 中，已知边长 $AB = 1$ ， $AC = 2$ ，且角 $B - C = \frac{2\pi}{3}$ 。我们需要求出这个三角形的面积。

步骤 1：确定角度关系

根据三角形的内角和公式：

A + B + C = \pi

已知 $B = C + \frac{2\pi}{3}$ ，代入上式得到：

A + (C + \frac{2\pi}{3}) + C = \pi \\ A + 2C = \pi - \frac{2\pi}{3} = \frac{\pi}{3} \\ A = \frac{\pi}{3} - 2C

步骤 2：应用正弦定理

根据正弦定理：

\frac{AB}{\sin C} = \frac{AC}{\sin B} \\ \frac{1}{\sin C} = \frac{2}{\sin (C + \frac{2\pi}{3})}

利用三角函数加法公式：

\sin\left(C + \frac{2\pi}{3}\right) = \sin C \cos \frac{2\pi}{3} + \cos C \sin \frac{2\pi}{3} = -\frac{1}{2}\sin C + \frac{\sqrt{3}}{2}\cos C

代入得：

-\frac{1}{2}\sin C + \frac{\sqrt{3}}{2}\cos C = 2\sin C \\ \Rightarrow \sqrt{3}\cos C = 5\sin C \\ \Rightarrow \tan C = \frac{\sqrt{3}}{5}

步骤 3：计算角 $A$ 的正弦值

设 $\tan C = \frac{\sqrt{3}}{5}$ ，则：

\sin C = \frac{\sqrt{3}}{\sqrt{3 + 25}} = \frac{\sqrt{3}}{2\sqrt{7}} \\ \cos C = \frac{5}{2\sqrt{7}}

利用倍角公式：

\sin(2C) = 2\sin C \cos C = \frac{5\sqrt{3}}{14} \\ \cos(2C) = 1 - 2\sin^2 C = \frac{11}{14}

因此：

\sin A = \sin\left(\frac{\pi}{3} - 2C\right) = \sin\frac{\pi}{3} \cos(2C) - \cos\frac{\pi}{3} \sin(2C) \\ = \frac{\sqrt{3}}{2} \cdot \frac{11}{14} - \frac{1}{2} \cdot \frac{5\sqrt{3}}{14} \\ = \frac{11\sqrt{3}}{28} - \frac{5\sqrt{3}}{28} = \frac{6\sqrt{3}}{28} = \frac{3\sqrt{3}}{14}

步骤 4：计算面积

三角形的面积公式为：

\text{面积} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A = \frac{1}{2} \cdot 1 \cdot 2 \cdot \sin A = \sin A

因此，面积为：

\frac{3\sqrt{3}}{14}

最终答案：

三角形的面积为 $\frac{3\sqrt{3}}{14}$ 。

질문

사고

답변

步骤 1：确定角度关系

步骤 2：应用正弦定理

步骤 3：计算角 AAA 的正弦值

步骤 4：计算面积

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步骤 3：计算角 $A$ 的正弦值