正弦定理zhèngxián dìnglǐ

Core Concept

The Law of Sines is an essential tool for solving triangles, describing the proportional relationship between side lengths and the sines of their opposite angles.

Theorem Statement

In $\triangle ABC$, let $a, b, c$ be the sides opposite to angles $A, B, C$ respectively, and $R$ be the circumradius:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

Theorem Proof

Method 1: Area Method

Let $S$ be the area of $\triangle ABC$:

$S = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C$

From the first two expressions: $bc\sin A = ac\sin B$ $\frac{a}{\sin A} = \frac{b}{\sin B}$

Other relationships can be proved similarly.

Method 2: Circumcircle Method

Let $R$ be the circumradius of $\triangle ABC$. Draw diameter $AD$ through point $A$, connect $BD$.

Since $\angle ABD = 90°$ (angle in semicircle) and $\angle D = \angle C$ (angles subtended by same arc):

$\sin C = \sin D = \frac{AB}{AD} = \frac{c}{2R}$

Therefore: $\frac{c}{\sin C} = 2R$

Other relationships follow similarly.

Applications

1. Given Two Angles and One Side (AAS/ASA)

Example: In $\triangle ABC$, $A = 60°$, $B = 45°$, $a = \sqrt{3}$, find $b$.

Solution: $\frac{b}{\sin B} = \frac{a}{\sin A}$ $b = \frac{a \cdot \sin B}{\sin A} = \frac{\sqrt{3} \cdot \sin 45°}{\sin 60°} = \sqrt{2}$

2. Given Two Sides and One Angle (SSA)

Example: In $\triangle ABC$, $a = 8$, $b = 7$, $A = 60°$, find $B$.

Solution: $\sin B = \frac{b \cdot \sin A}{a} = \frac{7 \cdot \sin 60°}{8} = \frac{7\sqrt{3}}{16}$

Note: This case may have two solutions (ambiguous case)

3. Side-Angle Conversion

From $\frac{a}{\sin A} = \frac{b}{\sin B}$: $a : b : c = \sin A : \sin B : \sin C$

Number of Solutions (SSA Case)

Given $a, b, A$, to find $B$ where $\sin B = \frac{b\sin A}{a}$:

1. No solution: $\frac{b\sin A}{a} > 1$
2. One solution:
   • $\frac{b\sin A}{a} = 1$ (when $B = 90°$)
   • $b \geq a$ (B is unique)
3. Two solutions: $\frac{b\sin A}{a} < 1$ and $b < a$ (one acute, one obtuse)

CSCA Practice Problems

💡 Note: The following problems are designed according to CSCA exam syllabus and Chinese standardized testing format.

[Example 1] Basic (Difficulty ★★★☆☆)

In $\triangle ABC$, $a = 3$, $b = 5$, $\sin A = \frac{1}{3}$, find $\sin B$.

Solution:

By the law of sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$

$\sin B = \frac{b \cdot \sin A}{a} = \frac{5 \cdot \frac{1}{3}}{3} = \frac{5}{9}$

Answer: $\sin B = \frac{5}{9}$

Common Misconceptions

❌ Misconception 1: Forgetting to check number of solutions

Wrong: Not checking if there are 0, 1, or 2 solutions in SSA case

Correct: Must analyze the conditions to determine number of solutions

❌ Misconception 2: Wrong circumradius formula

Wrong: $\frac{a}{\sin A} = R$

Correct: $\frac{a}{\sin A} = 2R$

Study Tips

1. ✅ Understand the essence: Sides are proportional to sines of opposite angles
2. ✅ Memorize formula: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$
3. ✅ Master applications: AAS, ASA, SSA cases
4. ✅ Check solution count: Especially for SSA case

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💡 Exam Tip: The Law of Sines is a core tool for solving triangles, mandatory in CSCA exams! Accounts for about 50% of triangle-solving problems.