📚 The Magic of 120° in IGCSE Mathematics | IGCSE数学中120°的奥秘
The number 120 appears again and again in IGCSE Mathematics – as an interior angle, a circle theorem angle, a trigonometric value, and a key part of many real-world calculations. This article explores every place where 120° might show up in your Edexcel IGCSE exams, with clear explanations and worked examples.
数字120在IGCSE数学中反复出现——它可以是多边形的内角、圆相关定理的角度、三角函数值，也是许多实际问题计算中的关键部分。本文将带你逐一梳理120°在Edexcel IGCSE考试中可能出现的所有场景，提供清晰的解释和例题。
1. Why 120°? | 为什么是120°？
120°). This makes it a natural angle in equilateral-like structures and regular polygons.
120°恰好是整个圆周的三分之一（360° ÷ 3 = 120°），也是平角的三分之二（180° × 2 ÷ 3 = 120°）。因此它在等边类结构和正多边形中自然出现。
Because 120° is between 90° and 180°, it is classified as an obtuse angle. In trigonometry, it lies in the second quadrant, where sine is positive and cosine and tangent are negative.
由于120°介于90°和180°之间，它属于钝角。在三角函数中，它位于第二象限，此时正弦为正，余弦和正切为负。
2. Interior Angles of Regular Polygons | 正多边形的内角
For a regular polygon with n sides, each interior angle is given by the formula:
Interior angle = (n − 2) × 180° ÷ n
Setting this equal to 120° gives:
(n − 2) × 180 ÷ n = 120
180n − 360 = 120n
60n = 360
n = 6
So a regular polygon with interior angle 120° is a regular hexagon. This is a classic IGCSE question: you must be able to work both forwards (from n to angle) and backwards (from angle to n).
令内角等于120°，解得n = 6，因此每个内角为120°的正多边形是正六边形。这是IGCSE经典题型：你必须既能从边数求角度，也能从角度反求边数。
3. Exterior Angles | 外角
The exterior angle of a regular polygon is 180° minus the interior angle. For a regular hexagon:
Exterior angle = 180° − 120° = 60°
e each exterior angle is 60°, the number of sides is 360 ÷ 60 = 6, confirming the hexagon.
正多边形的外角等于180°减去内角。对于正六边形，外角为60°。任意多边形的外角和为360°，因为每个外角为60°，所以边数为360 ÷ 60 = 6，再次验证是六边形。
- If interior angle = 120°, exterior angle = 60°.
- If interior angle = 120°, then n = 360 ÷ 60 = 6.
- 若内角 = 120°，则外角 = 60°。
- 若内角 = 120°，则边数 n = 360 ÷ 60 = 6。
4. Angles Around a Point | 周角问题
At a point, the total angle is 360°. Three equal angles of 120° exactly fill the space:
120° + 120° + 120° = 360°
This is why three regular hexagons meet at a point without gaps – a fact used in tiling and tessellation questions. In the IGCSE exam, you may be asked to explain why regular hexagons tessellate.
在一点处，总角度为360°。三个120°的角刚好填满空间。这就是为什么三个正六边形可以在一个顶点处无间隙地拼接——这是密铺问题中常用的事实。考试中可能会要求你解释正六边形为什么能密铺。
5. Circle Theorem: Angle at the Centre | 圆心角定理
In a circle, the angle at the centre is twice the angle at the circumference subtended by the same arc. If the angle at the circumference is 60°, then the angle at the centre is 120°.
在圆中，同弧所对的圆心角等于圆周角的两倍。若圆周角为60°，则圆心角为120°。
Angle at centre = 2 × 60° = 120°
Several circle theorem questions use this relationship. For example, if a chord subtends an angle of 60° at the circumference, the central angle is 120°, so the triangle formed by the two radii and the chord is an isosceles triangle with vertex angle 120°.
许多圆定理题目都运用这一关系。例如，若一条弦在圆周上张成60°角，则圆心角为120°，由两条半径和弦构成的三角形就是顶角为120°的等腰三角形。
6. Area of a Sector | 扇形面积
A sector with angle 120° is exactly one-third of a full circle. If the radius is r, the area of the sector is:
Sector area = (θ ÷ 360) × πr² = (120 ÷ 360) × πr² = (1 ÷ 3)πr²
ตัวอย่าง: If r = 6 cm, then sector area = (1/3) × π × 6² = 12π cm² ≈ 37.7 cm².
120°的扇形恰好是整个圆的三分之一。若半径为r，扇形面积为(θ/360)×πr² = (1/3)πr²。例如r = 6 cm时，扇形面积为12π cm² ≈ 37.7 cm²。
7. Arc Length | 弧长
The arc length of a sector is also proportional to the angle. For a radius r and angle 120°:
Arc length = (θ ÷ 360) × 2πr = (120 ÷ 360) × 2πr = (1 ÷ 3) × 2πr = (2πr) ÷ 3
e arc length is (2π × 9) ÷ 3 = 6π cm ≈ 18.8 cm. Remember to use the same units throughout.
弧长同样与角度成正比。半径为r、角度为120°时，弧长为(2πr)/3。若半径为9 cm，弧长为6π cm ≈ 18.8 cm。注意全程使用相同单位。
8. The Cosine Rule with 120° | 余弦定理中的120°
The cosine rule relates the sides of a triangle to an included angle:
c² = a² + b² − 2ab cos C
If C = 120°, then cos 120° = −0.5, so the rule becomes:
c² = a² + b² − 2ab × (−0.5) = a² + b² + ab
= 3 cm, b = 5 cm, and the included angle is 120°, then c² = 9 + 25 + 15 = 49, so c = 7 cm.
余弦定理将三角形的边与夹角联系起来：c² = a² + b² − 2ab cos C。当C = 120°时，cos 120° = −0.5，于是c² = a² + b² + ab。这个简化形式值得记住。例如a = 3 cm，b = 5 cm，夹角为120°时，c² = 9 + 25 + 15 = 49，所以c = 7 cm。
9. The Sine Rule and the Ambiguous Case | 正弦定理与模糊情况
The sine rule states:
a ÷ sin A = b ÷ sin B = c ÷ sin C
s as one of the possibilities. For example, if sin B = 0.866, then B could be 60° or 120°.
正弦定理为a/sin A = b/sin B = c/sin C。当已知两边和一非夹角时，可能存在两个三角形。钝角120°经常作为其中一个可能解出现。例如，若sin B = 0.866，则B可能是60°或120°。
| Angle | sin | cos | tan |
| 60° | √3/2 ≈ 0.866 | 0.5 | √3 ≈ 1.732 |
| 120° | √3/2 ≈ 0.866 | −0.5 | −√3 ≈ −1.732 |
why the ambiguous case exists: the same sine value corresponds to two different angles in the range 0° to 180°.
注意sin 120° = sin 60°。这就是模糊情形存在的原因：相同的正弦值在0°到180°范围内对应两个不同的角。
10. Trigonometric Exact Values | 精确三角函数值
You are expected to know the exact values for 120°. Using the unit circle:
sin 120° = √3/2
cos 120° = −1/2
tan 120° = −√3
These values appear in non-calculator papers. A common way to derive them is to use the reference angle 60° and then adjust the sign according to the quadrant.
考试要求掌握120°的精确三角函数值。单位圆中：sin 120° = √3/2，cos 120° = −1/2，tan 120° = −√3。这些值会出现在非计算器试卷中。常用方法是先求参考角60°的函数值，再根据象限调整符号。
11. Bearing and Direction Problems | 方位角与方向问题
20° clockwise from north. This is equivalent to 30° east of south (since 180° − 120° = 60°? Wait: 120° from north = South-East? Let' 180°. 120° lies between 90° and 180°, so it is south-east, specifically 180° − 120° = 60°? Actually from south, the angle is 180° − 120° = 60° west of south? No: 120° is 30° south of east (because 120° − 90° = 30°). So "120° bearing" outh of east. Check: East is 90°, so 120° is 30° towards south from east. That is correct.
方位角是从正北方向顺时针测量的角度。方位角120°表示从正北顺时针转120°，即南偏东30°（因为120° − 90° = 30°）。在IGCSE中，你可能会被要求绘制或解读这样的方位角，并结合余弦定理求距离。
In vector or trigonometry problems, bearings of 120° lead to angles of 60° between the direction and the east–west line, allowing you to use sine and cosine rules effectively.
在向量或三角函数问题中，方位角120°会导致方向与东西线之间形成60°角，从而可以有效地使用正弦和余弦定理。
12. Worked Exam Questions | 考试例题精讲
Here are two typical IGCSE questions involving 120°.
以下两道典型IGCSE题目都涉及120°。
Example 1: A regular polygon has interior angle 120°. How many sides does it have?
例题1：一个正多边形的内角为120°，它有几条边？
rior + exterior = 180°, exterior = 60°. Number of sides = 360° ÷ 60° = 6. Answer: 6 sides.
解：因为内角 + 外角 = 180°，所以外角 = 60°。边数 = 360° ÷ 60° = 6。答案：6条边。
Example 2: A sector has radius 8 cm and angle 120°. Find its area and arc length.
例题2：一个扇形半径为8 cm，圆心角为120°。求其面积和弧长。
Area = (120/360) × π × 8² = (1/3) × 64π = 64π/3 cm² ≈ 67.0 cm².
Arc length = (120/360) × 2π × 8 = (1/3) × 16π = 16π/3 cm ≈ 16.8 cm.
面积 = (120/360) × π × 8² = 64π/3 cm² ≈ 67.0 cm²；弧长 = (120/360) × 2π × 8 = 16π/3 cm ≈ 16.8 cm。
13. Common Mistakes to Avoid | 常见错误提醒
- Using 120° as an acute angle in right-angled trigonometry – it is not valid for SOH-CAH-TOA.
- Forgetting that cos 120° is negative.
- Mixing up the sector area formula with the arc length formula.
- Assuming the interior angle formula gives the exterior angle.
- 在直角三角函数中把120°当作锐角使用——它不适用于SOH-CAH-TOA。
- 忘记cos 120°是负数。
- 混淆扇形面积公式和弧长公式。
- 误把内角公式的结果当作外角。
Write down the quadrants: for 120°, sin positive, cos and tan negative. This small habit will save you many marks.
记住象限符号：对于120°，正弦为正，余弦和正切为负。这个小习惯能为你保住很多分数。
14. Practice Makes Perfect | 熟能生巧
Try these questions on your own:
请独立尝试以下题目：
- Find the exact value of tan 120°.
- A triangle has sides 5 cm and 7 cm with included angle 120°. Find the third side.
- Three regular hexagons meet at a point. Explain why they tessellate perfectly.
- 求tan 120°的精确值。
- 一个三角形的两条边分别为5 cm和7 cm，夹角为120°，求第三边。
- 三个正六边形在一个顶点相遇。解释它们为什么能完美密铺。
Answers: (1) −√3; (2) √25+49+35 terior angle is 120°, so 3 × 120° = 360°, exactly filling the full angle around the point.
答案：(1) −√3；(2) √25+49+35 = √109 cm ≈ 10.44 cm；(3) 每个内角为120°，三个内角之和为3 × 120° = 360°，正好填满一点周围的全部角度。
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