The Magic of 120° in IGCSE Mathematics | IGCSE数学中120°的奥秘

📚 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° is exactly one-third of a full turn (360° ÷ 3 = 120°). It is also two-thirds of a straight angle (180° × 2 ÷ 3 = 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°

The sum of exterior angles of any polygon is 360°. Since 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

If the radius is 9 cm, the 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

This simplified form is worth remembering. For example, if a = 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

When you are given two sides and a non-included angle, there may be two possible triangles. The obtuse angle 120° often appears 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

Notice that sin 120° = sin 60°. This is 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

A bearing is measured clockwise from north. A bearing of 120° means the direction is 120° clockwise from north. This is equivalent to 30° east of south (since 180° − 120° = 60°? Wait: 120° from north = South-East? Let’s calculate: North = 0°, East = 90°, South = 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” is 30° south 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？

Solution: Since interior + 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) = √109 cm ≈ 10.44 cm; (3) Each interior 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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The Magic of 120° in IGCSE Mathematics | IGCSE数学中120°的奥秘