Trigonometry: sine, cosine, tangent, Pythagoras, sine and cosine rules

Trigonometry – Cambridge IGCSE Mathematics 0580 (Geometry C6)

This note covers every core requirement of the IGCSE syllabus (C6.1 – C6.2) and clearly marks the optional Extended (E) material (E6.3 – E6.6). It is written for learners who have a basic knowledge of algebra and geometry.

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1. Right‑angled triangles – terminology

• Hypotenuse: the side opposite the right angle.
• Opposite side: the side opposite the acute angle you are working with.
• Adjacent side: the side that is next to the acute angle and is not the hypotenuse.

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2. Trigonometric ratios (SOH‑CAH‑TOA)

For any acute angle θ in a right‑angled triangle:

\[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \]

Ratio	Definition	Formula
Sine	Opposite ÷ Hypotenuse	\(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}\)
Cosine	Adjacent ÷ Hypotenuse	\(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\)
Tangent	Opposite ÷ Adjacent	\(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\)

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3. Pythagoras’ theorem (C6.1)

In any right‑angled triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[ a^{2}+b^{2}=c^{2} \]

3.1 Proof / verification activity (Core)

Given a triangle with sides \(5\;\text{cm},\;12\;\text{cm},\;13\;\text{cm}\):

• Calculate \(5^{2}+12^{2}=25+144=169\).
• Calculate \(13^{2}=169\).
• Since the two results are equal, the triangle is right‑angled (the side of length 13 cm is the hypotenuse).

Use this same test whenever you need to confirm whether a triangle is right‑angled.

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4. Using trigonometric ratios

4.1 Finding a missing side (Core)

Example 1 – Side from a known angle and hypotenuse

In right‑angled triangle \(ABC\), \(\angle B = 30^{\circ}\) and the hypotenuse \(c = 10\text{ cm}\). Find the opposite side \(a\) and the adjacent side \(b\).

\[ a = c\sin30^{\circ}=10\times0.5=5\text{ cm} \qquad b = c\cos30^{\circ}=10\times\frac{\sqrt3}{2}=5\sqrt3\text{ cm} \]

4.2 Finding a missing acute angle (Core)

Example 2 – Angle from two sides

In a right‑angled triangle the opposite side is \(5\;\text{cm}\) and the hypotenuse is \(13\;\text{cm}\). Find the acute angle θ opposite the 5 cm side.

\[ \sin\theta=\frac{5}{13}\;\Longrightarrow\; \theta=\sin^{-1}\!\left(\frac{5}{13}\right)\approx22.6^{\circ} \]

4.3 Solving a right‑angled triangle in one step (Core)

Example 3 – Height of a tree

A tree is 20 m away from a point on the ground. The angle of elevation to the top of the tree is \(38^{\circ}\). Find the height of the tree.

\[ \text{height}=20\tan38^{\circ}\approx20\times0.781=15.6\text{ m} \]

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5. Real‑world applications (C6.2)

5.1 Angles of elevation & depression

• Angle of elevation: measured upward from the horizontal.
• Angle of depression: measured downward from the horizontal.
• The horizontal line always forms the adjacent side of the right‑angled triangle.

Example 4 – Angle of depression from a cliff

The top of a cliff is 30 m high. From the top the angle of depression to a boat is \(22^{\circ}\). Find the horizontal distance from the base of the cliff to the boat.

\[ \tan22^{\circ}=\frac{30}{\text{distance}} \;\Longrightarrow\; \text{distance}=\frac{30}{\tan22^{\circ}}\approx74\text{ m} \]

5.2 Bearings (C6.2)

A bearing is measured clockwise from North (0° → 360°). To use trigonometric ratios you must convert the bearing to an acute angle relative to the appropriate cardinal direction.

Example 5 – Bearing conversion

A surveyor walks 150 m on a bearing of \(210^{\circ}\). What are the north‑south (N‑S) and east‑west (E‑W) components of the displacement?

• 210° lies in the South‑West quadrant. The acute angle with the South direction is \(210^{\circ}-180^{\circ}=30^{\circ}\).
• South component: \(150\cos30^{\circ}=150\times\frac{\sqrt3}{2}=129.9\text{ m (south)}\).
• West component: \(150\sin30^{\circ}=150\times0.5=75\text{ m (west)}\).

5.3 Other real‑life contexts (Core)

Example 6 – Navigation of a rescue helicopter

A helicopter is 8 km north of a stranded hiker. The pilot sees the hiker at an angle of elevation of \(12^{\circ}\). Assuming the helicopter flies at a constant altitude of 1 km, find the horizontal distance the pilot must travel to be directly above the hiker.

\[ \tan12^{\circ}=\frac{1}{\text{horizontal distance}} \;\Longrightarrow\; \text{horizontal distance}=\frac{1}{\tan12^{\circ}}\approx4.8\text{ km} \]

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6. Calculator reminder (Core)

All IGCSE questions use degrees. Before entering any trigonometric function, set your calculator to Degree mode. Forgetting this is a common source of error.

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7. Extended (E) – Sine & Cosine Rules (E6.3 – E6.6)

These rules are not required for the core syllabus but are useful for challenge questions and for the next level of study.

• Law of sines (any triangle with sides \(a,b,c\) opposite angles \(A,B,C\)): \[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \]
• Law of cosines (any triangle): \[ c^{2}=a^{2}+b^{2}-2ab\cos C \] (and the two cyclic permutations). This reduces to Pythagoras’ theorem when \(C=90^{\circ}\) because \(\cos90^{\circ}=0\).

Example 7 – Sine rule (Extended)

In \(\triangle PQR\), \(P=45^{\circ}\), \(Q=60^{\circ}\) and side \(p=8\text{ cm}\) (opposite \(P\)). Find side \(q\).

\[ \frac{p}{\sin P}=\frac{q}{\sin Q} \;\Longrightarrow\; q=8\frac{\sin60^{\circ}}{\sin45^{\circ}} =8\frac{\sqrt3/2}{\sqrt2/2}=8\sqrt{\frac{3}{2}}\approx9.8\text{ cm} \]

Example 8 – Cosine rule (Extended)

In \(\triangle XYZ\), \(x=5\text{ cm}\), \(y=7\text{ cm}\) and the included angle \(Z=120^{\circ}\). Find side \(z\).

\[ z^{2}=x^{2}+y^{2}-2xy\cos Z =25+49-2(5)(7)\bigl(-\tfrac12\bigr)=74+35=109 \] \[ z=\sqrt{109}\approx10.44\text{ cm} \]

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8. Quick reference sheet

Concept	Key formula	When to use
Sine, cosine, tangent (Core)	\(\sin\theta=\dfrac{\text{opp}}{\text{hyp}},\; \cos\theta=\dfrac{\text{adj}}{\text{hyp}},\; \tan\theta=\dfrac{\text{opp}}{\text{adj}}\)	Right‑angled triangles – find a missing side or acute angle.
Pythagoras (Core)	\(a^{2}+b^{2}=c^{2}\)	Find a side or confirm a right angle.
Angle of elevation / depression (Core)	\(\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}\)	Heights, distances, ladders, towers, etc.
Bearing problems (Core)	Convert bearing → acute angle, then use \(\sin,\cos,\tan\)	Navigation, map‑based questions.
Law of sines (Extended)	\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)	Any triangle – two angles & one side, or two sides & a non‑included angle.
Law of cosines (Extended)	\(c^{2}=a^{2}+b^{2}-2ab\cos C\) (cyclic)	Any triangle – two sides & included angle, or three sides to find an angle.

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9. Common pitfalls & how to avoid them

• Opposite vs. adjacent: Identify the side relative to the *specific* acute angle you are solving for.
• Calculator mode: Always set to degrees before using \(\sin,\cos,\tan\).
• Angle of elevation vs. depression: Both use \(\tan\theta\); the only difference is the direction in which the angle is measured.
• Bearing conversion: Reduce the bearing to an acute angle measured from the nearest cardinal direction (N, E, S, W) before applying trigonometric ratios.
• Pythagoras verification: Never assume a triangle is right‑angled; check \(a^{2}+b^{2}=c^{2}\) first.
• Extended material misuse: Apply the sine or cosine rule only when the triangle is *not* right‑angled, and match each side with its opposite angle.

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10. Summary of what you must know for the Core exam

1. Definitions of hypotenuse, opposite and adjacent.
2. SOH‑CAH‑TOA ratios and how to rearrange them to find missing sides *or* missing acute angles.
3. Pythagoras’ theorem, including a quick verification test.
4. Use of \(\tan\) for angles of elevation and depression.
5. How to convert a bearing to an acute angle and then apply trigonometric ratios.
6. Always work in degrees on the calculator.

Once these are mastered, you can safely attempt any Core C6 question and, if you wish, explore the optional Extended material on the sine and cosine rules.