CDS Mathematics · Triangles

Sine and Cosine Rules

The sides of a triangle are in the ratio of the sines of the opposite angles, and the cosine rule gives a side from the other two and the angle between them.

All Triangles notes CDS Mathematics strategy

Why this matters

Six PYQs, three of them HARD. The sine rule turns an angle ratio into a side ratio, the cosine rule handles a 60° or 120° angle, and splitting the area with ½ab sin C finds a line drawn from a vertex. These are the only places where CDS Triangles leans on trigonometry.

Concept 1 of 2: The sine rule

Draw the circumcircle. Every side is a chord, and a chord facing an angle

A

at the circle has length

2R\sin A

. So each side is

2R

times the sine of the angle opposite it.

Definition

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ , where $R$ is the circumradius.
So $a : b : c = \sin A : \sin B : \sin C$ .
Angles $30^\circ, 30^\circ, 120^\circ$ give sides $1 : 1 : \sqrt3$ ; angles $30^\circ, 60^\circ, 90^\circ$ give $1 : \sqrt3 : 2$ ; angles $36^\circ, 72^\circ, 72^\circ$ give base : leg $= \dfrac{\sqrt5 - 1}{2}$ .

Sine rule

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

Worked example

In triangle

ABC

A = 45^\circ

B = 60^\circ

and

a = 2\sqrt2

. Find

b

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — Elementary Mathematics · Q82Hard

Example 1 · Triangles · Sine and Cosine Rules

The angles of a triangle are in the ratio 1 : 1 : 4. If the perimeter of the triangle is k times its largest side, then what is the value of k ?

Sines, not angles

Angles in the ratio

1 : 2 : 3

give sides

\sin 30^\circ : \sin 60^\circ : \sin 90^\circ = 1 : \sqrt3 : 2

, not

1 : 2 : 3

Drill 2 more on the sine rule

Concept 2 of 2: The cosine rule and ½ab sin C

The cosine rule is Pythagoras with a correction for an angle that is not

90^\circ

: subtract when the angle is acute, add when it is obtuse. The area formula

\dfrac12 ab\sin C

lets a triangle be cut along a line from a vertex into two pieces whose areas add up.

Definition

$a^2 = b^2 + c^2 - 2bc\cos A$ .
$A = 60^\circ$ : $a^2 = b^2 + c^2 - bc$ . $A = 120^\circ$ : $a^2 = b^2 + c^2 + bc$ . $A = 90^\circ$ : Pythagoras.
Area: $\Delta = \dfrac12 bc\sin A$ . If a line $CD$ from $C$ splits $\angle C$ into two parts, the areas of the two pieces add to the whole, which gives an equation for $CD$ .

Cosine rule and area

a^2 = b^2 + c^2 - 2bc\cos A, \qquad \Delta = \tfrac12 bc\sin A

Worked example

In triangle

ABC

b = 5

c = 8

and

A = 60^\circ

. Find

a

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q64Moderate

Example 2 · Triangles · Sine and Cosine Rules

ABC is an equilateral triangle. The side BC is trisected at D such that BC = 3 BD. What is the ratio of

AD^2

AB^2

cos 120° is negative

120^\circ

the term

-2bc\cos A

becomes

+bc

, so the opposite side is longer than Pythagoras would give. Dropping the sign gives the

60^\circ

answer, which is always an option.

Drill 2 more on the cosine rule and ½ab sin c

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The sine rule
Sine rule
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$
The cosine rule and ½ab sin C
Cosine rule and area
$a^2 = b^2 + c^2 - 2bc\cos A, \qquad \Delta = \tfrac12 bc\sin A$

Watch out for (2)

Sines, not angles
→ The sine rule
cos 120° is negative
→ The cosine rule and ½ab sin C

Test yourself on Triangles

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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