AQA GCSE Maths

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(Sine and Cosine rule)

Sine Rule and Sine Area Rule

Sine Rule and Area Rule

 

The Sine Rule

The sine rule allows us to find missing sides or angles in any non-right-angled triangle.

For any triangle:

asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

\(a\}, \(b\), and cc are sides of the triangle

AA, BB, and CC are their opposite angles

 

diagram of labelled triangle for sine rule

 

When to Use It

You have a pair of an angle and its opposite side

You want to find a missing side or angle that also has a known opposite pair

 

Finding a Missing Side

To find side aa:

a=sinAbsinBa = \frac{\sin A \cdot b}{\sin B}

Finding a Missing Angle

To find angle AA:

sinA=asinBb\sin A = \frac{a \cdot \sin B}{b}

Then use:

A=sin1(asinBb)A = \sin^{-1}\left(\frac{a \cdot \sin B}{b}\right)

 

Ambiguous Case Warning

In some situations (especially when given two sides and an angle not between them), the triangle could be drawn in two different ways:

One angle is acute, the other is obtuse

Your calculator only returns the acute angle

Use: Obtuse angle=180°acute angle\text{Obtuse angle} = 180\degree - \text{acute angle}

 

Example: Using the Sine Rule

In triangle XYZ:

XY=10.5cmXY = 10.5 cm

YZ=14.2cmYZ = 14.2 cm

Angle XZY=31°XZY = 31\degree

 

diagram of triangle for example question

 

Step 1: Find angle X

Use:

sinX14.2=sin31°10.5 sinX=14.2sin(31)10.50.6993 X=sin1(0.6993)44.4°\frac{\sin X}{14.2} = \frac{\sin 31\degree}{10.5}  \\ \sin X = \frac{14.2 \cdot \sin(31)}{10.5} \approx 0.6993 \\  X = \sin^{-1}(0.6993) \approx 44.4\degree

Step 2: Find angle Y

Y=180°31°44.4°=104.6°Y = 180\degree - 31\degree- 44.4\degree = 104.6\degree

 

Area of a Triangle Using Sine

To find the area of any triangle (not just right-angled), use:

Area=12absinC\text{Area} = \frac{1}{2} ab \sin C

 

sine rule for calculating area of a triangle

 

Where:

  • aa and bb are two sides
  • CC is the included angle

When to Use This Formula

You know two sides and the angle between them

It avoids having to use height directly

 

Worked Example: Area of a Triangle

In triangle DEF:

DE=5.2mDE = 5.2 m

EF=7.1mEF = 7.1 m

D=64°\angle D = 64\degree

 

diagram of triangle for question

 

Calculate the Area

Area=125.27.1sin(64)\text{Area} = \frac{1}{2} \cdot 5.2 \cdot 7.1 \cdot \sin(64^\circ)

Area0.55.27.10.898816.6 m2 (3 s.f.)\text{Area} \approx 0.5 \cdot 5.2 \cdot 7.1 \cdot 0.8988 \approx 16.6 \text{ m}^2 \text{ (3 s.f.)}

 

 

Tuity Tip

Hover me!

Always label your triangle clearly.

Ensure angle and side pairs are used correctly.

Use inverse sine for angles.

Convert units if necessary before calculating area.

Round only at the final step unless otherwise asked.

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