Mathematics » Trigonometry » Area, Sine, And Cosine Rules

The Area Rule

Area, Sine, and Cosine Rules

There are three identities relating to the trigonometric functions that make working with triangles easier:

  1. the area rule

  2. the sine rule

  3. the cosine rule

The Area Rule

Optional Investigation: The area rule

  1. Consider \(\triangle ABC\):


    Complete the following:

    1. Area \(\triangle ABC\) = \(\cfrac{1}{2} \times \ldots \times AC\)

    2. \(\sin \hat{B} = \ldots\) and \(AC = \ldots \times \ldots\)

    3. Therefore area \(\triangle ABC\) = \(\ldots \times \ldots \times \ldots \times \ldots\)

  2. Consider \(\triangle A’B’C’\):


    Complete the following:

    1. How is \(\triangle A’B’C’\) different from \(\triangle ABC\)?

    2. Calculate area \(\triangle A’B’C’\).

  3. Use your results to write a general formula for determining the area of \(\triangle PQR\):


For any \(\triangle ABC\) with \(AB = c, BC = a\) and \(AC = b\), we can construct a perpendicular height (\(h\)) from vertex \(A\) to the line \(BC\):


In \(\triangle ABC\):

\begin{align*} \sin \hat{B}&= \cfrac{h}{c} \\ \therefore h &= c \sin \hat{B} \end{align*}

And we know that

\begin{align*} \text{Area } \triangle ABC &= \cfrac{1}{2} \times a \times h \\ &= \cfrac{1}{2} \times a \times c \sin \hat{B} \\ \therefore \text{Area } \triangle ABC &= \cfrac{1}{2} ac \sin \hat{B} \end{align*}

Alternatively, we could write that

\begin{align*} \sin \hat{C}&= \cfrac{h}{b} \\ \therefore h &= b \sin \hat{C} \end{align*}

And then we would have that

\begin{align*} \text{Area } \triangle ABC &= \cfrac{1}{2} \times a \times h \\ &= \cfrac{1}{2} ab \sin \hat{C} \end{align*}

Similarly, by constructing a perpendicular height from vertex \(B\) to the line \(AC\), we can also show that area \(\triangle ABC = \cfrac{1}{2} bc \sin \hat{A}\).

The area rule

In any \(\triangle ABC\):

\begin{align*} \text{Area } \triangle ABC &= \cfrac{1}{2} bc \sin \hat{A} \\ &= \cfrac{1}{2} ac \sin \hat{B} \\ &= \cfrac{1}{2} ab \sin \hat{C} \end{align*}



Find the area of \(\triangle ABC\) (correct to two decimal places):


Use the given information to determine unknown angles and sides

\[\begin{array}{rll} AB = AC &= 7 & (\text{given}) \\ \therefore \hat{B} = \hat{C} &= \text{50}\text{°} & (\angle \text{s opp. equal sides}) \\ \text{And } \hat{A} &= \text{180}\text{°} – \text{50}\text{°} – \text{50}\text{°} & (\angle \text{s sum of } \triangle ABC) \\ \therefore \hat{A} &= \text{80}\text{°} & \end{array}\]

Use the area rule to calculate the area of \(\triangle ABC\)

Notice that we do not know the length of side \(a\) and must therefore choose the form of the area rule that does not include this side of the triangle.

In \(\triangle ABC\):

\[\begin{array}{rll} \text{Area } &= \cfrac{1}{2} bc \sin \hat{A} & \\ &= \cfrac{1}{2} (7)(7) \sin \text{80}\text{°} & \\ &= \text{24.13} & \end{array}\]

Write the final answer

Area of \(\triangle ABC = \text{24.13}\) square units.



Show that the area of \(\triangle DEF = \cfrac{1}{2} df \sin \hat{E}\).


Construct a perpendicular height \(h\)

Draw \(DH\) such that \(DH \perp EF\) and let \(DH = h\), \(D\hat{E}F = \hat{E}_1\) and \(D\hat{E}H = \hat{E}_2\).

In \(\triangle DHE\):

\[\begin{array}{rll} \sin \hat{E}_2 &= \dfrac{h}{f} & \\ h &= f \sin (\text{180}\text{°} – \hat{E}_1) & (\angle \text{s on str. line})\\ &= f \sin \hat{E}_1 & \end{array}\]

Use the area rule to calculate the area of \(\triangle DEF\)

In \(\triangle DEF\):

\begin{align*} \text{Area } &= \cfrac{1}{2} d \times h \\ &= \cfrac{1}{2} df \sin \hat{E}_1 \end{align*}

The area rule

In any \(\triangle PQR\):


The area rule states that the area of any triangle is equal to half the product of the lengths of the two sides of the triangle multiplied by the sine of the angle included by the two sides.

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