Area of a triangle

In a nutshell

The trigonometric area of a triangle can be used to find the area of a triangle without being given the perpendicular height.

Area of a triangle formula

The trigonometric formula for the area of a triangle is used when you have two sides and an angle between them. This is sufficient information to find the area. The area is given to be:

$Area=\dfrac{1}{2}ab\sin(C)$

This is for a triangle with sides $a,b,c$ and corresponding angles $A,B,C$ .

Maths; Trigonometry; KS5 Year 12; Area of a triangle

Example 1

A triangle has two sides with lengths $2cm$ and $4cm$ and the angle between them is $24^\circ$ . Find the area of the triangle to two decimal places.

Sketch and label the sides and angles of this triangle first.

Maths; Trigonometry; KS5 Year 12; Area of a triangle

Now, use the formula:

$Area=\dfrac{1}{2}ab\sin(C)$

Substitute in known values:

$Area = \dfrac{1}{2}(2)(4)\sin(24)$

$Area =4\times \sin(24)=1.62694....$

$\underline{Area\,=1.63cm^2 \space (2\space d.p.)}$

Note: If a triangle is given as a sketch, the labels will not necessarily match the variables in the formula for the area of a triangle. In this case it is possible to adjust the formula to fit the situation.

Example 2

In $\triangle ABC$ , $BC= (x+3) \ cm$ , $AC = x \ cm$ and $\angle {BCA} = 130^{\circ}$ . The area of this triangle is $6cm^2$ . What is the value of $x$ to two decimal places?

$\angle BCA$ is between sides $AC$ and $BC$ .

Use the formula for area:

$Area=\dfrac{1}{2}ab\sin(C)$

Substitute in known values:

$6 = \dfrac12(x+3)(x)\sin(130)$

Solve for $x$ :

$\begin{aligned}6 &=\dfrac12(x+3)(x)\sin(130) \\ \\ 12 &= (x+3)(x)\sin(130) \\ \\ \dfrac{12}{\sin(130)} &= (x^2+3x) \\ \\ 0 &= x^2+3x - \dfrac{12}{\sin(130)} \end{aligned}$

Solve the quadratic:

$\dfrac{-(3)\pm\sqrt {(3)^2 - 4(1)(-\dfrac{12}{\sin(130)})}}{2(1)} = \dfrac{-3\pm\sqrt {9 +\dfrac{48}{\sin(130)}}}{2}$

$x= 2.73259... \ , -5.73259...$

A side length must be a positive value:

$\underline{x= 2.73 \ cm \space (2\space d.p.)}$ 