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 $$A,B,C$$.

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.

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:

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