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Area of a Triangle

Syllabus

determine the area of any triangle, given two sides and an included angle, by using the rule A=12absinCA=\frac{1}{2}ab\sin C, and solve related practical problems

For non-right angled trigonometry, we label the triangles as shown below:

Important

Sides are denoted with a lowercase letter and angles uppercase. A\angle A is opposite aa, B\angle B is opposite bb and C\angle C is opposite cc.

note

A labelled triangle is not on your reference sheet.

Deriving The Formula (Extension)

In stages 4 and 5.2, to calculate the area of a triangle you would have used the formula A=12bhA=\frac{1}{2}bh where bb is the length of the base and hh is the perpendicular height.

Using right-angled trigonometry we can express hh in terms of the side lengths and angles of the triangle.

sinC=opphypsinC=haasinC=hh=asinC\begin{aligned}\sin C&=\frac{\text{opp}}{\text{hyp}}\\\sin C&=\frac{h}{a}\\ a\sin C&=h\\h&=a\sin C\end{aligned}

Which can be substituted into A=12bhA=\frac{1}{2}bh to give A=12absinCA=\frac{1}{2}ab\sin C

Applying The Formula

Important

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

note
note

Don't forget that A=12bhA=\frac{1}{2}bh is still a valid way of calculating the area of a triangle.

Calculate the area of the triangle below. Answer correct to 3 significant figures.

Solution
A=12absinCA=12(8.2)(10.7)sin32(Substitute)A=23.247...A23.2 cm2(Rounded to appropraite numberof s.f. and include units)\begin{aligned} A&=\frac{1}{2}ab\sin C\\ A&=\frac{1}{2}(8.2)(10.7)\sin 32^\circ&&\text{(Substitute)}\\ A&=23.247...\\ A&\approx23.2\text{ cm}^2&&\text{(Rounded to appropraite number}\\ &&&\text{of s.f. and include units)} \end{aligned}