Area of Triangle in Terms of Two Sides and Angle

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Corollary to Area of Triangle in Terms of Side and Altitude

The area of a triangle $ABC$ is given by:

$\dfrac 1 2 a b \sin C$

where:

$a, b$ are two of the sides
$C$ is the angle of the vertex opposite its other side $c$.


Proof 1

TriangleAreaTwoSidesAngle.png
\(\displaystyle \map \Area {ABC}\) \(=\) \(\displaystyle \frac 1 2 h c\) Area of Triangle in Terms of Side and Altitude
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 h \paren {p + q}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 a b \paren {\frac p a \frac h b + \frac h a \frac q b}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 a b \paren {\sin \alpha \cos \beta + \cos \alpha \sin \beta}\) Definition of Sine of Angle and Definition of Cosine of Angle
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 a b \, \map \sin {\alpha + \beta}\) Sine of Sum
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 a b \sin C\)

$\blacksquare$


Proof 2

TriangleAreaTwoSidesAngle-2.png

By definition of sine:

$h = b \sin C$

From Area of Triangle in Terms of Side and Altitude:

$\map \Area {ABC} = \dfrac {a h} 2$

Substituting:

$\map \Area {ABC} = \dfrac {a b \sin C} 2$

$\blacksquare$


Sources