Area of Triangle in Terms of Two Sides and Angle

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$


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