Area of Triangle in Terms of Two Sides and Angle/Proof 2
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Theorem
The area of a triangle $ABC$ is given by:
- $\dfrac 1 2 a b \sin C$
where:
Proof
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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Area of the triangle