Sine of Angle of Triangle by Semiperimeter
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Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
- $\sin A = \dfrac 2 {b c} \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $\sin$ denotes sine and $s$ is the semiperimeter: $s = \dfrac {a + b + c} 2$.
Proof
Let $Q$ be the area of $\triangle ABC$.
From Area of Triangle in Terms of Two Sides and Angle:
- $Q = \dfrac {b c \sin A} 2$
From Heron's Formula:
- $Q = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
Equating the two:
- $\dfrac {b c \sin A} 2 = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
from which follows the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.95$