Sine of Angle of Triangle by Semiperimeter

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: $5.95$