Law of sines and cosines

General triangle. In general form of the triangle  $ \triangle ABC$, the sides are denoted by $ a = \overline{BC}$, $ b = \overline{CA}$, and $ c = \overline{AB}$, and the angles are denoted by $ \alpha = \angle BAC$, $ \beta = \angle CBA$, and $ \gamma = \angle ACB$.

\includegraphics{lec15d.ps}

Inscribed angle theorem. $ \angle ABC$ is called an inscribed angle, and $ \angle AOC$ is called a central angle. Then they have the following relationship:

$\displaystyle \angle ABC = \frac{1}{2} \angle AOC .
$

\includegraphics{lec15a.ps}

Law of sines. Let $ R$ is the radius of the circumscribed circle of $ \triangle ABC$, and let $ AB'$ be the diameter of the circle.

\includegraphics{lec15b.ps}

Since $ \angle ABC = \angle AB'C = \beta$ by the inscribed angle theorem, we have $ \overline{AC} = \overline{AB'} \sin\angle AB'C
= 2R \sin\beta$; thus, we obtain

$\displaystyle 2R = \frac{\overline{AC}}{\sin \angle ABC}
= \frac{b}{\sin \beta} .
$

Since the choice of $ \angle ABC$ is arbitrary, in general we have

$\displaystyle 2R = \frac{a}{\sin \alpha}
= \frac{b}{\sin \beta}
= \frac{c}{\sin \gamma}.
$

Law of cosines. Let $ BD$ be perpendicular to $ AC$. The square length  $ \overline{BD}^2$ can be expressed in terms of the sides $ a, b, c$ and the angle $ \alpha$ via the two different right triangles  $ \triangle ABD$ and  $ \triangle BCD$.

$\displaystyle \overline{BD}^2$ $\displaystyle = \overline{AB}^2 - \overline{AD}^2 = c^2 - (c \cos\alpha)^2$ (2-1)
  $\displaystyle = \overline{BC}^2 - \overline{DC}^2 = a^2 - (b - c \cos\alpha)^2$ (2-2)

\includegraphics{lec15c.ps}

Comparing the rightmost expressions in (2-1)-(2-2), we obtain $ a^2 = b^2 + c^2 - 2bc \cos \alpha$. In general we have

    $\displaystyle a^2 = b^2 + c^2 - 2bc \cos \alpha$  
    $\displaystyle b^2 = c^2 + a^2 - 2ca \cos \beta$  
    $\displaystyle c^2 = a^2 + b^2 - 2ab \cos \gamma$  

Heron's formula. Let $ S$ be the area of  $ \triangle ABC$, and let $ s = \dfrac{a + b + c}{2}$. By using $ \cos\alpha = \dfrac{b^2 + c^2 - a^2}{2bc}$ via the law of cosines, we obtain

$\displaystyle S = \frac{1}{2} bc \sin\alpha
= \sqrt{\frac{1}{4} b^2 c^2 \sin^2\...
... \sqrt{\frac{1}{4} b^2 c^2 (1 - \cos^2\alpha)}
= \sqrt{s(s - a)(s - b)(s - c)}
$



Department of Mathematics
Last modified: 2005-09-29