Law of Cosines/Proof 3/Obtuse Triangle

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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 $B$
$c$ is opposite $C$.

Let $\triangle ABC$ be an obtuse triangle such that $A$ is obtuse


Then:

$c^2 = a^2 + b^2 - 2a b \cos C$


Proof

Let $\triangle ABC$ be an obtuse triangle.

CosineRule-Proof3-obtuse.png

Let $AC$ be extended and $BD$ be dropped perpendicular to $AC$.

Let:

\(\ds h\) \(=\) \(\ds BD\)
\(\ds e\) \(=\) \(\ds CD\)
\(\ds f\) \(=\) \(\ds AD\)


We have that $\triangle CDB$ and $\triangle ADB$ are right triangles.

Hence:

\(\text {(1)}: \quad\) \(\ds c^2\) \(=\) \(\ds h^2 + f^2\) Pythagoras's Theorem
\(\text {(2)}: \quad\) \(\ds a^2\) \(=\) \(\ds h^2 + e^2\) Pythagoras's Theorem
\(\text {(3)}: \quad\) \(\ds e^2\) \(=\) \(\ds \paren {b + f}^2\)
\(\ds \) \(=\) \(\ds b^2 + f^2 + 2 b f\)
\(\text {(4)}: \quad\) \(\ds e\) \(=\) \(\ds a \cos C\) Definition of Cosine of Angle


Then:

\(\ds c^2\) \(=\) \(\ds h^2 + f^2\) from $(1)$
\(\ds \) \(=\) \(\ds a^2 - e^2 + f^2\) from $(2)$
\(\ds \) \(=\) \(\ds a^2 - b^2 - f^2 - 2 b f + f^2\) substituting for $e^2$ from $(3)$
\(\ds \) \(=\) \(\ds a^2 - b^2 - 2 b f + 2 b^2 - 2 b^2\) simplifying and adding and subtracting $2 b^2$
\(\ds \) \(=\) \(\ds a^2 + b^2 - 2 b \paren {b + f}\) rearranging
\(\ds \) \(=\) \(\ds a^2 + b^2 - 2 a b \cos C\) using $(4)$ to substitute $b + f = e$ with $a \cos C$

$\blacksquare$


Sources

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