Law of cosines: Pythagoras generalized to any triangle

The formula

In triangle $ABC$ (any triangle):

$a^2 = b^2 + c^2 - 2bc \cos(\widehat{A})$

with $a = BC$ (opposite $A$), $b = AC$, $c = AB$.

Similarly for $B, C$:

• $b^2 = a^2 + c^2 - 2ac \cos(\widehat{B})$
• $c^2 = a^2 + b^2 - 2ab \cos(\widehat{C})$

Special case: $\widehat{A} = 90°$

$\cos(90°) = 0$, so: $a^2 = b^2 + c^2$

Pythagoras! Cosine rule contains Pythagoras.

Two uses

Use 1: find a side from two sides and the included angle

$b = 5$, $c = 7$, $\widehat{A} = 60°$. Find $a$.

$a^2 = 25 + 49 - 2 \times 5 \times 7 \times 0.5 = 39$ $a \approx 6.24$.

Use 2: find an angle from 3 sides

Isolate $\cos$: $\cos(\widehat{A}) = \dfrac{b^2 + c^2 - a^2}{2bc}$

Example: $a = 4, b = 5, c = 6$. Find $\widehat{A}$.

$\cos A = (25 + 36 - 16)/60 = 0.75$ $\widehat{A} \approx 41.4°$.

Common mistakes

1. Confusing angle and side: side $a$ is opposite to angle $\widehat{A}$.
2. Forgetting the sign in "$- 2bc \cos A$".
3. Using Pythagoras on a non-right triangle.

?À toi de jouer

Triangle with b = 3, c = 4, angle A = 90°. Find a.

A.5 (cosine rule becomes Pythagoras)B.7C.√7D.12

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