Table of Contents

cos rules or Laws of cos

Cos rule or law of cos states that in A $\triangle$ ABC a, b and c are lengths of sides BC CA and AB. Then

$$a^2=b^2+c^2-2bc \ Cos \ A$$

$$b^2=a^2+c^2-2ac\ Cos\ B$$

$$c^2=a^2+b^2-2ab\ Cos\ C$$


$$cos\ B= \frac{a^2+c^2-b^2}{2ac}$$

$$cos\ C= \frac{b^2+a^2-c^2}{2ba}$$

$$cos\ A= \frac{b^2+c^2-a^2}{2bc}$$

It seems like a different form of Pythagoras theorem for every triangle doesn’t it?


Let’s consider a $\triangle$ ABC as shown in the figure with a, b and c being the lengths of the sides BC, CA and AB. we can represent the triangle vectorically as shown in the diagram. That means $\vec{AB}$, $\vec{BC}$ and $\vec{AC}$ are forming the triangle so we can understand that | $\vec{AB}$ | = c, | $\vec{BC}$ | = a and | $\vec{AC}$ | = b.

From the triangle we can write that


Apply mod on both sides

$$| \vec{AC} |=| \vec{AB} + \vec{BC} |$$

$$ b = | \vec{AB} + \vec{BC} |$$

From vectors we have $| \vec{x} + \vec{y} | = \sqrt {x^2+y^2+2xy\ cos(\vec{x},\vec{y})}$

To find the angle between $\vec{AB}$ and $\vec{BC}$ lets extend the $\vec{AB}$ so that both vector tails meet. Let $\alpha$ be the angle between those two vectors.

(note: we extended the vector to find the angle between two vectors by not considering B as angle, because angle between two vectors is angle made by vectors when both heads meet or tails meet but not one tail and one head)

From the above figure we can find $\alpha$

$$\alpha + B=180^0$$

$$\alpha =180^0 - B $$


$$ b = | \vec{AB} + \vec{BC} |$$

$$ b =\sqrt{ | \vec{AB}|^2 +| \vec{BC} |^2 +2 | \vec{AB} | |\vec{BC}| cos(\vec{AB} , \vec{BC})}$$

$$ b =\sqrt{ c^2 + a^2 +2 ca\ cos(\alpha)}$$

$$ b =\sqrt{ c^2 + a^2 +2 ca\ cos(180^0 - B)}$$

$$ b =\sqrt{ c^2 + a^2 -2 ca\ cos\ B}$$

$$b^2=a^2+c^2-2ac\ Cos\ B$$

Hence proved.

Now we can extend this proof by simplification and we can write

$$cos\ B= \frac{a^2+c^2-b^2}{2ac}$$

Similarly we can do the same for the other two angles and we can prove the other two equations of cos rule


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