Triangle

# The Projection Law

The projection law states that in any triangle $ABC$, $a=b\cos C+c\cos B$ $b=c\cos A+a\cos C$ $c=a\cos B+b\cos A$

These formulae express the algebraic sum of the projections of any two sides on the third side in terms of the third side.

Let $ABC$ be a triangle. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively.

If $R$ be the circum-radius of the triangle $ABC$, then from sine law, we have, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\therefore a=2R\sin A$ $\therefore b=2R\sin B$ $\therefore c=2R\sin C$

Now, $b\cos C+c\cos B$ $=2R\sin B\sin C+2R\sin B\cos B$ $=2R(\sin B\cos C+\cos C\sin B)$ $=2R\sin(B+C)$ $=2R\sin(π-A)\;\;\;(\because A+B+C=π)$ $=2R\sin A$ $=a$

We can also prove this by using the cosine law. From the cosine law, we have, $\cos A=\frac{b^2+c^2-a^2}{2bc}$ $\cos B=\frac{c^2+a^2-b^2}{2ca}$ $\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, $b\cos C+c\cos B$ $=b\left(\frac{a^2+b^2-c^2}{2ab}\right)+c\left(\frac{c^2+a^2-b^2}{2ca}\right)$ $=\frac{1}{2a}(a^2+b^2-c^2+c^2+a^2-b^2)$ $=\frac{1}{2a}\cdot 2a^2$ $=a$

In the same way, we can prove, $b=c\cos A+a\cos C$ $c=a\cos B+b\cos A$

[Also see: Vector Method to Prove the Projection Law]

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