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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