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The Law of Cosines or Cosine Rule

Category : JEE Main & Advanced

In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of their included angle, that is for a triangle ABC,



(1) \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A\Rightarrow \cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]



(2) \[{{b}^{2}}={{c}^{2}}+{{a}^{2}}-2ca\cos B\Rightarrow \cos B=\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ca}\]    



(3) \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\Rightarrow \cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\]



Combining with \[\sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}\]



We have by division, \[\tan A=\frac{abc}{R({{b}^{2}}+{{c}^{2}}-{{a}^{2}})},\]



\[\tan B=\frac{abc}{R({{c}^{2}}+{{a}^{2}}-{{b}^{2}})},\,\,\tan C=\frac{abc}{R({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}\]




where, R be the radius of the circum-circle of the triangle ABC.

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