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Question Number 12170 by indreshpatelindresh@435gmail.i last updated on 15/Apr/17
In any △ABC,   2(bc cos A+ca cos B+ab cos C) =
$$\mathrm{In}\:\mathrm{any}\:\bigtriangleup{ABC}, \\ $$$$\:\mathrm{2}\left({bc}\:\mathrm{cos}\:{A}+{ca}\:\mathrm{cos}\:{B}+{ab}\:\mathrm{cos}\:{C}\right)\:= \\ $$
Answered by mrW1 last updated on 15/Apr/17
a^2 =b^2 +c^2 −2bccos A  ⇒2bccos A=b^2 +c^2 −a^2   similarily  ⇒2cacos B=c^2 +a^2 −b^2   ⇒2abcos C=a^2 +b^2 −c^2      ⇒2(bc cos A+ca cos B+ab cos C) =a^2 +b^2 +c^2
$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:{A} \\ $$$$\Rightarrow\mathrm{2}{bc}\mathrm{cos}\:{A}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${similarily} \\ $$$$\Rightarrow\mathrm{2}{ca}\mathrm{cos}\:{B}={c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{ab}\mathrm{cos}\:{C}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$ \\ $$$$\:\Rightarrow\mathrm{2}\left({bc}\:\mathrm{cos}\:{A}+{ca}\:\mathrm{cos}\:{B}+{ab}\:\mathrm{cos}\:{C}\right)\:={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$

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