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Question-22714




Question Number 22714 by selestian last updated on 22/Oct/17
Answered by ajfour last updated on 22/Oct/17
λ =a+b+c   let   l=bcos^2 ((C/2))+ccos^2 ((B/2))  ⇒    2l=b(1+cos C)+c(1+cos B)  ⇒    2l=b+c+(bcos C+ccos B)  or    2l=b+c+a    ⇒   l= =(((a+b+c))/2) = (λ/2) .
$$\lambda\:={a}+{b}+{c}\: \\ $$$${let}\:\:\:{l}={b}\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{C}}{\mathrm{2}}\right)+{c}\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{B}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{l}={b}\left(\mathrm{1}+\mathrm{cos}\:{C}\right)+{c}\left(\mathrm{1}+\mathrm{cos}\:{B}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{l}={b}+{c}+\left({b}\mathrm{cos}\:{C}+{c}\mathrm{cos}\:{B}\right) \\ $$$${or}\:\:\:\:\mathrm{2}{l}={b}+{c}+{a} \\ $$$$\:\:\Rightarrow\:\:\:{l}=\:=\frac{\left({a}+{b}+{c}\right)}{\mathrm{2}}\:=\:\frac{\lambda}{\mathrm{2}}\:. \\ $$
Commented by selestian last updated on 22/Oct/17
thAnk you so much sir...
$${thAnk}\:{you}\:{so}\:{much}\:{sir}… \\ $$

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