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