Question Number 144662 by mnjuly1970 last updated on 27/Jun/21
$$\:\:\:\:\:\:\:………..\:\:\mathrm{Calculus}……….. \\ $$$$\:\mathrm{In}\:\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\:: \\ $$$$\hat {\mathrm{B}}\:=\:\mathrm{2}\:\hat {\mathrm{C}}\:\:\:\:,\:\:{a}\:\:=\:\lambda\:{b}\:\:\:{then}\:{specify} \\ $$$$\:{the}\:\:{limits}\:{of}\:{the}\:{changes}\:\:\:'\:\:\lambda\:\:'\:\:: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by mindispower last updated on 27/Jun/21
![(a/b)=λ (a/(sin(A)))=(b/(sin(B)))=(c/(sin(C))) λ=((sin(π−(3c)))/(sin(2c)))=((sin(3c))/(sin(c))) sin(3c)=3sin(c)−4sin^2 (c) λ=3−4sin^2 (c) A+B+C=π A+3C=π A=π−3C>0⇒0<C<(π/3) sin^2 (C)≤sin^2 ((π/3))=(3/4) λ≥3−((4.3)/4)=0 λ∈]0,3[](https://www.tinkutara.com/question/Q144670.png)
$$\frac{{a}}{{b}}=\lambda \\ $$$$\frac{{a}}{{sin}\left({A}\right)}=\frac{{b}}{{sin}\left({B}\right)}=\frac{{c}}{{sin}\left({C}\right)} \\ $$$$\lambda=\frac{{sin}\left(\pi−\left(\mathrm{3}{c}\right)\right)}{{sin}\left(\mathrm{2}{c}\right)}=\frac{{sin}\left(\mathrm{3}{c}\right)}{{sin}\left({c}\right)} \\ $$$${sin}\left(\mathrm{3}{c}\right)=\mathrm{3}{sin}\left({c}\right)−\mathrm{4}{sin}^{\mathrm{2}} \left({c}\right) \\ $$$$\lambda=\mathrm{3}−\mathrm{4}{sin}^{\mathrm{2}} \left({c}\right) \\ $$$${A}+{B}+{C}=\pi \\ $$$${A}+\mathrm{3}{C}=\pi \\ $$$${A}=\pi−\mathrm{3}{C}>\mathrm{0}\Rightarrow\mathrm{0}<{C}<\frac{\pi}{\mathrm{3}} \\ $$$${sin}^{\mathrm{2}} \left({C}\right)\leqslant{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\lambda\geqslant\mathrm{3}−\frac{\mathrm{4}.\mathrm{3}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left.\lambda\in\right]\mathrm{0},\mathrm{3}\left[\right. \\ $$