Question Number 196261 by sonukgindia last updated on 21/Aug/23
Answered by mr W last updated on 22/Aug/23
Commented by mr W last updated on 22/Aug/23
$${MD}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{2}{a}\right)^{\mathrm{2}} −\mathrm{2}×{a}×\mathrm{2}{a}×\mathrm{cos}\:\left(\pi−\beta\right)=\mathrm{5}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \mathrm{cos}\:\beta \\ $$$$\Rightarrow\frac{{MD}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$${MC}^{\mathrm{2}} ={c}^{\mathrm{2}} +\left(\mathrm{2}{c}\right)^{\mathrm{2}} −\mathrm{2}×{c}×\mathrm{2}{c}×\mathrm{cos}\:\beta=\mathrm{5}{c}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \mathrm{cos}\:\beta \\ $$$$\Rightarrow\frac{{MC}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$$\frac{{a}}{{b}}=\frac{{MD}}{{a}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{{MD}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$${similarly} \\ $$$$\Rightarrow\frac{{c}^{\mathrm{2}} }{{d}^{\mathrm{2}} }=\frac{{MC}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{d}^{\mathrm{2}} }=\mathrm{2}×\mathrm{5}=\mathrm{10}\:\checkmark \\ $$
Commented by MM42 last updated on 22/Aug/23
$$\mathscr{VERY}\:\:\:\mathscr{NICE} \\ $$
Commented by mr W last updated on 22/Aug/23
$$\mathfrak{thanks}\:\mathfrak{alot}! \\ $$