Question-196261

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^2 =a^2 +(2a)^2 −2×a×2a×cos (π−β)=5a^2 +4a^2 cos β ⇒((MD^2 )/a^2 )=5+4 cos β MC^2 =c^2 +(2c)^2 −2×c×2c×cos β=5c^2 −4c^2 cos β ⇒((MC^2 )/c^2 )=5−4 cos β (a/b)=((MD)/a) ⇒(a^2 /b^2 )=((MD^2 )/a^2 )=5+4 cos β similarly ⇒(c^2 /d^2 )=((MC^2 )/c^2 )=5−4 cos β ⇒(a^2 /b^2 )+(c^2 /d^2 )=2×5=10 ✓

$${MD}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{2}{a}\right)^{\mathrm{2}} −\mathrm{2}×{a}×\mathrm{2}{a}×\mathrm{cos}\:\left(\pi−\beta\right)=\mathrm{5}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \mathrm{cos}\:\beta \\ $$$$\Rightarrow\frac{{MD}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$${MC}^{\mathrm{2}} ={c}^{\mathrm{2}} +\left(\mathrm{2}{c}\right)^{\mathrm{2}} −\mathrm{2}×{c}×\mathrm{2}{c}×\mathrm{cos}\:\beta=\mathrm{5}{c}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \mathrm{cos}\:\beta \\ $$$$\Rightarrow\frac{{MC}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$$\frac{{a}}{{b}}=\frac{{MD}}{{a}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{{MD}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$${similarly} \\ $$$$\Rightarrow\frac{{c}^{\mathrm{2}} }{{d}^{\mathrm{2}} }=\frac{{MC}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{d}^{\mathrm{2}} }=\mathrm{2}×\mathrm{5}=\mathrm{10}\:\checkmark \\ $$

Commented by MM42 last updated on 22/Aug/23

$$\mathscr{VERY}\:\:\:\mathscr{NICE} \\ $$

Commented by mr W last updated on 22/Aug/23

$$\mathfrak{thanks}\:\mathfrak{alot}! \\ $$

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