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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

$${MD}^2=a^2+{\left(2a\right)}^2-2\times a\times 2a\times \cos (\pi -\beta )=5a^2+4a^2\cos \beta$$$$\Rightarrow \frac{{MD}^2}{a^2}=5+4\cos \beta$$$${MC}^2=c^2+{\left(2c\right)}^2-2\times c\times 2c\times \cos \beta =5c^2-4c^2\cos \beta$$$$\Rightarrow \frac{{MC}^2}{c^2}=5-4\cos \beta$$$$\frac{a}{b}=\frac{MD}{a}$$$$\Rightarrow \frac{a^2}{b^2}=\frac{{MD}^2}{a^2}=5+4\cos \beta$$$$similarly$$$$\Rightarrow \frac{c^2}{d^2}=\frac{{MC}^2}{c^2}=5-4\cos \beta$$$$\Rightarrow \frac{a^2}{b^2}+\frac{c^2}{d^2}=2\times 5=10✓$$

Commented by MM42 last updated on 22/Aug/23

$$\mathcal{VERY}\mathcal{NICE}$$

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

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

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