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Question Number 212630 by efronzo1 last updated on 19/Oct/24

Commented by efronzo1 last updated on 19/Oct/24

$$\:\: \\ $$

Answered by mr W last updated on 19/Oct/24

Commented by mr W last updated on 19/Oct/24

tan α=((h−(b/2))/(b/2))=((2h−b)/b) tan 2α=((a−(b/2))/h)=((2a−b)/(2h)) ((2a−b)/(2h))=((2×((2h−b)/b))/(1−(((2h−b)/b))^2 )) ((2a−b)/(2h))=((2×((2h−b)/b))/(1−(((2h−b)/b))^2 )) let λ=((2h)/b) ((2((a/b))−1)/λ)=((2(λ−1))/(1−(λ−1)^2 )) λ=((4((a/b)))/(2((a/b))+1))=((2h)/b) ⇒h=((2ab)/(2a+b)) ✓

$$\mathrm{tan}\:\alpha=\frac{{h}−\frac{{b}}{\mathrm{2}}}{\frac{{b}}{\mathrm{2}}}=\frac{\mathrm{2}{h}−{b}}{{b}} \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{{a}−\frac{{b}}{\mathrm{2}}}{{h}}=\frac{\mathrm{2}{a}−{b}}{\mathrm{2}{h}} \\ $$$$\frac{\mathrm{2}{a}−{b}}{\mathrm{2}{h}}=\frac{\mathrm{2}×\frac{\mathrm{2}{h}−{b}}{{b}}}{\mathrm{1}−\left(\frac{\mathrm{2}{h}−{b}}{{b}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}{a}−{b}}{\mathrm{2}{h}}=\frac{\mathrm{2}×\frac{\mathrm{2}{h}−{b}}{{b}}}{\mathrm{1}−\left(\frac{\mathrm{2}{h}−{b}}{{b}}\right)^{\mathrm{2}} } \\ $$$${let}\:\lambda=\frac{\mathrm{2}{h}}{{b}} \\ $$$$\frac{\mathrm{2}\left(\frac{{a}}{{b}}\right)−\mathrm{1}}{\lambda}=\frac{\mathrm{2}\left(\lambda−\mathrm{1}\right)}{\mathrm{1}−\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\lambda=\frac{\mathrm{4}\left(\frac{{a}}{{b}}\right)}{\mathrm{2}\left(\frac{{a}}{{b}}\right)+\mathrm{1}}=\frac{\mathrm{2}{h}}{{b}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{2}{ab}}{\mathrm{2}{a}+{b}}\:\:\:\checkmark \\ $$

Commented by Spillover last updated on 20/Oct/24

great

$${great} \\ $$

Answered by Spillover last updated on 19/Oct/24

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