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Question-216471

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Question Number 216471 by Ismoiljon_008 last updated on 08/Feb/25
Commented by Ismoiljon_008 last updated on 08/Feb/25
help, please
$$\:\:\:{help},\:{please} \\ $$
Commented by mr W last updated on 08/Feb/25
a solution is given in Q207018. an other solution see below.
$${a}\:{solution}\:{is}\:{given}\:{in}\:{Q}\mathrm{207018}. \\ $$$${an}\:{other}\:{solution}\:{see}\:{below}. \\ $$
Commented by Ismoiljon_008 last updated on 08/Feb/25
thank you
$$\:\:\:{thank}\:{you} \\ $$
Answered by mr W last updated on 08/Feb/25
Commented by mr W last updated on 08/Feb/25
α+β=(π/2) p=2r sin α, q=2r sin β (p/(sin θ))=(4/(sin α)) (q/(sin θ))=(5/(sin β)) (p/q)=((4 sin β)/(5 sin α)) ⇒ ((sin α)/(sin β))=((4 sin β)/(5 sin α)) ⇒5 sin^2 α=4 sin^2 β=4 cos^2 α ⇒tan^2 α=(4/5) ⇒tan α=(2/( (√5))) ⇒sin α=(2/3) ((2r sin α)/(sin θ))=(4/(sin α)) ⇒sin θ=((r sin^2 α)/2)=(r/2)×((2/3))^2 =((2r)/9) (2r)^2 =10^2 +9^2 −2×10×9 cos (π−2θ) 4r^2 =181+180 cos (2θ) 4r^2 =181+180(1−2×((4r^2 )/(81))) ⇒r=((57)/(14)) ✓
$$\alpha+\beta=\frac{\pi}{\mathrm{2}} \\ $$$${p}=\mathrm{2}{r}\:\mathrm{sin}\:\alpha,\:{q}=\mathrm{2}{r}\:\mathrm{sin}\:\beta \\ $$$$\frac{{p}}{\mathrm{sin}\:\theta}=\frac{\mathrm{4}}{\mathrm{sin}\:\alpha} \\ $$$$\frac{{q}}{\mathrm{sin}\:\theta}=\frac{\mathrm{5}}{\mathrm{sin}\:\beta} \\ $$$$\frac{{p}}{{q}}=\frac{\mathrm{4}\:\mathrm{sin}\:\beta}{\mathrm{5}\:\mathrm{sin}\:\alpha}\:\Rightarrow\:\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}=\frac{\mathrm{4}\:\mathrm{sin}\:\beta}{\mathrm{5}\:\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\mathrm{5}\:\mathrm{sin}^{\mathrm{2}} \:\alpha=\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\beta=\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\alpha \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \:\alpha=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{\mathrm{2}{r}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta}=\frac{\mathrm{4}}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{r}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}}=\frac{{r}}{\mathrm{2}}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{2}{r}}{\mathrm{9}} \\ $$$$\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{9}\:\mathrm{cos}\:\left(\pi−\mathrm{2}\theta\right) \\ $$$$\mathrm{4}{r}^{\mathrm{2}} =\mathrm{181}+\mathrm{180}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right) \\ $$$$\mathrm{4}{r}^{\mathrm{2}} =\mathrm{181}+\mathrm{180}\left(\mathrm{1}−\mathrm{2}×\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{81}}\right) \\ $$$$\Rightarrow{r}=\frac{\mathrm{57}}{\mathrm{14}}\:\checkmark \\ $$

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