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Question-168324

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Question Number 168324 by leicianocosta last updated on 07/Apr/22
Answered by mr W last updated on 08/Apr/22
Commented by mr W last updated on 08/Apr/22
c=x cos θ a=12−x cos θ b=18−x cos θ R=x sin θ 2(tan^(−1) ((12−x cos θ)/(x sin θ))+tan^(−1) ((18−x cos θ)/(x sin θ)))=π−∠C=2θ tan^(−1) ((12−x cos θ)/(x sin θ))+tan^(−1) ((18−x cos θ)/(x sin θ))=θ ((((12−x cos θ)/(x sin θ))+((18−x cos θ)/(x sin θ)))/(1−((12−x cos θ)/(x sin θ))×((18−x cos θ)/(x sin θ))))=tan θ x^2 =12×18 ⇒x=6(√6)
$${c}={x}\:\mathrm{cos}\:\theta \\ $$$${a}=\mathrm{12}−{x}\:\mathrm{cos}\:\theta \\ $$$${b}=\mathrm{18}−{x}\:\mathrm{cos}\:\theta \\ $$$${R}={x}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{2}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{12}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{18}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}\right)=\pi−\angle{C}=\mathrm{2}\theta \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{12}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{18}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}=\theta \\ $$$$\frac{\frac{\mathrm{12}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}+\frac{\mathrm{18}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}}{\mathrm{1}−\frac{\mathrm{12}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}×\frac{\mathrm{18}−{x}\:\mathrm{cos}\:\theta}{{x}\:\mathrm{sin}\:\theta}}=\mathrm{tan}\:\theta \\ $$$${x}^{\mathrm{2}} =\mathrm{12}×\mathrm{18} \\ $$$$\Rightarrow{x}=\mathrm{6}\sqrt{\mathrm{6}} \\ $$
Commented by Tawa11 last updated on 08/Apr/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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