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Question Number 147768 by aliibrahim1 last updated on 23/Jul/21

Answered by liberty last updated on 23/Jul/21

$$LetBC=x\Rightarrow \tan 44°=\frac{CD}{x}$$$$\Rightarrow CD=x\tan 44°$$$$\mathrm{L}etAC=y\Rightarrow \tan 48°=\frac{CD}{y}$$$$\Rightarrow CD=y\tan 48°$$$$wherey=\sqrt{{30}^2-x^2}=\sqrt{900-x^2}$$$$(\bullet )CD=CD$$$$\Rightarrow x\tan 44°=\sqrt{900-x^2}\tan 48°$$$$\Rightarrow x^2\tan {}^{2}44°=(900-x^2)\tan {}^{2}48°$$$$\Rightarrow x^2(\tan {}^{2}44°+{\tan }^{2}48°)=900\times \tan {}^{2}48°$$$$\Rightarrow x=\sqrt{\frac{900\times {\tan }^{2}48°}{\tan {}^{2}44°+\tan {}^{2}48°}}\approx 22.64$$$$\therefore CD=22.64\times \tan 44°\approx 21.87$$

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