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Question Number 217783 by malwan last updated on 21/Mar/25

Commented by malwan last updated on 21/Mar/25

Commented by malwan last updated on 21/Mar/25

$$Icantsendaphoto\underset{―}{\underbrace{⋚}}$$

Commented by mr W last updated on 21/Mar/25

$$youseemstilltohavetheold$$$$problem.$$

Commented by Tinku Tara last updated on 21/Mar/25

$$\mathrm{Maybe}\mathrm{your}\mathrm{image}\mathrm{is}\mathrm{too}\mathrm{big}.$$$$\mathrm{Try}\mathrm{uploading}\mathrm{a}\mathrm{smaller}\mathrm{image}$$

Commented by malwan last updated on 22/Mar/25

$$Iwilltrysir$$

Commented by malwan last updated on 22/Mar/25

Commented by malwan last updated on 22/Mar/25

Commented by mr W last updated on 23/Mar/25

$$onlyifitisalsogiventhatPX\mathrm{\perp }XY.$$

Commented by mr W last updated on 23/Mar/25

Commented by mr W last updated on 23/Mar/25

$$\tan \alpha =\frac{XY}{XP}=\frac{2r}{r}=2$$$$\Rightarrow \alpha ={\tan }^{-1}2$$$$\beta =\frac{\pi }{2}-\alpha =\frac{\pi }{2}-{\tan }^{-1}2$$$$\mathrm{\angle }PXQ=2\beta =2(\frac{\pi }{2}-{\tan }^{-1}2)=\pi -2{\tan }^{-1}2$$

Commented by malwan last updated on 23/Mar/25

$$fantastic,$$$$thankyousomuchsir\cancel{⋚}$$

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