Question Number 217783 by malwan last updated on 21/Mar/25
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Commented by malwan last updated on 21/Mar/25
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Commented by malwan last updated on 21/Mar/25
$${I}\:{cant}\:{send}\:{a}\:{photo}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by mr W last updated on 21/Mar/25
$${you}\:{seem}\:{still}\:{to}\:{have}\:{the}\:{old} \\ $$$${problem}. \\ $$
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
$${I}\:{will}\:{try}\:{sir} \\ $$
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
$${only}\:{if}\:{it}\:{is}\:{also}\:{given}\:{that}\:{PX}\bot{XY}. \\ $$
Commented by mr W last updated on 23/Mar/25
Commented by mr W last updated on 23/Mar/25
$$\mathrm{tan}\:\alpha=\frac{{XY}}{{XP}}=\frac{\mathrm{2}{r}}{{r}}=\mathrm{2} \\ $$$$\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\alpha=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$$\angle{PXQ}=\mathrm{2}\beta=\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right)=\pi−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$
Commented by malwan last updated on 23/Mar/25
$${fantastic}\:, \\ $$$${thank}\:{you}\:{so}\:{much}\:{sir}\:\:\cancel{\lesseqgtr} \\ $$