Question Number 72841 by indalecioneves last updated on 03/Nov/19
$${Could}\:{someone}\:{help}\:{me}\:{on}\:{this}\:{question}? \\ $$$${Knowing}\:{that}\:{the}\:{area}\:{of}\:{a}\:{circle}\:{segment}\:{is}\:{given}\:{by}\:{A}={R}^{\mathrm{2}} \left(\theta−{sin}\theta\right)/\mathrm{2}.\:{Where}\:{A}=\mathrm{7}{m}^{\mathrm{2}} ;\:{R}^{\mathrm{2}} =\frac{\mathrm{28}}{\pi}. \\ $$$${What}\:{is}\:{the}\:{best}\:{answer}\:{for}\:{the}\:{angle}\:{value}\:\left({degree}\right) \\ $$$$\left.{a}\right)\:\mathrm{85}°<\theta<\mathrm{90}° \\ $$$$\left.{b}\right)\:\mathrm{95}°<\theta<\mathrm{100}° \\ $$$$\left.{c}\right)\:\mathrm{105}°<\theta<\mathrm{110}° \\ $$$$\left.{d}\right)\:\mathrm{115}°<\theta<\mathrm{120}° \\ $$$$\left.{e}\right)\:\mathrm{125}°<\theta<\mathrm{135}° \\ $$
Commented by MJS last updated on 04/Nov/19
$$\mathrm{there}'\mathrm{s}\:\mathrm{information}\:\mathrm{missing} \\ $$$$\mathrm{what}\:\mathrm{is}\:{m}?\:\mathrm{if}\:{m}\:\mathrm{is}\:\mathrm{just}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{factor}: \\ $$$$\mathrm{7}{m}^{\mathrm{2}} =\frac{\mathrm{28}}{\mathrm{2}\pi}\left(\theta−\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{is}\:{R}^{\mathrm{2}} \pi=\mathrm{28} \\ $$$$\Rightarrow \\ $$$$\mathrm{0}\leqslant\mathrm{7}{m}^{\mathrm{2}} \leqslant\mathrm{28} \\ $$$$\mathrm{0}\leqslant{m}^{\mathrm{2}} \leqslant\mathrm{4} \\ $$$$\mathrm{area}\geqslant\mathrm{0}\:\Rightarrow\:{m}\geqslant\mathrm{0} \\ $$$$\mathrm{0}\leqslant{m}\leqslant\mathrm{2} \\ $$$$\Rightarrow \\ $$$$\mathrm{0}\leqslant\theta−\mathrm{sin}\:\theta\leqslant\mathrm{2}\pi \\ $$$$\Rightarrow \\ $$$$\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$$$\mathrm{in}\:\mathrm{degree}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{360} \\ $$
Commented by indalecioneves last updated on 09/Nov/19
$${Hello},\:{Sir}! \\ $$$${Thanks}\:{for}\:{attention}. \\ $$$${You}\:{can}\:{only}\:{considerer}\:{it}\:\mathrm{1}=\frac{\mathrm{2}}{\pi}.\left(\theta−{sin}\theta\right),\:{because}\:{m}^{\mathrm{2}} \:{means}\:{only}\:{meters}\:{per}\:{square}. \\ $$$$ \\ $$
Commented by MJS last updated on 09/Nov/19
$$\mathrm{you}\:\mathrm{should}\:\mathrm{have}\:\mathrm{written}\:{R}^{\mathrm{2}} =\frac{\mathrm{28}}{\pi}{m}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{it} \\ $$$$\mathrm{would}\:\mathrm{have}\:\mathrm{been}\:\mathrm{clear} \\ $$
Answered by MJS last updated on 09/Nov/19
$$\left(\mathrm{1}\right)\:\:\mathrm{7}=\frac{\mathrm{28}}{\mathrm{2}\pi}\left(\theta−\mathrm{sin}\:\theta\right) \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{generally}\:\mathrm{0}\leqslant\theta<\mathrm{2}\pi\:\mathrm{and}\:−\mathrm{1}\leqslant\mathrm{sin}\:\theta\:\leqslant\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\:\mathrm{the}\:\mathrm{area}\:{A}=\mathrm{7}\:\mathrm{is}\:\mathrm{a}\:\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{is}\:{R}^{\mathrm{2}} \pi=\mathrm{28} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\:\Rightarrow\:\mathrm{sin}\:\theta\:=\theta−\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:−\mathrm{1}\leqslant\theta−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{1}\:\Leftrightarrow\:\frac{\pi}{\mathrm{2}}−\mathrm{1}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{but}\:\mathrm{0}\leqslant\theta\:\Rightarrow\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\:\Rightarrow\:\frac{\pi}{\mathrm{2}}<\theta<\pi \\ $$$$\:\:\:\:\:\mathrm{but}\:\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1}\:\Rightarrow\:\frac{\pi}{\mathrm{2}}<\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{or}\:\mathrm{90}°<\theta\leqslant\mathrm{147}.\mathrm{30}° \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{segment} \\ $$$$\mathrm{with}\:\theta=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\left(=\mathrm{120}°\right) \\ $$$$\frac{\mathrm{28}}{\mathrm{2}\pi}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{\mathrm{28}}{\mathrm{3}}−\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\pi}\approx\mathrm{5}.\mathrm{47} \\ $$$$\Rightarrow\:\theta>\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}\pi}{\mathrm{3}}<\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1}\:\mathrm{or}\:\mathrm{120}°<\theta\leqslant\mathrm{147}.\mathrm{30}° \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{options}\:\left({e}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{one} \\ $$
Commented by indalecioneves last updated on 12/Nov/19
$${Thank}\:{you},\:{Sir}. \\ $$$${God}\:{bless}\:{you}! \\ $$