Could-someone-help-me-on-this-question-Knowing-that-the-area-of-a-circle-segment-is-given-by-A-R-2-sin-2-Where-A-7m-2-R-2-28-pi-What-is-the-best-answer-for-the-angle-value-degree-a-85

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Question Number 72841 by indalecioneves last updated on 03/Nov/19

$$Couldsomeonehelpmeonthisquestion?$$$$KnowingthattheareaofacirclesegmentisgivenbyA=R^2(\theta -sin\theta )/2.WhereA=7m^2;R^2=\frac{28}{\pi }.$$$$Whatisthebestanswerfortheanglevalue\left(degree\right)$$$$a)85°<\theta <90°$$$$b)95°<\theta <100°$$$$c)105°<\theta <110°$$$$d)115°<\theta <120°$$$$e)125°<\theta <135°$$

Commented by MJS last updated on 04/Nov/19

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{information}\mathrm{missing}$$$$\mathrm{what}\mathrm{is}m?\mathrm{if}m\mathrm{is}\mathrm{just}\mathrm{a}\mathrm{constant}\mathrm{factor}:$$$$7m^2=\frac{28}{2\pi }(\theta -\sin \theta )$$$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{is}{R}^2\pi =28$$$$\Rightarrow$$$$0⩽7m^2⩽28$$$$0⩽m^2⩽4$$$$\mathrm{area}⩾0\Rightarrow m⩾0$$$$0⩽m⩽2$$$$\Rightarrow$$$$0⩽\theta -\sin \theta ⩽2\pi$$$$\Rightarrow$$$$0⩽\theta ⩽2\pi$$$$\mathrm{in}\mathrm{degree}0⩽\theta ⩽360$$

Commented by indalecioneves last updated on 09/Nov/19

$$Hello,Sir!$$$$Thanksforattention.$$$$Youcanonlyconsidererit1=\frac{2}{\pi }.(\theta -sin\theta ),because{m}^{2}meansonlymeterspersquare.$$$$$$

Commented by MJS last updated on 09/Nov/19

$$\mathrm{you}\mathrm{should}\mathrm{have}\mathrm{written}{R}^2=\frac{28}{\pi }{m}^2\mathrm{then}\mathrm{it}$$$$\mathrm{would}\mathrm{have}\mathrm{been}\mathrm{clear}$$

Answered by MJS last updated on 09/Nov/19

$$\left(1\right)7=\frac{28}{2\pi }(\theta -\sin \theta )$$$$\left(2\right)\mathrm{generally}0⩽\theta <2\pi \mathrm{and}-1⩽\sin \theta ⩽1$$$$\left(3\right)\mathrm{the}\mathrm{area}A=7\mathrm{is}\mathrm{a}\mathrm{quarter}\mathrm{of}\mathrm{the}\mathrm{area}\mathrm{of}$$$$\mathrm{the}\mathrm{circle}\mathrm{which}\mathrm{is}{R}^2\pi =28$$$$$$$$\left(1\right)\Rightarrow \sin \theta =\theta -\frac{\pi }{2}$$$$\left(2\right)-1⩽\theta -\frac{\pi }{2}⩽1\iff \frac{\pi }{2}-1⩽\theta ⩽\frac{\pi }{2}+1$$$$\mathrm{but}0⩽\theta \Rightarrow 0⩽\theta ⩽\frac{\pi }{2}+1$$$$\left(3\right)\Rightarrow \frac{\pi }{2}<\theta <\pi$$$$\mathrm{but}\theta ⩽\frac{\pi }{2}+1\Rightarrow \frac{\pi }{2}<\theta ⩽\frac{\pi }{2}+1$$$$\mathrm{or}90°<\theta ⩽147.30°$$$$$$$$\mathrm{we}\mathrm{can}\mathrm{easily}\mathrm{calculate}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{segment}$$$$\mathrm{with}\theta =\frac{2\pi }{3}(=120°)$$$$\frac{28}{2\pi }(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})=\frac{28}{3}-\frac{7\sqrt{3}}{\pi }\approx 5.47$$$$\Rightarrow \theta >\frac{2\pi }{3}$$$$\Rightarrow \frac{2\pi }{3}<\theta ⩽\frac{\pi }{2}+1\mathrm{or}120°<\theta ⩽147.30°$$$$\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{options}\left(e\right)\mathrm{is}\mathrm{the}\mathrm{right}\mathrm{one}$$

Commented by indalecioneves last updated on 12/Nov/19

$$Thankyou,Sir.$$$$Godblessyou!$$

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