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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^2 (θ−sinθ)/2. Where A=7m^2 ; R^2 =((28)/π). What is the best answer for the angle value (degree) a) 85°<θ<90° b) 95°<θ<100° c) 105°<θ<110° d) 115°<θ<120° e) 125°<θ<135°

$${Could}\:{someone}\:{help}\:{me}\:{on}\:{this}\:{question}? \\ $$$${Knowing}\:{that}\:{the}\:{area}\:{of}\:{a}\:{circle}\:{segment}\:{is}\:{given}\:{by}\:{A}={R}^{\mathrm{2}} \left(\theta−{sin}\theta\right)/\mathrm{2}.\:{Where}\:{A}=\mathrm{7}{m}^{\mathrm{2}} ;\:{R}^{\mathrm{2}} =\frac{\mathrm{28}}{\pi}. \\ $$$${What}\:{is}\:{the}\:{best}\:{answer}\:{for}\:{the}\:{angle}\:{value}\:\left({degree}\right) \\ $$$$\left.{a}\right)\:\mathrm{85}°<\theta<\mathrm{90}° \\ $$$$\left.{b}\right)\:\mathrm{95}°<\theta<\mathrm{100}° \\ $$$$\left.{c}\right)\:\mathrm{105}°<\theta<\mathrm{110}° \\ $$$$\left.{d}\right)\:\mathrm{115}°<\theta<\mathrm{120}° \\ $$$$\left.{e}\right)\:\mathrm{125}°<\theta<\mathrm{135}° \\ $$

Commented by MJS last updated on 04/Nov/19

there′s information missing what is m? if m is just a constant factor: 7m^2 =((28)/(2π))(θ−sin θ) the area of the circle is R^2 π=28 ⇒ 0≤7m^2 ≤28 0≤m^2 ≤4 area≥0 ⇒ m≥0 0≤m≤2 ⇒ 0≤θ−sin θ≤2π ⇒ 0≤θ≤2π in degree 0≤θ≤360

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

Commented by indalecioneves last updated on 09/Nov/19

Hello, Sir! Thanks for attention. You can only considerer it 1=(2/π).(θ−sinθ), because m^2 means only meters per square.

$${Hello},\:{Sir}! \\ $$$${Thanks}\:{for}\:{attention}. \\ $$$${You}\:{can}\:{only}\:{considerer}\:{it}\:\mathrm{1}=\frac{\mathrm{2}}{\pi}.\left(\theta−{sin}\theta\right),\:{because}\:{m}^{\mathrm{2}} \:{means}\:{only}\:{meters}\:{per}\:{square}. \\ $$$$ \\ $$

Commented by MJS last updated on 09/Nov/19

you should have written R^2 =((28)/π)m^2 then it would have been clear

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

Answered by MJS last updated on 09/Nov/19

(1) 7=((28)/(2π))(θ−sin θ) (2) generally 0≤θ<2π and −1≤sin θ ≤1 (3) the area A=7 is a quarter of the area of the circle which is R^2 π=28 (1) ⇒ sin θ =θ−(π/2) (2) −1≤θ−(π/2)≤1 ⇔ (π/2)−1≤θ≤(π/2)+1 but 0≤θ ⇒ 0≤θ≤(π/2)+1 (3) ⇒ (π/2)<θ<π but θ≤(π/2)+1 ⇒ (π/2)<θ≤(π/2)+1 or 90°<θ≤147.30° we can easily calculate the area of a segment with θ=((2π)/3) (=120°) ((28)/(2π))(((2π)/3)−((√3)/2))=((28)/3)−((7(√3))/π)≈5.47 ⇒ θ>((2π)/3) ⇒ ((2π)/3)<θ≤(π/2)+1 or 120°<θ≤147.30° of the given options (e) is the right one

$$\left(\mathrm{1}\right)\:\:\mathrm{7}=\frac{\mathrm{28}}{\mathrm{2}\pi}\left(\theta−\mathrm{sin}\:\theta\right) \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{generally}\:\mathrm{0}\leqslant\theta<\mathrm{2}\pi\:\mathrm{and}\:−\mathrm{1}\leqslant\mathrm{sin}\:\theta\:\leqslant\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\:\mathrm{the}\:\mathrm{area}\:{A}=\mathrm{7}\:\mathrm{is}\:\mathrm{a}\:\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{is}\:{R}^{\mathrm{2}} \pi=\mathrm{28} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\:\Rightarrow\:\mathrm{sin}\:\theta\:=\theta−\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:−\mathrm{1}\leqslant\theta−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{1}\:\Leftrightarrow\:\frac{\pi}{\mathrm{2}}−\mathrm{1}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{but}\:\mathrm{0}\leqslant\theta\:\Rightarrow\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\:\Rightarrow\:\frac{\pi}{\mathrm{2}}<\theta<\pi \\ $$$$\:\:\:\:\:\mathrm{but}\:\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1}\:\Rightarrow\:\frac{\pi}{\mathrm{2}}<\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{or}\:\mathrm{90}°<\theta\leqslant\mathrm{147}.\mathrm{30}° \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{segment} \\ $$$$\mathrm{with}\:\theta=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\left(=\mathrm{120}°\right) \\ $$$$\frac{\mathrm{28}}{\mathrm{2}\pi}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{\mathrm{28}}{\mathrm{3}}−\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\pi}\approx\mathrm{5}.\mathrm{47} \\ $$$$\Rightarrow\:\theta>\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}\pi}{\mathrm{3}}<\theta\leqslant\frac{\pi}{\mathrm{2}}+\mathrm{1}\:\mathrm{or}\:\mathrm{120}°<\theta\leqslant\mathrm{147}.\mathrm{30}° \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{options}\:\left({e}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{one} \\ $$

Commented by indalecioneves last updated on 12/Nov/19

Thank you, Sir. God bless you!

$${Thank}\:{you},\:{Sir}. \\ $$$${God}\:{bless}\:{you}! \\ $$

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