Menu Close

Question-108107




Question Number 108107 by otchereabdullai@gmail.com last updated on 14/Aug/20
Commented by rexfordattacudjoe last updated on 16/Aug/20
By similarities,(r/(24))=((10)/(20))  r=((10×24)/(20))  r=12cm
$${By}\:{similarities},\frac{{r}}{\mathrm{24}}=\frac{\mathrm{10}}{\mathrm{20}} \\ $$$${r}=\frac{\mathrm{10}×\mathrm{24}}{\mathrm{20}} \\ $$$${r}=\mathrm{12}{cm} \\ $$
Commented by rexfordattacudjoe last updated on 16/Aug/20
length of an arc=(θ/(360))×2Πr  20=(θ/(360))×2(((22)/7))(24)  50400=1056θ  θ=47.7°
$${length}\:{of}\:{an}\:{arc}=\frac{\theta}{\mathrm{360}}×\mathrm{2}\Pi{r} \\ $$$$\mathrm{20}=\frac{\theta}{\mathrm{360}}×\mathrm{2}\left(\frac{\mathrm{22}}{\mathrm{7}}\right)\left(\mathrm{24}\right) \\ $$$$\mathrm{50400}=\mathrm{1056}\theta \\ $$$$\theta=\mathrm{47}.\mathrm{7}° \\ $$
Answered by JDamian last updated on 14/Aug/20
(a)  20 = 24 ∙ θ  ⇒ θ = (5/6) rad  10 = r ∙ θ  ⇒ r = ((10)/θ) = ((10)/(5/6)) = 12 cm.  (b)   20 + 10 + 2∙(24−12) = 54 cm.    (c)  from (a):  θ = (5/6) rad =  = (5/6) × ((180)/π) = ((150)/π) ≈ 47.75°
$$\left({a}\right) \\ $$$$\mathrm{20}\:=\:\mathrm{24}\:\centerdot\:\theta\:\:\Rightarrow\:\theta\:=\:\frac{\mathrm{5}}{\mathrm{6}}\:{rad} \\ $$$$\mathrm{10}\:=\:{r}\:\centerdot\:\theta\:\:\Rightarrow\:{r}\:=\:\frac{\mathrm{10}}{\theta}\:=\:\frac{\mathrm{10}}{\frac{\mathrm{5}}{\mathrm{6}}}\:=\:\mathrm{12}\:{cm}. \\ $$$$\left({b}\right)\:\:\:\mathrm{20}\:+\:\mathrm{10}\:+\:\mathrm{2}\centerdot\left(\mathrm{24}−\mathrm{12}\right)\:=\:\mathrm{54}\:{cm}. \\ $$$$ \\ $$$$\left({c}\right)\:\:{from}\:\left({a}\right):\:\:\theta\:=\:\frac{\mathrm{5}}{\mathrm{6}}\:{rad}\:= \\ $$$$=\:\frac{\mathrm{5}}{\mathrm{6}}\:×\:\frac{\mathrm{180}}{\pi}\:=\:\frac{\mathrm{150}}{\pi}\:\approx\:\mathrm{47}.\mathrm{75}° \\ $$
Commented by otchereabdullai@gmail.com last updated on 15/Aug/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *