Find-the-number-of-sides-of-two-regular-polygons-that-their-sides-has-a-ratio-5-4-and-of-9-as-a-difference-between-their-angles-

Question Number 182474 by Acem last updated on 10/Dec/22

$${Find}\:{the}\:{number}\:{of}\:{sides}\:{of}\:{two}\:{regular}\:{polygons} \\ $$$$\:{that}\:{their}\:{sides}\:{has}\:{a}\:{ratio}\:\mathrm{5}:\mathrm{4}\:{and}\:{of}\:\mathrm{9}°\:{as}\:{a} \\ $$$$\:{difference}\:{between}\:{their}\:{angles}. \\ $$

Answered by mr W last updated on 10/Dec/22

n−sided regular polygon: angle θ_n =180−((360)/n) say two regular polygons with x and y sides respectively. (x/y)=(5/4) θ_x −θ_y =((360)/y)−((360)/x)=9 ⇒(1/y)−(1/x)=(1/(40)) ⇒(1/y)−(4/(5y))=(1/(40)) ⇒y=8 ⇒x=10 i.e. their number of sides is 10 and 8 respectively.

$${n}−{sided}\:{regular}\:{polygon}: \\ $$$${angle}\:\theta_{{n}} =\mathrm{180}−\frac{\mathrm{360}}{{n}} \\ $$$${say}\:{two}\:{regular}\:{polygons}\:{with}\:{x}\:{and} \\ $$$${y}\:{sides}\:{respectively}. \\ $$$$\frac{{x}}{{y}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\theta_{{x}} −\theta_{{y}} =\frac{\mathrm{360}}{{y}}−\frac{\mathrm{360}}{{x}}=\mathrm{9} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{\mathrm{40}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{y}}−\frac{\mathrm{4}}{\mathrm{5}{y}}=\frac{\mathrm{1}}{\mathrm{40}}\:\Rightarrow{y}=\mathrm{8}\:\Rightarrow{x}=\mathrm{10} \\ $$$${i}.{e}.\:{their}\:{number}\:{of}\:{sides}\:{is}\:\mathrm{10}\:{and}\:\mathrm{8} \\ $$$${respectively}. \\ $$

Commented by Acem last updated on 10/Dec/22

$${Thanks}\:{Sir}! \\ $$

Answered by som(math1967) last updated on 10/Dec/22

let sides are 5x,4x ⇒((360)/(4x)) −((360)/(5x))=9 ⇒((90−72)/x)=9 ⇒x=((18)/9)=2 ∴ sides are 10,8

$${let}\:{sides}\:{are}\:\mathrm{5}{x},\mathrm{4}{x} \\ $$$$\:\Rightarrow\frac{\mathrm{360}}{\mathrm{4}{x}}\:−\frac{\mathrm{360}}{\mathrm{5}{x}}=\mathrm{9} \\ $$$$\Rightarrow\frac{\mathrm{90}−\mathrm{72}}{{x}}=\mathrm{9} \\ $$$$\Rightarrow{x}=\frac{\mathrm{18}}{\mathrm{9}}=\mathrm{2} \\ $$$$\therefore\:{sides}\:{are}\:\mathrm{10},\mathrm{8} \\ $$

Commented by Acem last updated on 10/Dec/22

$${Thanksss}\:{Sir}! \\ $$

Commented by som(math1967) last updated on 10/Dec/22

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Commented by Acem last updated on 10/Dec/22

$${Coeur}\:\:\left(:\right. \\ $$

Answered by Acem last updated on 10/Dec/22

ϕ_2 −ϕ_1 = ((360)/n_2 ) − ((360 ((4/5)))/n_2 ) ; ϕ: external angle 9= ((72)/n_2 ) , n_2 = 8, n_1 = 10

$$\varphi_{\mathrm{2}} −\varphi_{\mathrm{1}} =\:\frac{\mathrm{360}}{{n}_{\mathrm{2}} }\:−\:\frac{\mathrm{360}\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{{n}_{\mathrm{2}} }\:\:\:\:;\:\:\:\varphi:\:{external}\:{angle} \\ $$$$\:\mathrm{9}=\:\frac{\mathrm{72}}{{n}_{\mathrm{2}} }\:,\:\:\:\:\:\:{n}_{\mathrm{2}} =\:\mathrm{8},\:{n}_{\mathrm{1}} =\:\mathrm{10} \\ $$$$\: \\ $$

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