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Question Number 147706 by aliibrahim1 last updated on 22/Jul/21

Answered by mr W last updated on 22/Jul/21

Commented by aliibrahim1 last updated on 22/Jul/21

sorry sir i tried to build on that before sending it didnt work with me

$${sorry}\:{sir}\:{i}\:{tried}\:{to}\:{build}\:{on}\:{that}\:{before}\:{sending}\:{it}\:{didnt}\:{work}\:{with}\:{me} \\ $$

Commented by mr W last updated on 22/Jul/21

say A(p,p^2 )  s=2p  β=((180−108)/2)=36°  α=108−36=72°  x_B =p+2p cos 72°  y_B =p^2 +2p sin 72°  p^2 +2p sin 72°=p^2 (1+2 cos 72°)^2   sin 72°=2pcos 72°(1+cos 72°)  p=((tan 72°)/(2(1+cos 72°)))  y_C =y_B +2p sin 36°  y_C =p^2 +2p(sin 72°+sin 36°)  y_C =((tan^2  72°)/(4(1+cos 72°)^2 ))+((tan 72°)/((1+cos 72°)))(sin 72°+sin 36°)  =5

$${say}\:{A}\left({p},{p}^{\mathrm{2}} \right) \\ $$$${s}=\mathrm{2}{p} \\ $$$$\beta=\frac{\mathrm{180}−\mathrm{108}}{\mathrm{2}}=\mathrm{36}° \\ $$$$\alpha=\mathrm{108}−\mathrm{36}=\mathrm{72}° \\ $$$${x}_{{B}} ={p}+\mathrm{2}{p}\:\mathrm{cos}\:\mathrm{72}° \\ $$$${y}_{{B}} ={p}^{\mathrm{2}} +\mathrm{2}{p}\:\mathrm{sin}\:\mathrm{72}° \\ $$$${p}^{\mathrm{2}} +\mathrm{2}{p}\:\mathrm{sin}\:\mathrm{72}°={p}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\mathrm{72}°\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{72}°=\mathrm{2}{p}\mathrm{cos}\:\mathrm{72}°\left(\mathrm{1}+\mathrm{cos}\:\mathrm{72}°\right) \\ $$$${p}=\frac{\mathrm{tan}\:\mathrm{72}°}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{72}°\right)} \\ $$$${y}_{{C}} ={y}_{{B}} +\mathrm{2}{p}\:\mathrm{sin}\:\mathrm{36}° \\ $$$${y}_{{C}} ={p}^{\mathrm{2}} +\mathrm{2}{p}\left(\mathrm{sin}\:\mathrm{72}°+\mathrm{sin}\:\mathrm{36}°\right) \\ $$$${y}_{{C}} =\frac{\mathrm{tan}^{\mathrm{2}} \:\mathrm{72}°}{\mathrm{4}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{72}°\right)^{\mathrm{2}} }+\frac{\mathrm{tan}\:\mathrm{72}°}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{72}°\right)}\left(\mathrm{sin}\:\mathrm{72}°+\mathrm{sin}\:\mathrm{36}°\right) \\ $$$$=\mathrm{5} \\ $$

Commented by aliibrahim1 last updated on 23/Jul/21

thanks sir appreciate it

$${thanks}\:{sir}\:{appreciate}\:{it} \\ $$

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