Question Number 215094 by ajfour last updated on 28/Dec/24
Commented by ajfour last updated on 28/Dec/24
A question discussed Statics+Geometry
https://youtu.be/vktsHDGgBlw?si=z6XC7xw6G8m6XjrS
Commented by mr W last updated on 28/Dec/24
Commented by mr W last updated on 28/Dec/24
$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\theta\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{sin}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right)} \\ $$$$\Rightarrow\mathrm{tan}\:\left(\alpha+\theta\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow\alpha+\theta=\frac{\pi}{\mathrm{2}}−\alpha\:\Rightarrow\theta=\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$
Commented by Frix last updated on 28/Dec/24
$$\frac{{R}}{{r}}=\mathrm{12}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$
Commented by ajfour last updated on 28/Dec/24
Explain me the first line itself sir.
Commented by mr W last updated on 28/Dec/24
$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\theta\right)}{\mathrm{sin}\:\alpha}=\frac{{CD}}{{AC}}=\frac{{CD}}{{BC}}=\frac{\mathrm{sin}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right)} \\ $$
Commented by ajfour last updated on 28/Dec/24
$${Yeah} \\ $$$$\:\theta_{{ajfour}} +\theta_{{mrw}} =\mathrm{0}\:\:\:\: \\ $$$${we}\:{have}\:{just}\:{marked}\:{differently}. \\ $$$${we}\:{have}\:{same}\:{situation}.\:{for}\:{example} \\ $$$${check}\:{for}\:\alpha=\mathrm{60}°\:. \\ $$
Commented by ajfour last updated on 28/Dec/24
$${Thanks}\:{sir}\:{Frix},\:{you}\:{came}\:{back}\: \\ $$$${after}\:{really}\:{some}\:{time}. \\ $$