Question Number 181987 by Acem last updated on 03/Dec/22
Answered by Frix last updated on 03/Dec/22
$$\left.\frac{\mathrm{sin}\:\mathrm{45}°}{{BC}}=\frac{\mathrm{sin}\:\mathrm{105}°}{{AC}}=\frac{\mathrm{sin}\:\mathrm{30}°}{{AB}}\right) \\ $$$$\Rightarrow\frac{{BC}}{{AC}}=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$${AC}+{BC}=\mathrm{14}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{get} \\ $$$${AC}=\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge{BC}=\frac{\mathrm{14}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}}\:\Rightarrow\:{AB}=\frac{\mathrm{7}\sqrt{\mathrm{2}}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}} \\ $$
Commented by Acem last updated on 03/Dec/22
$${Thanks}\:{Sir}! \\ $$
Answered by mr W last updated on 03/Dec/22
Commented by mr W last updated on 03/Dec/22
$${AD}=\mathrm{14} \\ $$$${AB}=\frac{\mathrm{sin}\:\mathrm{15}}{\mathrm{sin}\:\mathrm{120}}×\mathrm{14}=\frac{\mathrm{7}\left(\mathrm{3}\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}\right)}{\:\mathrm{3}} \\ $$$${BD}=\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\mathrm{120}}×\mathrm{14}=\frac{\mathrm{14}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$${BC}=\frac{{BD}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{15}}=\frac{\mathrm{14}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}} \\ $$$${AC}=\mathrm{14}−\frac{\mathrm{14}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}}=\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
Commented by Acem last updated on 03/Dec/22
$${Very}\:{well}\:{Sir}!\:{Thanks}.\:{Just}\:{fix}\:\mathrm{2}\:{terms}\:{inAB}\:;\:\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}} \\ $$
Answered by Acem last updated on 03/Dec/22
$${Four}\:{methods}\:{to}\:{solve} \\ $$$$\:{Our}\:{fiends}\:{mr}.\:{Frix}\:{and}\:{mr}.\:{W} \\ $$$$\:{beside}\:{to}\:{two}\:{important}\:{formulas} \\ $$$$\:\mathrm{3}\bullet\:\frac{{a}+\:{b}}{{c}}=\:\frac{\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\alpha−\beta\right)}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\:\gamma}\:\Rightarrow{c}=\:\mid{AB}\mid=\:\mathrm{4}.\mathrm{184}=\:\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{3}}\:\left(\mathrm{3}−\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{4}\bullet\:\frac{{a}−{b}}{{a}+{b}}=\:\frac{\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\alpha−\beta\right)}{\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\alpha+\beta\right)}\:\Rightarrow\:{a}−{b}=\:−\mathrm{2}.\mathrm{165}=\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$\:{hence}\:{we}\:{find}\:{a}=\mid{BC}\mid\:,\:{b}=\:\mid{AC}\mid \\ $$$$\:{then}\:{compt}\:{the}\:{rest}\:{by}\:{sine}\:{laws} \\ $$$$ \\ $$$$\circledast\:{I}\:{prefer}\:{extend}\:{AC}\:{method}\:{although}\:{the}\:\mathrm{3}{rd}\:{one} \\ $$$$\:{is}\:{the}\:{fasted}. \\ $$$$ \\ $$$$\:{Thanks}\:{my}\:{brothers}\:{for}\:{sharing}! \\ $$