Question Number 186667 by cherokeesay last updated on 08/Feb/23
Answered by som(math1967) last updated on 08/Feb/23
$${cos}\mathrm{30}=\frac{{QR}^{\mathrm{2}} +{QT}^{\mathrm{2}} −{RT}^{\mathrm{2}} }{\mathrm{2}×{QR}×{QT}} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{16}+\mathrm{16}−{RT}^{\mathrm{2}} }{\mathrm{2}×\mathrm{16}} \\ $$$$\Rightarrow{RT}=\sqrt{\mathrm{32}−\mathrm{16}\sqrt{\mathrm{3}}}=\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${RT}={RU}=\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${Area}\:{of}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\mathrm{4}×{sin}\mathrm{30}=\mathrm{4}{squnit} \\ $$$$\angle{SRU}=\mathrm{360}−\left(\mathrm{90}+\mathrm{90}+\mathrm{75}\right)=\mathrm{105} \\ $$$${Area}\:{of}\:{B}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}×{sin}\mathrm{105} \\ $$$$=\mathrm{8}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{8}×\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{2}×\mathrm{2}=\mathrm{4}{sq}\:{unit} \\ $$$$\therefore\frac{{A}}{{B}}=\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1} \\ $$
Commented by som(math1967) last updated on 08/Feb/23
Answered by mr W last updated on 08/Feb/23
Commented by mr W last updated on 08/Feb/23
$${for}\:{any}\:{case}: \\ $$$${A}=\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${B}=\frac{{ab}\:\mathrm{sin}\:\left(\pi−\theta\right)}{\mathrm{2}}=\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}}={A} \\ $$$$\Rightarrow\frac{{A}}{{B}}=\mathrm{1}\:\checkmark \\ $$