Question Number 221637 by ASAD7391 last updated on 08/Jun/25
Answered by fantastic last updated on 09/Jun/25
$${let}\:\angle{A}=\theta\:\therefore\angle{B}=\mathrm{2}\theta\: \\ $$$${but}\:\angle{A}+\angle{B}+\angle{C}=\mathrm{180}^{\mathrm{0}} \\ $$$${or}\:\theta+\mathrm{2}\theta=\mathrm{180}^{\mathrm{0}} −\mathrm{90}^{\mathrm{0}} =\mathrm{90}^{\mathrm{0}} \\ $$$${or}\:\theta=\frac{\mathrm{90}^{\mathrm{0}} }{\mathrm{3}}=\mathrm{30}^{\mathrm{0}} \\ $$$${So}\:\angle{A}=\mathrm{30}^{\mathrm{0}} {and}\:\angle{B}=\mathrm{2}×\mathrm{30}^{\mathrm{0}} =\mathrm{60}^{\mathrm{0}} \\ $$$$ \\ $$$$\mathrm{cos}\:\mathrm{60}^{\mathrm{0}} =\frac{{BC}}{{AB}} \\ $$$${or}\:\frac{{BC}}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${So}\:{AB}=\mathrm{2}{BC}\checkmark\:\left({proved}\right) \\ $$