Question Number 71233 by mr W last updated on 13/Oct/19
Commented by ajfour last updated on 13/Oct/19
Commented by ajfour last updated on 13/Oct/19
$$\mathrm{cos}\:\mathrm{60}°=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\frac{\mathrm{7}{b}^{\mathrm{2}} }{\mathrm{9}}}{\mathrm{2}{ab}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Also}\:\:\mathrm{cos}\:\mathrm{60}°=\frac{{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} −\frac{\mathrm{28}{b}^{\mathrm{2}} }{\mathrm{9}}}{\mathrm{4}{ab}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{9}}{b}^{\mathrm{2}} ={ab}\:\&\:{a}^{\mathrm{2}} +\frac{\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{2}{ab} \\ $$$$\:\:\:\mathrm{2}{b}^{\mathrm{2}} −\mathrm{9}{ab}+\mathrm{9}{a}^{\mathrm{2}} =\mathrm{0}\:\:\& \\ $$$$\:\:\mathrm{8}{b}^{\mathrm{2}} −\mathrm{18}{ab}+\mathrm{9}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:{b}=\left(\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{72}}}{\mathrm{4}}\right){a}\:\:\& \\ $$$$\:\:\:\:\:{b}=\left(\frac{\mathrm{18}\pm\sqrt{\mathrm{324}−\mathrm{288}}}{\mathrm{16}}\right){a} \\ $$$$\Rightarrow\:\:{b}=\mathrm{3}{a},\:\frac{\mathrm{3}{a}}{\mathrm{2}}\:\:\:\&\:\:{b}=\frac{\mathrm{3}{a}}{\mathrm{4}}\:,\:\frac{\mathrm{3}{a}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{b}=\frac{\mathrm{3}{a}}{\mathrm{2}} \\ $$$$\:\Rightarrow\:{BC}={b}\sqrt{\mathrm{7}}\:=\:\mathrm{150}\sqrt{\mathrm{7}}\:. \\ $$
Commented by mr W last updated on 13/Oct/19
$${nice}\:{method}!\:{thanks}\:{sir}! \\ $$
Answered by john santuy last updated on 22/Dec/19
$${let}\:{AC}\:={t}.\:{AD}\:={y}.\:{using}\:{cosine}\: \\ $$$${rules}\:.\:\left(\mathrm{3}{y}\right)^{\mathrm{2}} ={t}^{\mathrm{2}} +\:\mathrm{4}{t}^{\mathrm{2}} −\mathrm{2}{t}×\mathrm{2}{t}×{cos}\:\mathrm{120}^{{o}} \\ $$$${then}\:{BC}\:=\mathrm{150}\sqrt{\mathrm{7}} \\ $$$$ \\ $$