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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

$$\cos 60°=\frac{a^2+b^2-\frac{7b^2}{9}}{2ab}=\frac{1}{2}$$$$Also\cos 60°=\frac{a^2+4b^2-\frac{28b^2}{9}}{4ab}=\frac{1}{2}$$$$\Rightarrow a^2+\frac{2}{9}{b}^2=ab\mathrm{\&}{a}^2+\frac{8b^2}{9}=2ab$$$$2b^2-9ab+9a^2=0\mathrm{\&}$$$$8b^2-18ab+9a^2=0$$$$\Rightarrow b=\left(\frac{9\pm \sqrt{81-72}}{4}\right)a\mathrm{\&}$$$$b=\left(\frac{18\pm \sqrt{324-288}}{16}\right)a$$$$\Rightarrow b=3a,\frac{3a}{2}\mathrm{\&}b=\frac{3a}{4},\frac{3a}{2}$$$$\Rightarrow b=\frac{3a}{2}$$$$\Rightarrow BC=b\sqrt{7}=150\sqrt{7}.$$

Commented by mr W last updated on 13/Oct/19

$$nicemethod!thankssir!$$

Answered by john santuy last updated on 22/Dec/19

$$letAC=t.AD=y.usingcosine$$$$rules.{\left(3y\right)}^2=t^2+4t^2-2t\times 2t\times cos{120}^o$$$$thenBC=150\sqrt{7}$$$$$$

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