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Question Number 68425 by mr W last updated on 10/Sep/19

Commented by Prithwish sen last updated on 10/Sep/19

$$\mathrm{AB}=2\mathrm{x}\mathrm{AC}=\mathrm{x}$$$$3\mathrm{x}\times 100\times \frac{\sqrt{3}}{2}={2\mathrm{x}}^2\times \frac{\sqrt{3}}{2}$$$${2\mathrm{x}}^2-300\mathrm{x}=0\Rightarrow \mathrm{x}=150$$$$\mathrm{AB}=300\mathrm{AC}=150$$$$\mathrm{BC}=\sqrt{{300}^2+{150}^2-2\times 300\times 150\times \cos 120}$$$$=\sqrt{90000+22500+45000}$$$$=\sqrt{157500}=150\sqrt{7}$$

Commented by mr W last updated on 10/Sep/19

$$thanksalotsir!$$

Answered by ajfour last updated on 10/Sep/19

$$letAB=2AC=2p$$$$BD=x,DC=y$$$$\cos 120°=\frac{p^2+4p^2-{(x+y)}^2}{4p^2}=-\frac{1}{2}$$$$\Rightarrow {(x+y)}^2=7p^2$$$$\frac{BD}{DC}=\frac{x}{y}=\frac{2p}{p}=2$$$$\Rightarrow 3y=p\sqrt{7}ory=\frac{p\sqrt{7}}{3},x=\frac{2p\sqrt{7}}{3}$$$$considering△ADC$$$$\cos 60°=\frac{{AD}^2+{AC}^2-{CD}^2}{2\left(AD\right)\left(AC\right)}=\frac{1}{2}$$$$sayAD=a$$$$\frac{a^2+p^2-y^2}{2ap}=\frac{1}{2}$$$$a^2+p^2-\frac{7p^2}{9}=ap$$$$\frac{2p^2}{9}-ap+a^2=0$$$$\Rightarrow 2p^2-9ap+9a^2=0$$$$2p^2-6ap-3ap+9a^2=0$$$$2p(p-3a)-3a(p-3a)=0$$$$p=\frac{3a}{2},orp=3a\left(ishouldreject\right)$$$$BC=x+y=p\sqrt{7}=\left(\frac{3\sqrt{7}}{2}\right)a$$$$\mathit{B}\mathit{C}=150\sqrt{7}.$$

Commented by mr W last updated on 10/Sep/19

$$thanksalotsir!$$

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