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Question Number 49060 by Pk1167156@gmail.com last updated on 02/Dec/18

Answered by mr W last updated on 02/Dec/18

$$way1:$$$$\frac{3}{\sin C}=\frac{7}{\sin A}=\frac{x}{\sin B}=\frac{x}{\sin (A+C)}$$$$\sin C=\frac{3\sin A}{7}=\frac{3\sin 120°}{7}=\frac{3\sqrt{3}}{14}$$$$\cos C=\frac{\sqrt{{14}^2-9\times 3}}{14}=\frac{13}{14}$$$$$$$$x=\frac{7\sin (A+C)}{\sin A}=7(\cos C+\frac{\sin C}{\tan A})$$$$=7(\frac{13}{14}-\frac{3\sqrt{3}}{14\times \sqrt{3}})=5$$$$$$$$way2:$$$$7^2=3^2+x^2-2\times 3x\times \cos 120°$$$$49=9+x^2+3x$$$$x^2+3x-40=0$$$$(x-5)(x+8)=0$$$$\Rightarrow x=5or-8\left(notwhatweneed\right)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18

$$cos{120}^0=\frac{3^2+x^2-7^2}{2\times 3\times x}$$$$\frac{-1}{2}=\frac{-40+x^2}{6x}$$$$-3x=-40+x^2$$$$x^2+3x-40=0$$$$(x+8)(x-5)=0$$$$x=5AC=5$$$$$$

Commented by Pk1167156@gmail.com last updated on 02/Dec/18

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