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Question Number 167043 by mnjuly1970 last updated on 05/Mar/22

Answered by mr W last updated on 05/Mar/22

$$say\mathrm{\angle }B=\beta$$$$\tan \beta =\frac{3\sin 2\alpha }{4-3\cos 2\alpha }=\frac{6\sin \alpha \cos \alpha }{1+6{\sin }^2\alpha }$$$$\frac{\sin \beta }{6}=\frac{\sin \alpha }{4}\Rightarrow \sin \beta =\frac{3\sin \alpha }{2}$$$${\left(\frac{6\sin \alpha \cos \alpha }{1+6{\sin }^2\alpha }\right)}^2=\frac{{\left(\frac{3\sin \alpha }{2}\right)}^2}{1-{\left(\frac{3\sin \alpha }{2}\right)}^2}$$$$64{\sin }^2\alpha =15$$$$\sin \alpha =\frac{\sqrt{15}}{8}\Rightarrow \cos \alpha =\frac{7}{8}$$$$3^2=6^2+x^2-2\times 6\times x\times \cos \alpha$$$$2x^2-21x+54=0$$$$(2x-9)(x-6)=0$$$$\Rightarrow x=\frac{9}{2},x=6\left(rejected\right)$$

Commented by mnjuly1970 last updated on 05/Mar/22

$$SirW:thankyousomuch$$$${\left(grateful\right)}^{\mathrm{\infty }}$$

Commented by Tawa11 last updated on 05/Mar/22

$$\mathrm{Great}\mathrm{sir}$$

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