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Question Number 69272 by TawaTawa last updated on 22/Sep/19

Answered by $@ty@m123 last updated on 22/Sep/19

$$Let\mathrm{\angle }DPC=\theta$$$$\Rightarrow \mathrm{\angle }Q=\theta -30$$$$LetCP=x$$$$In△DCP,$$$$\tan \theta =\frac{\sqrt{3}}{x}...\left(1\right)$$$$In△DCQ,$$$$\tan (\theta -30)=\frac{\sqrt{3}}{x+2}$$$$\frac{\tan \theta -\tan 30}{1+\tan \theta .\tan 30}=\frac{\sqrt{3}}{x+2}$$$$\frac{\frac{\sqrt{3}}{x}-\frac{1}{\sqrt{3}}}{1+\frac{1}{x}}=\frac{\sqrt{3}}{x+2}$$$$\frac{\frac{3-x}{\sqrt{3}x}}{\frac{x+1}{x}}=\frac{\sqrt{3}}{x+2}$$$$\frac{3-x}{\sqrt{3}(x+1)}=\frac{\sqrt{3}}{x+2}$$$$(3-x)(x+2)=3(x+1)$$$$3x-x^2+6-2x=3x+3$$$$x^2+2x-3=0$$$$(x+3)(x-1)=0$$$$x=-3,1$$$$xcannotbenegative.$$$$\therefore x=1$$

Commented by TawaTawa last updated on 22/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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