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Question Number 172915 by mnjuly1970 last updated on 03/Jul/22

Commented by infinityaction last updated on 03/Jul/22

$$6?$$

Answered by mr W last updated on 03/Jul/22

Commented by mr W last updated on 03/Jul/22

$$\cos \gamma =\frac{4-1}{4+1}=\frac{3}{5}$$$${BD}^2=4^2+5^2+2\times 4\times 5\times \frac{3}{5}=65$$$$EB=\sqrt{{BD}^2-1^2}=8$$$$\tan \beta =\frac{1}{8}$$$$\tan \alpha =\frac{4}{7}$$$$\tan B=\tan (\alpha +\beta )=\frac{\frac{4}{7}+\frac{1}{8}}{1-\frac{4}{7}\times \frac{1}{8}}=\frac{39}{52}$$$$AC=AB\times \tan B=8\times \frac{39}{52}=6✓$$

Commented by Tawa11 last updated on 03/Jul/22

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

Answered by som(math1967) last updated on 04/Jul/22

Commented by som(math1967) last updated on 04/Jul/22

$$letAC=x$$$$\mathrm{\angle }DCO=\mathrm{\angle }OCB=\theta$$$$\therefore \mathrm{\angle }ACB=2\theta$$$$AE=4-1=3,OE=\sqrt{5^2-3^2}=4$$$$\therefore CD=x-4$$$$tan\theta =\frac{1}{x-4},tan2\theta =\frac{8}{x}$$$$tan2\theta =\frac{2tan\theta }{1-{tan}^2\theta }$$$$\frac{8}{x}=\frac{\frac{2}{x-4}}{\frac{{(x-4)}^2-1}{{(x-4)}^2}}$$$$\frac{8}{x}=\frac{2(x-4)}{x^2-8x+15}$$$$4x^2-32x+60=x^2-4x$$$$3x^2-28x+60=0$$$$3x^2-18x-10x+60=0$$$$(x-6)(3x-10)=0$$$$(x-6)=0[\because x>4\therefore x>\frac{10}{3},(3x-10)\ne 0]$$$$\therefore x=6$$

Commented by Tawa11 last updated on 04/Jul/22

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

Answered by infinityaction last updated on 04/Jul/22

Commented by infinityaction last updated on 04/Jul/22

$$DB=\sqrt{4^2+7^2}=\sqrt{65}$$$$△DEB$$$$BE=\sqrt{{\left(\sqrt{65}\right)}^2-1^2}=8$$$$△ABC$$$${(x+4)}^2+8^2={(x+8)}^2$$$$\cancel{x^2}+16+8x+\cancel{64}=\cancel{x^2}+\cancel{64}+16x$$$$8x=16\Rightarrow x=2$$$$AC=4+x=4+2=6$$

Commented by Tawa11 last updated on 04/Jul/22

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

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