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Question Number 71964 by mr W last updated on 22/Oct/19

Commented by mr W last updated on 22/Oct/19

$$findx=?$$

Answered by behi83417@gmail.com last updated on 24/Oct/19

$${\mathrm{y}}^2={12}^2+{10}^2-2\times 12\times 10\times \frac{1}{2}=$$$$=144+100-120=124\Rightarrow \mathrm{y}=\sqrt{124}$$$${\mathrm{m}}^2={12}^2+{\mathrm{x}}^2-24\mathbf{x}\frac{\sqrt{3}}{2}=144+{\mathbf{x}}^2-12\sqrt{3}.\mathbf{x}$$$${\mathrm{n}}^2={10}^2+{\mathrm{x}}^2-2\times 10\times \mathrm{x}\frac{\sqrt{3}}{2}=100+{\mathrm{x}}^2-10\sqrt{3}.\mathrm{x}$$$$\mathbf{m}.10+\mathbf{n}\times 12=\mathbf{x}.\mathbf{y}\Rightarrow$$$$10{(144+{\mathrm{x}}^2-12\sqrt{3}\mathrm{x})}^{\frac{1}{2}}+12{(100+{\mathrm{x}}^2-10\sqrt{3}\mathrm{x})}^{\frac{1}{2}}=\mathrm{x}\times \sqrt{124}$$$$\Rightarrow \mathbf{x}\approx 12.703.$$

Answered by mind is power last updated on 22/Oct/19

$$\mathrm{AB}=12,\mathrm{AC}=10$$$$\mathrm{AD}=\mathrm{x}$$$${\mathrm{BC}}^2=144+100-240\cos \left(60\right)=244-120=124$$$$\mathrm{BC}=\sqrt{124}=2\sqrt{31}$$$$\frac{\mathrm{AB}.\mathrm{BC}.\mathrm{AC}}{{2\mathrm{S}}_{\mathrm{ABC}}}=2\mathrm{R}$$$${\mathrm{S}}_{\mathrm{ABC}}=\frac{10.\sin \left(60\right).12}{2}=30\sqrt{3}$$$$2\mathrm{R}=\frac{10.12.2\sqrt{31}}{2.30\sqrt{3}}=\frac{4\sqrt{31}}{\sqrt{3}}$$$$\mathrm{R}=2\sqrt{\frac{31}{3}}$$$${\mathrm{BD}}^2={12}^2+{\mathrm{x}}^2-24\mathrm{xcos}\left(30\right)=144+{\mathrm{x}}^2-12\mathrm{x}\sqrt{3}$$$${\mathrm{S}}_{\mathrm{ABD}}=\frac{12\mathrm{xsin}\left(30\right)}{2}=3\mathrm{x}$$$$\Rightarrow 4\sqrt{\frac{31}{3}}=\frac{\mathrm{x}.\sqrt{({\mathrm{x}}^2-12\mathrm{x}\sqrt{3}+144)}.12}{6\mathrm{x}}=2\sqrt{{\mathrm{x}}^2-12\mathrm{x}\sqrt{3}+144}$$$$\Rightarrow 4.\frac{31}{3}={\mathrm{x}}^2-12\mathrm{x}\sqrt{3}+144$$$$\Rightarrow 124={3\mathrm{x}}^2-36\sqrt{3}\mathrm{x}+432$$$$\Rightarrow {3\mathrm{x}}^2-36\sqrt{3}\mathrm{x}+308=0$$$$(3.{36}^2)-12.308=3.1296-3696=3888-3696=192$$$${\mathrm{x}}_1=\frac{36\sqrt{3}-\sqrt{192}}{6}=\frac{36\sqrt{3}-8\sqrt{3}}{6}=4\sqrt{3}$$$${\mathrm{X}}_2=\frac{36\sqrt{3}+\sqrt{192}}{6}=\frac{44\sqrt{3}}{6}=\frac{22\sqrt{3}}{3}$$$$\mathrm{x}⩾12\Rightarrow \mathrm{x}=\frac{22\sqrt{3}}{3}$$$$$$

Commented by mr W last updated on 23/Oct/19

$$thankssir!correctanswer.$$

Commented by mind is power last updated on 23/Oct/19

$$\mathrm{y},\mathrm{re}\mathrm{welcom}$$

Answered by mr W last updated on 23/Oct/19

$$\frac{\sin \alpha }{x}=\frac{\sin (\alpha +30)}{10}$$$$\frac{\sin (180-\alpha )}{x}=\frac{\sin (180-\alpha +30)}{12}$$$$\Rightarrow \frac{\sin (\alpha +30)}{10}=\frac{\sin (\alpha -30)}{12}$$$$\Rightarrow \frac{\sin \alpha \times \frac{\sqrt{3}}{2}+\cos \alpha \times \frac{1}{2}}{10}=\frac{\sin \alpha \times \frac{\sqrt{3}}{2}-\cos \alpha \times \frac{1}{2}}{12}$$$$\Rightarrow {\cot }^{-1}\alpha =-\frac{\sqrt{3}}{11}$$$$\frac{\sin \alpha }{x}=\frac{\sin \alpha \times \frac{\sqrt{3}}{2}+\cos \alpha \times \frac{1}{2}}{10}$$$$\frac{1}{x}=\frac{\sqrt{3}+{\cot }^{-1}\alpha }{20}=\frac{\sqrt{3}-\frac{\sqrt{3}}{11}}{20}=\frac{\sqrt{3}}{22}$$$$\Rightarrow x=\frac{22}{\sqrt{3}}$$

Commented by mr W last updated on 23/Oct/19

$$alternativeway:$$$$x^2+{10}^2-2\times 10x\cos 30=x^2+{12}^2-2\times 12x\cos 30$$$$2(12-10)x\cos 30={12}^2-{10}^2=44$$$$\Rightarrow x=\frac{11}{\cos 30}=\frac{22}{\sqrt{3}}$$

Commented by ajfour last updated on 23/Oct/19

$$thebest!$$

Commented by mind is power last updated on 23/Oct/19

$$\mathrm{nice}\mathrm{sir}$$

Answered by ajfour last updated on 23/Oct/19

Commented by ajfour last updated on 23/Oct/19

$$A(6\sqrt{3},6)B(5\sqrt{3},-5)$$$$leteq.ofcircle{x}^2+y^2+2gx+2fy=0$$$$Asatisfies,Bsatisfies$$$$144+12\sqrt{3}g+12f=0$$$$100+10\sqrt{3}g-10f=0$$$$\Rightarrow 120\sqrt{3}g=-1320$$$$\Rightarrow g=-\frac{11}{\sqrt{3}}$$$$fory=0$$$$x=0,-2g=\frac{22}{\sqrt{3}}=x.$$

Commented by mind is power last updated on 23/Oct/19

$$\mathrm{nice}\mathrm{solution}\mathrm{sir}$$

Answered by john santu last updated on 19/Jan/20

$$\frac{10}{12}=\frac{t}{h}\Rightarrow t=\frac{5}{6}h$$$$\left(i\right)\frac{25h^2}{36}=100+x^2-10\sqrt{3}x$$$$\left(ii\right)h^2=144+x^2-12\sqrt{3}x$$$$\frac{25}{36}(144+x^2-12x\sqrt{3})=100+x^2-10\sqrt{3}x$$$$25x^2-300x\sqrt{3}+3600=3600+36x^2-360x\sqrt{3}$$$$11x^2-60x\sqrt{3}=0$$$$x=\frac{60\sqrt{3}}{11}$$

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