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Question Number 199770 by Rupesh123 last updated on 09/Nov/23

Answered by aleks041103 last updated on 09/Nov/23

$$C=\pi -3\alpha$$$$\Rightarrow \frac{5}{sin\left(2\alpha \right)}=\frac{6}{sin(\pi -3\alpha )}=\frac{6}{sin\left(3\alpha \right)}$$$$\Rightarrow 6sin\left(2\alpha \right)=5sin\left(3\alpha \right)$$$$sin\left(2\alpha \right)=2sin\left(\alpha \right)cos\left(\alpha \right)$$$$sin\left(3\alpha \right)=sin\left(2\alpha \right)cos\left(\alpha \right)+sin\left(\alpha \right)cos\left(2\alpha \right)=$$$$=2sin\left(\alpha \right){cos}^2\left(\alpha \right)+sin\left(\alpha \right)(1-2{sin}^2\left(\alpha \right))=$$$$=2sin\left(\alpha \right)(1-{sin}^2\left(\alpha \right))+sin\left(\alpha \right)(1-2{sin}^2\left(\alpha \right))=$$$$=2sin\left(\alpha \right)-2{sin}^3\left(\alpha \right)+sin\left(\alpha \right)-2{sin}^3\left(\alpha \right)=$$$$=sin\left(\alpha \right)(3-4{sin}^2\alpha )$$$$\Rightarrow 6\times 2sin\left(\alpha \right)cos\left(\alpha \right)=5sin\left(\alpha \right)(3-4{sin}^2\left(\alpha \right))$$$$\Rightarrow 12cos\left(\alpha \right)=5(4(1-{sin}^2\left(\alpha \right))-1)$$$$\Rightarrow 12cos\left(\alpha \right)=20{cos}^2\left(\alpha \right)-5$$$$z={cos}^2\left(\alpha \right)$$$$\Rightarrow 20z^2-12z-5=0$$$$z_{1,2}=\frac{12\pm \sqrt{144+400}}{40}=\frac{12\pm \sqrt{544}}{40}=$$$$=\frac{12\pm \sqrt{16\times 34}}{40}=\frac{3\pm \sqrt{34}}{10}$$$$C=\pi -3\alpha >0\Rightarrow 0<\alpha <\pi /3\Rightarrow z=cos\left(\alpha \right)>0$$$$\Rightarrow z=\frac{3+\sqrt{34}}{10}<1$$$$X^2=5^2+6^2-2\times 5\times 6\times cos\left(\alpha \right)=$$$$=25+36-60z=$$$$=61-6(3+\sqrt{34})=$$$$=61-18-6\sqrt{34}=$$$$=43-6\sqrt{34}=$$$$=9+34-6\sqrt{34}=$$$$={\left(3\right)}^2-2\times \left(3\right)\times \left(\sqrt{34}\right)+{\left(\sqrt{34}\right)}^2=$$$$={(\sqrt{34}-3)}^2$$$$\Rightarrow X=\sqrt{34}-3$$

Commented by Mingma last updated on 09/Nov/23

Nice, solution

Answered by mr W last updated on 09/Nov/23

Commented by mr W last updated on 09/Nov/23

$$\underset{―}{MethodI}$$$$\cos \alpha =\frac{6-\frac{6-x}{2}}{5}=\frac{3}{5}+\frac{x}{10}$$$$\cos 2\alpha =\frac{\frac{6-x}{2}}{x}=\frac{3}{x}-\frac{1}{2}$$$$\frac{3}{x}-\frac{1}{2}=2{(\frac{3}{5}+\frac{x}{10})}^2-1$$$$x^3+12x^2+11x-150=0$$$$(x+6)(x^2+6x-25)=0$$$$\Rightarrow x=-6,-\sqrt{34}-3\Rightarrow rejected$$$$\Rightarrow x=\sqrt{34}-3✓$$

Commented by aleks041103 last updated on 09/Nov/23

$$Veruniceandelegant!Bravo!!$$

Commented by mr W last updated on 09/Nov/23

$$thankssir!$$

Commented by Mingma last updated on 09/Nov/23

Nice solution!

Answered by mr W last updated on 09/Nov/23

Commented by mr W last updated on 09/Nov/23

$$\underset{―}{MethodII}$$$$\mathrm{\Delta }ABC\sim \mathrm{\Delta }BDC$$$$\frac{x}{5}=\frac{5}{x+6}$$$$\Rightarrow x^2+6x-25=0$$$$\Rightarrow x=-3+\sqrt{34}✓$$

Commented by Mingma last updated on 09/Nov/23

Nice solution!

Answered by ajfour last updated on 09/Nov/23

Commented by ajfour last updated on 09/Nov/23

$$2p+x=6$$$$\Rightarrow 2p=6-x$$$${CD}^2=x^2-p^2=25-{(x+p)}^2$$$$\Rightarrow 2x^2+x\left(2p\right)=25$$$$2x^2+x(6-x)=25$$$${(x+3)}^2=25+9$$$$x=\sqrt{34}-3$$

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