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Question Number 160522 by mnjuly1970 last updated on 01/Dec/21

Answered by mr W last updated on 03/Dec/21

Commented by mr W last updated on 03/Dec/21

$${BE}^2={25}^2+3^2-2\times 25\times 3\cos (\pi -2\alpha )$$$${BE}^2=634+150\cos \left(2\alpha \right)$$$${BE}^2=484+300{\cos }^2\alpha$$$$$$$$BC=2\times 25\cos \alpha =50\cos \alpha$$$$$$$${BD}^2=5^2+{\left(50\cos \alpha \right)}^2-2\times 5\times 50\cos \alpha \cos \alpha$$$${BD}^2=25+2000{\cos }^2\alpha$$$$$$$${ED}^2={BE}^2+{BD}^2-2\times BE\times BD\cos 45°$$$${17}^2=484+300{\cos }^2\alpha +25+2000{\cos }^2\alpha -\sqrt{2(484+300{\cos }^2\alpha )(25+2000{\cos }^2\alpha )}$$$$220+2300{\cos }^2\alpha =\sqrt{2(484+300{\cos }^2\alpha )(25+2000{\cos }^2\alpha )}$$$${220}^2+{2300}^2{\cos }^4\alpha +2\times 220\times 2300{\cos }^2\alpha =2\times 484\times 25+2\times (300\times 25+484\times 2000){\cos }^2\alpha +2\times 300\times 2000{\cos }^4\alpha$$$$20450{\cos }^4\alpha -4695{\cos }^2\alpha +121=0$$$${\cos }^2\alpha =\frac{4695\pm \sqrt{{4695}^2-4\times 20450\times 121}}{2\times 20450}=\frac{1}{5},\frac{121}{4090}$$$$\tan \alpha =\sqrt{\frac{1}{{\cos }^2\alpha }-1}=\sqrt{5-1}=2$$$$\tan \alpha =\sqrt{\frac{4090}{121}-1}=\frac{63}{11}$$$$$$$$\alpha ={\tan }^{-1}2\approx 63.435°,{\tan }^{-1}\frac{63}{11}\approx 80.956°$$

Commented by mr W last updated on 04/Dec/21

Commented by mr W last updated on 04/Dec/21

$$thisdiagramshowswhytwosolutions$$$$exist.$$

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