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Question Number 187716 by yaslm last updated on 20/Feb/23

Answered by Ar Brandon last updated on 20/Feb/23

$$\mathrm{\Sigma }{F}_x:300\sin \beta =200\sin \alpha$$$$\Rightarrow \sin \beta =\frac{2\sin \alpha }{3}$$$$\mathrm{\Sigma }{F}_{\mathrm{y}}:300\cos \beta +200\cos \alpha =400$$$$\Rightarrow 3\sqrt{1-{\sin }^2\beta }+2\cos \alpha =4$$$$\Rightarrow 3\sqrt{1-\frac{{4\sin }^2\alpha }{9}}=4-2\cos \alpha$$$$\Rightarrow 9(1-\frac{{4\sin }^2\alpha }{9})=16-16\cos \alpha +{4\cos }^2\alpha$$$$\Rightarrow 16\cos \alpha =7+4({\cos }^2\alpha +{\sin }^2\alpha )$$$$\Rightarrow \cos \alpha =\frac{11}{16}\Rightarrow \alpha =\mathrm{arcos}\left(\frac{11}{16}\right)$$$$\Rightarrow \sin \beta =\frac{2}{3}\sqrt{1-{\cos }^2\alpha }=\frac{2}{3}\sqrt{1-\frac{121}{256}}$$$$=\frac{2}{3}\sqrt{\frac{135}{256}}=\frac{\sqrt{135}}{24}$$$$\Rightarrow \beta =\arcsin \left(\frac{\sqrt{135}}{24}\right)$$

Answered by mr W last updated on 21/Feb/23

Commented by mr W last updated on 21/Feb/23

$$\cos \alpha =\frac{2^2+4^2-3^2}{2\times 2\times 4}=\frac{11}{16}$$$$\Rightarrow \alpha ={\cos }^{-1}\frac{11}{16}\approx 46.567°$$$$\cos \beta =\frac{3^2+4^2-2^2}{2\times 3\times 4}=\frac{7}{8}$$$$\Rightarrow \beta ={\cos }^{-1}\frac{7}{8}\approx 28.955°$$

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