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Question Number 66522 by mr W last updated on 16/Aug/19

Answered by MJS last updated on 16/Aug/19

$$q^2+x^2=16$$$${(x-p)}^2+x^2=25$$$${(x-q)}^2+p^2=9$$$$$$$$\mathrm{solving}\mathrm{leads}\mathrm{to}$$$$p=\frac{3\sqrt{17}}{17}$$$$q=\frac{4\sqrt{17}}{17}$$$$x=\frac{16\sqrt{17}}{17}$$

Commented by MJS last updated on 16/Aug/19

$$...\mathrm{leading}\mathrm{to}\mathrm{an}\mathrm{equation}\mathrm{of}{4}^{\mathrm{th}}\mathrm{degree}\mathrm{for}v,$$$$\mathrm{or}{2}^{\mathrm{nd}}\mathrm{degree}\mathrm{for}{v}^2(v=\mathrm{any}\mathrm{of}p,q\mathrm{or}x)\mathrm{if}\mathrm{you}$$$$\mathrm{do}\mathrm{it}\mathrm{right}...$$

Answered by MJS last updated on 16/Aug/19

$$4\sin \alpha =q$$$$4\cos \alpha =x$$$$3\sin \alpha =p$$$$3\cos \alpha =x-q$$$$\Rightarrow$$$$\frac{q}{4}=\frac{p}{3}$$$$\frac{x}{4}=\frac{x-q}{3}$$$$\Rightarrow$$$$p=\frac{3x}{16}$$$$q=\frac{x}{4}$$$$$$$$q^2+x^2=16$$$$\frac{x^2}{16}+x^2=16$$$$\Rightarrow x=\frac{16\sqrt{17}}{17}$$

Commented by mr W last updated on 16/Aug/19

$$thankssir!$$

Commented by MJS last updated on 16/Aug/19

$$\mathrm{easier}...$$

Answered by mr W last updated on 16/Aug/19

$$\sqrt{4^2-x^2}+\frac{3}{4}x=x$$$$\sqrt{4^2-x^2}=\frac{x}{4}$$$$4^2-x^2=\frac{x^2}{16}$$$$4^2=\frac{17x^2}{16}$$$$\Rightarrow x=\frac{16}{\sqrt{17}}$$

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