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Question Number 139105 by mnjuly1970 last updated on 22/Apr/21

Answered by bramlexs22 last updated on 22/Apr/21

$$\frac{\mathrm{b}}{\sin \theta }=\frac{\mathrm{a}}{\sin 2\theta }\Rightarrow \mathrm{b}=\frac{\mathrm{a}}{2\cos \theta };\cos \theta =\frac{\mathrm{a}}{2\mathrm{b}}$$$$\measuredangle \mathrm{C}=\gamma =180°-3\theta$$$${\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\cos \gamma$$$$(\bullet )\cos \gamma =-\cos 3\theta$$$$\mathrm{c}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{abcos}3\theta }$$$$\cos 3\theta =\cos 2\theta \cos \theta -\sin 2\theta \sin \theta$$$$=\cos \theta (2\cos {}^2\theta -1)-2\cos \theta (1-\cos {}^2\theta )$$$$=4\cos {}^3\theta -3\cos \theta$$$$=4\left(\frac{{\mathrm{a}}^3}{{8\mathrm{b}}^3}\right)-3\left(\frac{\mathrm{a}}{2\mathrm{b}}\right)=\frac{{\mathrm{a}}^3}{{2\mathrm{b}}^3}-\frac{{3\mathrm{ab}}^2}{{2\mathrm{b}}^3}=\frac{{\mathrm{a}}^3-{3\mathrm{ab}}^2}{{2\mathrm{b}}^3}$$$$\mathrm{c}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\left(\frac{{\mathrm{a}}^3-{3\mathrm{ab}}^2}{{2\mathrm{b}}^3}\right)}$$$$=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+\frac{\mathrm{a}({\mathrm{a}}^3-{3\mathrm{ab}}^2)}{{\mathrm{b}}^2}}$$$$=\frac{\sqrt{{\mathrm{a}}^2{\mathrm{b}}^2+{\mathrm{b}}^4+{\mathrm{a}}^4-{3\mathrm{a}}^2{\mathrm{b}}^2}}{\mathrm{b}}$$$$=\frac{\sqrt{{\mathrm{a}}^4+{\mathrm{b}}^4-{2\mathrm{a}}^2{\mathrm{b}}^2}}{\mathrm{b}}=\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{\mathrm{b}}$$

Commented by mnjuly1970 last updated on 22/Apr/21

$$thanksalotmaster...$$

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