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Question Number 12928 by tawa last updated on 07/May/17

Answered by sandy_suhendra last updated on 09/May/17

$$\mathrm{cosine}\mathrm{rule}\mathrm{in}\mathrm{\Delta }\mathrm{ABC}:$$$${\mathrm{x}}^2={\mathrm{d}}^2+{\mathrm{d}}^2-2\mathrm{d}.\mathrm{d}.\cos (180-2\theta )$$$${\mathrm{x}}^2={2\mathrm{d}}^2-{2\mathrm{d}}^2(-\cos 2\theta )\Rightarrow \mathrm{use}\cos 2\theta ={2\cos }^2\theta -1$$$${\mathrm{x}}^2={2\mathrm{d}}^2-{2\mathrm{d}}^2(-{2\cos }^2\theta +1)$$$${\mathrm{x}}^2={2\mathrm{d}}^2+{4\mathrm{d}}^2{\cos }^2\theta -{2\mathrm{d}}^2$$$${\mathrm{x}}^2={4\mathrm{d}}^2{\cos }^2\theta$$$$\mathrm{x}=2\mathrm{d}\cos \theta$$$$$$$$\mathrm{sine}\mathrm{rule}\mathrm{in}\mathrm{\Delta }\mathrm{BCD}:$$$$\frac{\mathrm{k}}{\sin \theta }=\frac{\mathrm{x}}{\sin [180-(\theta +\beta )]}$$$$\frac{\mathrm{k}}{\sin \theta }=\frac{2\mathrm{d}\cos \theta }{\sin (\theta +\beta )}$$$$\mathrm{k}=\frac{2\mathrm{d}\cos \theta \sin \theta }{\sin (\theta +\beta )}\Rightarrow \mathrm{use}\sin 2\theta =2\sin \theta \cos \theta$$$$\mathrm{k}=\frac{2\mathrm{d}(0.5\sin 2\theta )}{\sin (\alpha +\beta )}$$$$\mathrm{k}=\frac{\mathrm{d}\sin 2\theta }{\sin (\alpha +\beta )}$$

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