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Question Number 6758 by FilupSmith last updated on 23/Jul/16

$$\mathrm{A}\mathrm{circle}\mathrm{of}\mathrm{radius}r\mathrm{has}\mathrm{a}\mathrm{point}O\mathrm{as}\mathrm{its}$$$$\mathrm{centre}.\mathrm{Points}A\mathrm{and}B\mathrm{are}\mathrm{points}\mathrm{on}\mathrm{the}$$$$\mathrm{circumference}.$$$$$$$$\mathrm{For}△OAB,\overline{OA}=\overline{OB}=r,\overline{AB}=d,\mathrm{\angle }AOB=\theta .$$$$\mathrm{What}\mathrm{is}\frac{r}{d}?$$

Commented by FilupSmith last updated on 23/Jul/16

$${\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{used}\mathrm{the}\mathrm{cosine}\mathrm{rule}\mathrm{and}\mathrm{pythagorus}$$$$\mathrm{and}\mathrm{have}\mathrm{gotten}\mathrm{two}\mathrm{different}\mathrm{results}\mathrm{for}$$$$\mathrm{the}\mathrm{variable}d.$$

Commented by Rasheed Soomro last updated on 23/Jul/16

$$Bycosinelaw$$$$d^2=r^2+r^2-2r^{2}co\theta =2r^2(1-cos\theta )$$$$\frac{r^2}{d^2}=\frac{1}{2(1-cos\theta )}$$$$\frac{r}{d}=+\sqrt{\frac{1}{2(1-cos\theta )}}[Sincer,d>0,\frac{r}{d}>0]$$$$=\frac{1}{\sqrt{2(1-cos\theta )}}........................A$$$$---------$$$$ByPythagorustheorm$$$$sin\frac{\theta }{2}=\frac{d/2}{r}=\frac{d}{2r}$$$$\frac{d}{r}=2sin\frac{\theta }{2}$$$$Butsin\frac{\theta }{2}=\pm \sqrt{\frac{1-cos\theta }{2}}$$$$Hence\frac{d}{r}=2(+\sqrt{\frac{1-cos\theta }{2}})[Sincer,d>0,\frac{r}{d}>0]$$$$\frac{r}{d}=\frac{1}{2}\left(\frac{\sqrt{2}}{\sqrt{1-cos\theta }}\right)=\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{1-cos\theta }}\right)$$$$=\frac{1}{\sqrt{2(1-cos\theta )}}.......................B$$$$Bothresultsaresame.$$$$$$

Commented by FilupSmith last updated on 24/Jul/16

$$\mathrm{Thank}\mathrm{you}!\mathrm{I}\mathrm{failed}\mathrm{to}\mathrm{realise}\mathrm{that}$$$$\frac{d}{r}=2\sin \left(\frac{\theta }{2}\right)$$

Commented by sandy_suhendra last updated on 25/Jul/16

$$\frac{r}{d}=\frac{1}{\sqrt{2(1-cos\theta )}}$$$$\frac{d}{r}=\sqrt{2(1-cos\theta )}\Rightarrow 1-cos\theta =2{sin}^2\left(\frac{\theta }{2}\right)$$$$\frac{d}{r}=\sqrt{4{sin}^2\left(\frac{\theta }{2}\right)}=2sin\left(\frac{\theta }{2}\right)$$

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