Question Number 6758 by FilupSmith last updated on 23/Jul/16
$$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:{r}\:\mathrm{has}\:\mathrm{a}\:\mathrm{point}\:{O}\:\mathrm{as}\:\mathrm{its} \\ $$$$\mathrm{centre}.\:\mathrm{Points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{circumference}. \\ $$$$ \\ $$$$\mathrm{For}\:\bigtriangleup{OAB},\:\overline {{OA}}=\overline {{OB}}={r},\:\overline {{AB}}={d},\:\angle{AOB}=\theta. \\ $$$$\mathrm{What}\:\mathrm{is}\:\frac{{r}}{{d}}? \\ $$
Commented by FilupSmith last updated on 23/Jul/16
$$\mathrm{I}'\mathrm{ve}\:\mathrm{used}\:\mathrm{the}\:\mathrm{cosine}\:\mathrm{rule}\:\mathrm{and}\:\mathrm{pythagorus} \\ $$$$\mathrm{and}\:\mathrm{have}\:\mathrm{gotten}\:\mathrm{two}\:\mathrm{different}\:\mathrm{results}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{variable}\:{d}. \\ $$
Commented by Rasheed Soomro last updated on 23/Jul/16
![By cosine law d^( 2) =r^2 +r^2 −2r^2 coθ=2r^2 (1−cosθ) (r^2 /d^( 2) )=(1/(2(1−cosθ))) (r/d)=+(√(1/(2(1−cosθ)))) [Since r,d>0 , (r/d)>0] =(1/( (√(2(1−cosθ)))))........................A −−−−−−−−− By Pythagorus theorm sin(θ/2)=((d/2)/r)=(d/(2r)) (d/r)=2sin(θ/2) But sin(θ/2)=±(√((1−cosθ)/2)) Hence (d/r)=2(+(√((1−cosθ)/2))) [Since r,d>0 , (r/d)>0] (r/d)=(1/2)(((√2)/( (√(1−cosθ)))))=(1/( (√2)))((1/( (√(1−cosθ))))) =(1/( (√(2(1−cosθ))))).......................B Both results are same.](https://www.tinkutara.com/question/Q6760.png)
$${By}\:{cosine}\:{law} \\ $$$${d}^{\:\mathrm{2}} ={r}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} {co}\theta=\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}−{cos}\theta\right) \\ $$$$\frac{{r}^{\mathrm{2}} }{{d}^{\:\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{cos}\theta\right)} \\ $$$$\frac{{r}}{{d}}=+\sqrt{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{cos}\theta\right)}}\:\:\:\left[{Since}\:\:{r},{d}>\mathrm{0}\:,\:\:\frac{{r}}{{d}}>\mathrm{0}\right] \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\left(\mathrm{1}−{cos}\theta\right)}}……………………{A} \\ $$$$−−−−−−−−− \\ $$$${By}\:{Pythagorus}\:{theorm} \\ $$$${sin}\frac{\theta}{\mathrm{2}}=\frac{{d}/\mathrm{2}}{{r}}=\frac{{d}}{\mathrm{2}{r}} \\ $$$$\frac{{d}}{{r}}=\mathrm{2}{sin}\frac{\theta}{\mathrm{2}} \\ $$$${But}\:\:{sin}\frac{\theta}{\mathrm{2}}=\pm\sqrt{\frac{\mathrm{1}−{cos}\theta}{\mathrm{2}}} \\ $$$${Hence}\:\:\:\frac{{d}}{{r}}=\mathrm{2}\left(+\sqrt{\frac{\mathrm{1}−{cos}\theta}{\mathrm{2}}}\right)\:\:\left[{Since}\:\:{r},{d}>\mathrm{0}\:,\:\:\frac{{r}}{{d}}>\mathrm{0}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{{d}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1}−{cos}\theta}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{cos}\theta}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\left(\mathrm{1}−{cos}\theta\right)}}…………………..{B} \\ $$$${Both}\:{results}\:{are}\:{same}. \\ $$$$ \\ $$
Commented by FilupSmith last updated on 24/Jul/16
$$\mathrm{Thank}\:\mathrm{you}!\:\mathrm{I}\:\mathrm{failed}\:\mathrm{to}\:\mathrm{realise}\:\mathrm{that} \\ $$$$\frac{{d}}{{r}}=\mathrm{2sin}\left(\frac{\theta}{\mathrm{2}}\right) \\ $$
Commented by sandy_suhendra last updated on 25/Jul/16
$$\frac{{r}}{{d}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\:\left(\mathrm{1}−{cos}\:\theta\right)}}\: \\ $$$$\frac{{d}}{{r}}\:=\:\sqrt{\mathrm{2}\:\left(\mathrm{1}−{cos}\:\theta\right)}\:\:\:\:\Rightarrow\:\mathrm{1}−{cos}\:\theta\:=\:\mathrm{2}\:{sin}^{\mathrm{2}} \:\left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\frac{{d}}{{r}}\:=\:\sqrt{\mathrm{4}\:{sin}^{\mathrm{2}} \:\left(\frac{\theta}{\mathrm{2}}\right)}\:\:=\:\mathrm{2}\:{sin}\:\left(\frac{\theta}{\mathrm{2}}\right) \\ $$