In odot(O, r),minor widehat{ACB} subtends an | vedclass

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(A) The length of an arc subtending an angle $	heta$ at the centre is given by $L = \frac{	heta}{360^{\circ}} 	imes 2\pi r$.The circumference of th

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$1:5$

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(A) The length of an arc subtending an angle $	heta$ at the centre is given by $L = \frac{	heta}{360^{\circ}} 	imes 2\pi r$.The circumference of the circle is $C = 2\pi r$.Therefore,the ratio of the length of the minor arc $\widehat{ACB}$ to the circumference is $\frac{L}{C} = \frac{(\frac{	heta}{360^{\circ}}) 	imes 2\pi r}{2\pi r} = \frac{	heta}{360^{\circ}}$.Given $	heta = 72^{\circ}$,the ratio is $\frac{72^{\circ}}{360^{\circ}} = \frac{1}{5}$.Thus,the ratio is $1:5$.

In $\odot(O, r)$,minor $\widehat{ACB}$ subtends an angle of measure $72^{\circ}$ at the centre. Then,the ratio of the length of minor $\widehat{ACB}$ and the circumference of the circle is ............

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