In odot(O, r),the length of minor widehat{ACB | vedclass

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Question

(B) The circumference of a circle is given by $C = 2\pi r$,where $r$ is the radius.The length of an arc $L$ subtending an angle $	heta$ at the centre

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The circumference of a circle is given by $C = 2\pi r$,where $r$ is the radius.The length of an arc $L$ subtending an angle $	heta$ at the centre is given by $L = \frac{	heta}{360^\circ} 	imes 2\pi r$.According to the problem,the length of the arc is one-eighth of the circumference:$L = \frac{1}{8} 	imes 2\pi r$.Equating the two expressions for $L$:$\frac{	heta}{360^\circ} 	imes 2\pi r = \frac{1}{8} 	imes 2\pi r$.Dividing both sides by $2\pi r$:$\frac{	heta}{360^\circ} = \frac{1}{8}$.Solving for $	heta$:$	heta = \frac{360^\circ}{8} = 45^\circ$.Thus,the measure of the angle subtended at the centre is $45^\circ$.

In $\odot(O, r)$,the length of minor $\widehat{ACB}$ is one-eighth of the circumference of the circle. Then,the measure of the angle subtended at the centre by that arc is $\ldots \ldots \ldots \ldots$ (in $^\circ$)

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