(A) The slant height $l$ of the cone is equal to the radius of the sector,which is $l = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.The cir

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(A) The slant height $l$ of the cone is equal to the radius of the sector,which is $l = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.The circumference of the base of the cone is equal to the arc length of the sector.Base circumference of the cone $= 2 \pi r = 2 \pi (5) = 10 \pi$.Arc length of the sector $= l \theta = 13 \theta$.Equating the two,we get $13 \theta = 10 \pi$.Therefore,$\theta = \frac{10 \pi}{13}$.

Class 11 Mathematics 10-1.Circle and System of Circles Locus Related Problem

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$A$ piece of paper in the shape of a sector of a circle (see $Fig. 1$) is rolled up to form a right-circular cone (see $Fig. 2$). The value of the angle $\theta$ is

KVPY 2011 All KVPY

A
$\frac{10 \pi}{13}$
B
$\frac{9 \pi}{13}$
C
$\frac{5 \pi}{13}$
D
$\frac{6 \pi}{13}$

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10-1.Circle and System of Circles

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$A$ piece of paper in the shape of a sector of a circle (see $Fig. 1$) is rolled up to form a right-circular cone (see $Fig. 2$). The value of the angle $\theta$ is

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Let $C$ be the centre and $A$ be one end of a diameter of the circle $x^2+y^2-2x-4y-20=0$. If $P$ is a point such that $A$ divides $CP$ in the ratio $2:3$,then the locus of $P$ is

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