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 $θ$ is

\frac{10 π}{13}

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\frac{9 π}{13}

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\frac{5 π}{13}

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\frac{6 π}{13}

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Solution

The correct option is A $\frac{10 π}{13}$
Slant height $= \sqrt{12^{2} + 5^{2}} = 13$
$S$ is the perimeter of base of a cone and $θ$ is the angle
$θ = \frac{S}{r} \Rightarrow S = r θ 2 π (5) = 13 θ \Rightarrow θ = \frac{10 π}{13}$
Hence, option $A$ is correct.

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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 θ is

The correct option is A 10π13Slant height=√122+52=13S is the perimeter of base of a cone and θ is the angleθ=Sr⇒S=rθ2π(5)=13θ⇒θ=10π13Hence, option A is correct.

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 $θ$ is

The correct option is A $\frac{10 π}{13}$
Slant height $= \sqrt{12^{2} + 5^{2}} = 13$
$S$ is the perimeter of base of a cone and $θ$ is the angle
$θ = \frac{S}{r} \Rightarrow S = r θ 2 π (5) = 13 θ \Rightarrow θ = \frac{10 π}{13}$
Hence, option $A$ is correct.