Cone Problem

**BobBali** · December 11th 2011, 11:48 PM

Hi All,

The problem below shows a sector of a circle on the right hand side and when it is folded it makes the cone on the left. Find the arc length.

I started of by finding the slant height or radius:

$\sqrt{10^2 + 6^2} = 11.7$

To find arc length i have the formula:
$\frac{\theta}{360} \times 2\pi r$

But, How do i find the angle in the unfolded sector?

I tried finding the the angle of the top part of the cone using:
$tan\theta = \frac{6}{10} = 31 \times 2 = 62$

Then, 360 - 62 = 329 But, this input into the arc-length formula is not the
right answer.

**earboth** · December 12th 2011, 12:25 AM

Originally Posted by BobBali

Hi All,

The problem below shows a sector of a circle on the right hand side and when it is folded it makes the cone on the left. Find the arc length.

I started of by finding the slant height or radius:

$\sqrt{10^2 + 6^2} = 11.7$

To find arc length i have the formula:
$\frac{\theta}{360} \times 2\pi r$

But, How do i find the angle in the unfolded sector?

I tried finding the the angle of the top part of the cone using:
$tan\theta = \frac{6}{10} = 31 \times 2 = 62$

Then, 360 - 62 = 329 But, this input into the arc-length formula is not the
right answer.

Obviously the arc length equals the circumference of the base area:

$c_{base} = 2 \cdot \pi \cdot 6$

If (and only if) you want to to know the central angle of the sector you have to determine the length of s using Pythagorean theorem. Then use the proportion:

$\frac{\theta}{360^\circ}=\frac{2 \cdot \pi \cdot 6}{2 \cdot \pi \cdot s}$

Solve for $\theta$ .

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