suppose you cut and throw away a 144 degree notch from a circular disk of paper

Question

suppose you cut and throw away a 144 degree notch from a circular disk of paper 10 inches in diameter. If you fold the remaining section . How tall will the cone be? what will be the radius of the cone base? what is the sine of the cone angle? what is the cone angle rounded to the nearest degree? At what latitude rounded to the nearest degree will the cone make contact with an upright sphere of diameter 4inches ?at what latitude rounded to the nearest degree will the cone make contact with an upright sphere of diameter less than 4inches ?

Explanation / Answer

I realize I am going slightly out of order, hopefully my steps make it clear why I am doing this.

Radius of the Cone Base -

We cut off 144 degrees of the circle, and what was left of the circumference became the circumference of the new cone's base. The circumference of a circle is:

C = d

Where C is the Circumference and d is the diameter

So the circumference of the original circle is:

C = d = 10" = 10"

Now, the original circle uses 360º, but the new cone has a base of (360º - 144º = 216º). So the ratio of the angles is:

216º / 360º = 3/5

So the ratio of the circumferences is the same.

Ccircle = 10"

Cbase = 10" (3/5) = 6"

Using our circumference formula above, we get:

Cbase = 6" = 2 rbase

6" = 2 rbase

rbase = 3"

Height of Cone -

So let's ignore the fact it's a cone for now, and just imagine a right triangle with one leg as the radius of the bottom circle, one leg as the slanted slope up the slide, and another leg as the height of the cone.Bottom:

Horizontal: Radius base, Vertical: Height of Cone, Slanted: Slanted slope to tip of cone

The definition of the sine is:

sin() = (Opposite Side)/(Hypotenuse)

The "cone angle" is the top angle, so the opposite side is the bottom leg which is 3", and the hypotenuse is 5"

sin() = (3")/(5") = 3/5

Cone Angle -

The opposite of the sine function is the arcsine function.

sin() = 3/5

arcsin(3/5) =

Plugging this into our calculator gives us: 36.87º

Rounded to the nearest degree, this gives: 37º

These two parts of the question make little sense to me -- when the diameter is less than four inches for the sphere, it should intersect at two sections on the cone, not just one. You can solve this by use of euclidean space, but that seems a little too advance for your level. Another option is to use some confusing geometry by drawing right triangles connecting the intersection point to the middle of the sphere, seeing that length is the radius r, then trying to work out the angle via picturing the situation in 2D space as just a triangle and a circle. Sorry if that wasn't much of a help.

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