law of sine/word problem

[viet]

You have a cone shaped bag. At the bottom of the bag is an orange with a radius of 2 inches. On top of the orange is a melon with a radius of 6 inches. It touches the orange and fits snugly in the bag, touching it in a ring around the orange. Its top is at the same level as the top of the bag.

Find height of the cone and its radius.

i drew the picture, i have half of the cone angles as 90 degrees, 60 degrees and 30 degrees. I dont know what to do next, can someone help?

 

[soroban]

Hello, viet!

Your angles are incorrect . . .

You have a cone shaped bag.
At the bottom of the bag is an orange with a radius of 2 inches.
On top of the orange is a melon with a radius of 6 inches.
It touches the orange and fits snugly in the bag, touching it in a ring around the orange.
Its top is at the same level as the top of the bag.

Find height of the cone and its radius.

    B                     D                     C
   o - - - - - - - - - - *o* - - - - - - - - - - o
    \              *      |     *            /
     \                    |                    /
      \       *          6|            *      /
       \    *             |             *    /
        \  *              |              *  /
         \ *             Ho              * /
          \*              |    \ 6       */
           \*             |        \    */
            \*           6|            o/J
             \ *          |          * /
              \     *     |G    *     /
               \         *o*         /
                \      *  |  *      /
                 \    *   2|   *    /
                  \  *    |    *  /
                   \ *   Fo    * /
                    \*    |\2  */
                    \*   2| \ */
                      \*  |  o/I
                       \ *o* /
                        \y|E/
                         \|/
                          o
                          A

$\displaystyle ABC$ is the cone.

$\displaystyle F$ is the center of the two-inch orange.
. . $\displaystyle FE\,=\,FG\,=\,FI\,=\,2$

$\displaystyle H$ is the center of the six-inch melon.
. . $\displaystyle HD\,=\,HG\,=\,HJ\,=\,6$

Let $\displaystyle EA\,=\,y.$

In right triangle $\displaystyle FIA:\;AF\,=\,y+2,\;FI\,=\,2$

In right triangle $\displaystyle HJA:\;AH\,=\,y+10,\;HJ\,=\,6$

Since $\displaystyle \Delta FIA\,\sim\,\Delta HJA:\;\frac{y+2}{2}\:=\:\frac{y+10}{6}\;\;\Rightarrow\;\;y\,=\,2$

. . Therefore: . $\displaystyle h\,=\,DA\,=\,18$


In right triangle $\displaystyle FIA:\;AI^2\,+\,FI^2\:=\:AF^2\;\;\Rightarrow\;\;AI^2\,+\,2^2\:=\:4^2\;\;\Rightarrow\;\;AI\,=\,2\sqrt{3}$


Let $\displaystyle DC\,=\,r$ (radius of cone).

Since $\displaystyle \Delta CDA\,\sim\,\Delta FIA:\;\frac{CD}{DA}\,=\,\frac{FI}{AI}\;\;\Rightarrow\;\;\frac{r}{18}\,=\,\frac{2}{2\sqrt{3}}$

. . Therefore: .$\displaystyle r\,=\,6\sqrt{3}$

 

[viet]

thanks for your help, great explanation

 

[Denis]

Today being Sunday the day of the Lord, after reciting the Hoooooly Rosary
a few times while begging for forgiveness of my week's sins, I experimented
with this geometrical intricacy, and would like to hereby partake with youze
of the following underhanded observations and notes:

Now, WHY in **** bring in 2 fruits in a conic bag causing the student to wonder if the
bag will have a half circular bottom (wrapped around the bottom of the orange!),
when the purpose is to have the student handle a triangle with sides tangent to 2
circles and the 2 circles tangent to each other ?
The teacher that made that up should be assigned to Lucifer's abode right away,
with no access to no Hoooooly Rosary.

Here's a little formula to handle this, given the radius of both circles:

R = radius of cone
b = bottom circle radius
t = top circle radius (t < b : my triangle points up, not down like Soroban's!)
x = portion from top of small circle to tip of triangle (Soroban's mistakedly called it y)
h = height of triangle

R = ht / sqrt[x(x + 2t)]
where:
x = 2t^2 / (b - t)
h = x + 2(b + t)

By the way, there are cases (only few, 11 with b<100) where all above
variables are integers; smallest has area of triangle = 768:
I'll leave it for someone to find