# word problem involving law of sine

• Nov 19th 2005, 07:33 PM
viet
word problem involving law of sine
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 with the radius of the orange and the melon. can someone help?
• Nov 19th 2005, 11:35 PM
ticbol
Quote:

Originally Posted by 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 with the radius of the orange and the melon. can someone help?

So you have the figure now there. Good.
If you have not done it yet, draw these:
>>>a vertical line from the bottom of the cone to the top of the cone. This line passes through the centers of the orange and melon.
>>>a radius of the orange from the center of the orange to the wall of the cone. This radius is perpendicular to the wall of the cone.
>>>a radius of the melon from the center of the melon to the wall of the cone. This radius is perpendicular to the wall of the cone.

The two radii are on the same side of the vertical line.

Call the line segment from the bottom of the orange to the bottom of the cone as "y".
Call the angle included by the vertical line and the side of the cone as theta.
Call the radius of the cone as R.
Call the height of the cone as H.

In the small right triangle formed by the radius of the orange, a portion of the wall of the cone and a portion of the vertical line,
sin(theta) = 2 / (2+y) --------(1)

In the medium right triangle formed by the radius of the melon, a portion of the wall of the cone and a portion of the vertical line,
sin(theta) = 6 / (6+2+2+y)
sin(theta) = 6 / (10+y) --------(2)

sin(theta) = sin(theta),
2 / (2+y) = 6 / (10+y)
Cross multiply,
2(10+y) = (2+y)(6)
20 +2y = 12 +6y
20 -12 = 6y -2y
8 = 4y
y = 8/4 = 2 inches. ------***
Substitute that into (1),
sin(theta) = 2/(2+2) = 2/4 = 1/2.
So, theta = arcsin(1/2) = 30 degrees. -------***

In the big right triangle formed by the horizontal top (half of it) of the cone (which is the radius of the cone, or R), the whole wall of the cone, and the whole vertical line (which is the height of the cone, or H):

a) H = (diameter of melon) +(diameter of orange) +y
H = (6+6) +(2+2) +2 = 18 inches ----------answer.

b) tan(theta) = R/H
tan(30deg) = R/18
R = 18*tan(30deg)
R = 18*(1/sqrt(3))
R = (6*3)/sqrt(3) = 6*sqrt(3) or 10.39 inches -------answer.

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Sorry, I was not able to use the Law of Sines.