SOLUTION: In the diagram attached below, Circles M and N are tangent to each other, and to Line AB and Line BC. If Angle ABC=120°, what is the ratio of the radius of Circle M to the radius
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-> SOLUTION: In the diagram attached below, Circles M and N are tangent to each other, and to Line AB and Line BC. If Angle ABC=120°, what is the ratio of the radius of Circle M to the radius
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Question 1189710: In the diagram attached below, Circles M and N are tangent to each other, and to Line AB and Line BC. If Angle ABC=120°, what is the ratio of the radius of Circle M to the radius of circle N
Diagram: https://imgur.com/a/4dfUycL Found 2 solutions by greenestamps, math_tutor2020:Answer by greenestamps(13200) (Show Source):
I'll use the letter notation he has set up.
Here's what the drawing looks like with those points mentioned.
Let circle N have a radius of 1. This means EN = ND = 1.
Goal: Find the length of FM
This will effectively give the ratio FM/EN, aka the ratio of the two radii
Note how FM/EN = FM/1 = FM
As mentioned by the other tutor, the 120 degree angle ABC is bisected into two smaller 60 degree angles
angle ABM = angle MBC = 60
This leads to triangles BEN and BFM being 30-60-90 triangles
The points of tangency E and F have the 90 degree angles at those locations.
The useful formulas for any 30-60-90 triangle are that
hypotenuse = 2*(short leg)
long leg = sqrt(3)*(short leg)
That second formula rearranges to
short leg = (long leg)/sqrt(3) = (long leg)*sqrt(3)/3
For triangle BEN we have EB as the short leg and EN as the long leg
short leg = (long leg)*sqrt(3)/3
EB = (EN)*sqrt(3)/3
EB = (1)*sqrt(3)/3
EB = sqrt(3)/3
And,
hypotenuse = 2*(short leg)
BN = 2*EB
BN = 2*sqrt(3)/3 is the hypotenuse of triangle BEN
Let x be the radius of circle M
FM = x
DM = x as well since they're both radii of the same circle M
Now focus on triangle BFM
Since it is also a 30-60-90 triangle we can say
short leg = (long leg)*sqrt(3)/3
BF = (FM)*sqrt(3)/3
BF = x*sqrt(3)/3
and also
hypotenuse = 2*(short leg)
BM = 2*(BF)
BM = 2x*sqrt(3)/3