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October 1, 2014

October 1, 2014

Posted by **Connie** on Friday, April 25, 2008 at 2:54pm.

Can anyone please give me some ideas how to do it without graphing it out? THANKS A LOT!

- MATH -
**Damon**, Friday, April 25, 2008 at 3:32pmThe centers are at (3,5) and at (7,5) along the horizontal line y = 5 and 7-3 = 4 apart

left radius = 5

right radius = 3

so

the points lie on vertical line x = something an equal distance above and below y = 5

so

triangle

base = 4 along y = 5 line

left leg = 5

right leg = 3

let's use law of cosines to find the angle on the left, between the y = 5 line and the long side length 5

3^2 = 5^2 + 4^2 - 2*5*4 cos A

9 = 25 + 16 - 40 cos A

32 = 40 cos A

A = 36.87 degrees

now the distance from x = 3 to our desired vertical line is

5 cos A which is 5 (32/40) = 4

so our line is

x = 3 + 4 = 7

which we could have said earlier seeing that our triangle is

3, 4, 5 and therefore a right triangle

- MATH -
**Jordan**, Friday, April 25, 2008 at 9:32pmWhat do you mean?

- MATH -
- MATH -
**Jordan**, Friday, April 25, 2008 at 9:32pmWhat do youm mean?

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