Find radiuses of circles, if OO1=20cm and 3AK=2BK

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Explanation:

The letter O looks very similar to the number zero, so I'm going to use other letters to avoid confusion. For the smaller circle on the left, it'll have center P. The circle on the right will have center Q. See the diagram below.

Based on that, we can say that PQ = 20 cm.

Let x = length of PK

PK + KQ = PQ

KQ = PQ - PK

KQ = 20 - x

We'll keep this in mind for later. I'll refer to this as "Fact 1" when it does come up later.

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We're given this equation

3AK=2BK

Let's get the numbers to one side and the letters to the other side like so:

3AK=2BK

3AK/BK = 2

AK/BK = 2/3

We'll use this later as well.  I'll refer to this as "Fact 2" when it does come up later.

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Next, draw a segment from P to A to form triangle PAK (also shown in the diagram below). Do the same for points Q and B to form triangle QBK.

The segments PA and PK are radii of the smaller circle, so PA = PK which leads to triangle PAK being isosceles. Triangle QBK is also isosceles for similar reasoning.

The angles AKP and QKB are vertical angles, so they are congruent.

The facts mentioned so far boil down to the two triangles being similar triangles (use the angle angle method to prove).

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Now that we know the triangles are similar, we can use a proportion to solve for x.

PK/KQ = AK/KB

PK/KQ = 2/3  .... use fact 2

x/(20-x) = 2/3 .... use fact 1

3x = 2(20-x) .... cross multiply

3x = 40 - 2x

3x+2x = 40

5x = 40

x = 40/5

x = 8

So,

The radius of the smaller circle on the left is 8 cm.

The radius of the larger circle on the right is 12 cm.

here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

=  √(225 +64)

=  √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.

okay wait before i answer, is this for your points on a graph?