- Thread starter Rinn 🌸
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I am not seeing this one. What's next?Knowing that a radius of a circle which meets a tangent to the circle is perpendicular to that tangent, can you begin by giving the measures of the 4 interior angles of the quadrilateral ABYX?

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I actually did that before reading your post. This is where I am getting stuck.Next, I would draw the line segment BX and a segment from X perpendicular to BY which gives us two triangles with which to work...

Maybe it is too late for me to be doing math. I guess I will see this in the morning.

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I haven't actually solved the problem, I was just posting suggestions I felt would lead to the solution.I actually did that before reading your post. This is where I am getting stuck.

Maybe it is too late for me to be doing math. I guess I will see this in the morning.

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I would draw XC parallel to AB, and consider triangle XYC. One approach is to name the radii x and y; then XY = x+y and DY = y-x. You can solve for y/x, and use that to find angle BXC. I haven't carried it out, but it should work.Hello! This is a problem I've been struggling with in my Geometry class and I was hoping someone could help walk me through the steps for my upcoming test.

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Thank you so much in advance! :]

-Rinn

On the other hand, this is from geometry, not trig. There might be another way ... no, there is always another way.

EDIT: Corrected error, replacing AB with XY.

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What is next is same...I was only wondering why circles are used...If I agree with that, then what is next?

To Rinn: do you know if the solution involves geometry only?

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So the problem could be stated this shorter way:

Quadrilateral ABYX with angles A=90, B=90, Y=75, X=105.

Calculate angle ABX.

Y'all agree?

We also know that XY = AX + BY, in addition to the angles. With only the angles, there would not be enough information.What is next is same...I was only wonderingwhy circles are used...

Ah So! Dumb me missed that...with trig, I now get angle BXY = 35We also know that XY = AX + BY, in addition to the angles. With only the angles, there would not be enough information.

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I agree, it seems unlikely this can be solved by mere geometry. It would be nice to hear back from Rinn.

I'm not including the 15 degrees: 35.989984... if I did; so we agree...I got a little less than 36 degrees by trig, which agrees with a drawing.