"..If m(arc AC) = 105˚, what is the measure of ∠BXY?"

Rinn 🌸

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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.
11264
Thank you so much in advance! :]
-Rinn
 

MarkFL

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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?
 

Jomo

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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?
I am not seeing this one. What's next?
 

MarkFL

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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...
 

Jomo

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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...
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.
 

MarkFL

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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.
I haven't actually solved the problem, I was just posting suggestions I felt would lead to the solution. :)
 

Denis

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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?

Edit: also given is XY = AX + BY (see DrP's post#11
 
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Jomo

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If I agree with that, then what is next?
 

Dr.Peterson

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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.
View attachment 11264
Thank you so much in advance! :]
-Rinn
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.

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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Denis

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If I agree with that, then what is next?
What is next is same...I was only wondering why circles are used...
To Rinn: do you know if the solution involves geometry only?
 

Dr.Peterson

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

Denis

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

Denis

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Think I finally got this straight!
Code:
A          b             B


a                        a


X          b             C
 
                         c

                         Y
angleCXY = 105 - 90 = 15, so angleCYX = 75
XY = 2a + c, so b = sqrt[(2a + c)^2 - c^2] (thanks DrP!)
 

Denis

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Well, played around with this: keep getting angleBXY = 20.989984....
Looks the the given 105 degrees was a rounded result.
Guess that eliminates a non-trig solution...
 

Dr.Peterson

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I got a little less than 36 degrees by trig, which agrees with a drawing.

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

Denis

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