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Points X and Y are on a circle centered at O, and point Q is outside the circle such that line QX and line QY are tangent to the circle. If angle XQO = 32, then what is the measure of minor arc XY, in degrees?

 Jan 15, 2020
 #1
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Arc XY is 142 degrees.

 Jan 15, 2020
 #2
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I'm sorry, could you include a solution? I'd like to understand how to solve the problem on my own.

Guest Jan 15, 2020
 #3
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See the following image

 

 

Note that OXQY  forms a quadrilateral

Since  QX and QY   are tangent to the circle, then angles  QXO  and QYO   = 90 °

And by symmetry,  angle   XQY  = 2(32)    = 64°

 

And the interior angles of a quadrilateral sum to 360°

 

So

 

QXO + QYO + XQY  + XOY  =  360

 

  90  +   90   +   64  + XOY  =  360

 

244 + XQY  = 360         subtract  244 from both sides

 

XOY  = 116°

 

And since  XOY  is a central angle........then minor arc  XY has the same measure  =   116°

 

 

cool cool cool

 Jan 15, 2020
edited by CPhill  Jan 15, 2020
edited by CPhill  Jan 15, 2020
 #4
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Thanks, I was actually able to solve it right before you lol!

Didn't properly understand the question so I didn't draw a proper diagram; once I did the solution was obvious.

Guest Jan 15, 2020

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