anonymous
  • anonymous
How to find angles 1,2,3,4,5,6,7,8 picture shown below.!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
I can't see how one can figure out those angles without some other properties of the lines. For example, angle 4 could be 90 degrees if P is center and line HD is a tangent to circle. But, that is not "given" by just looking at the picture.
anonymous
  • anonymous
Find the measure of the following numbered angles in circle P when arc AE = 53°, arc BA= 68°, and arc CB = 72°.

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anonymous
  • anonymous
Sorry!
anonymous
  • anonymous
Angle 7 = arc AB + arc AE = 53 + 68 = 131 degrees. Triangle PBE = isosceles triangle => angle 3 = (180-angle7)/2 Keep going like that.
anonymous
  • anonymous
would angle 8 be 90?
anonymous
  • anonymous
No. You need to calculate based on arc and triangle properties.
anonymous
  • anonymous
well, since BA=68, wouldn't angle 1 be half?
anonymous
  • anonymous
No. Angle 1 can be only obtained after you figure out: 1) Angle 4 2) Angle 2 3) Angle 6 in that order.
anonymous
  • anonymous
okay..
anonymous
  • anonymous
Angle 4 is 90 degrees if line HD looks like a tangent. It appears more information is needed. Also, it is nearly impossible to keep referring to the image in a different browser window and rationally respond.
anonymous
  • anonymous
I know angle 4=90 so wouldn't angle 2=45?
anonymous
  • anonymous
Yeah, I know. I don't know any other way to post the problem.
anonymous
  • anonymous
Angle 2 is greater than angle 4. So, it must be 90 plus something. To get that something, you need to form triangle PBA first. It will be an isosceles triangle. Given arc BA = 68 degrees, angles A and B in that triangle PBA = (180-68)/2 = 59 degrees. Angle B in that triangle PBA = angle 3 + angle ABF. Because you know angle 3 and angle PBA, you can get angle ABF. From that, you use the fact that triangle BAD forms right triangle to find angle ADB. You add 90 degrees to that to get angle 2. Very complex. :)
anonymous
  • anonymous
wow, I'll have to work on that, thanks!

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