anonymous
  • anonymous
Find the value of x. FG perpendicular to OP, RS perpendicular to OQ, FG=20, RS=24, OP=14
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
@Hunus @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
let r = radius of the circle

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jim_thompson5910
  • jim_thompson5910
by the pythagorean theorem, we can say (FP)^2 + (OP)^2 = (OF)^2 10^2 + 14^2 = r^2 solve for r and tell me what you get
anonymous
  • anonymous
100 + 196 = 296^2. Now what?
jim_thompson5910
  • jim_thompson5910
r^2 = 296 so r = ???
anonymous
  • anonymous
296
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
r^2 is 296, not r
anonymous
  • anonymous
87, 616
jim_thompson5910
  • jim_thompson5910
don't square it though
jim_thompson5910
  • jim_thompson5910
how do you undo squaring something
anonymous
  • anonymous
sqrt?
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
16.9?
jim_thompson5910
  • jim_thompson5910
you did sqrt(296) right?
anonymous
  • anonymous
Yea
jim_thompson5910
  • jim_thompson5910
it's close, but not correct
jim_thompson5910
  • jim_thompson5910
sqrt(296) = 17.2046505340852 round to whatever place you need to
anonymous
  • anonymous
17.2 isn't one of the answers. Only 16.9
jim_thompson5910
  • jim_thompson5910
but remember, we want x, not r
jim_thompson5910
  • jim_thompson5910
we use r to get x
anonymous
  • anonymous
the possible answers are: 16.9, 14, 12.3, and 12..
jim_thompson5910
  • jim_thompson5910
\[\large r = \sqrt{296} \rightarrow r^2 = 296\] \[\large (OQ)^2 + (RQ)^2 = (OR)^2\] \[\large (OQ)^2 + (RQ)^2 = r^2\] \[\large x^2 + 12^2 = 296\] \[\large x^2 + 144 = 296\] ... ... ... \[\large x = ???\]
anonymous
  • anonymous
152
anonymous
  • anonymous
12.3
jim_thompson5910
  • jim_thompson5910
\[\large x^2 = 152\] so \[\large x = ???\]
jim_thompson5910
  • jim_thompson5910
good, x is roughly 12.3
anonymous
  • anonymous
What about this one?
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jim_thompson5910
  • jim_thompson5910
what about it? what do you want to find? and what are you given?
anonymous
  • anonymous
Oh. Lol. Forgot.\ AB is a diameter and AB is perpendicular to CD.
jim_thompson5910
  • jim_thompson5910
alright so that's all the given info?
anonymous
  • anonymous
I know the angles are all 90 degrees.
jim_thompson5910
  • jim_thompson5910
what do you need to find exactly?
anonymous
  • anonymous
Find BD for AC= 58 DEGREES
jim_thompson5910
  • jim_thompson5910
AC is a segment, not an angle
anonymous
  • anonymous
Okay..
jim_thompson5910
  • jim_thompson5910
so is it ACP or ACB that's 58 degrees
anonymous
  • anonymous
This is exactly what it says. Find m BD for m AC= 58 degrees.
jim_thompson5910
  • jim_thompson5910
oh arc AC, gotcha
jim_thompson5910
  • jim_thompson5910
angle APC = (1/2)*(arc AC + arc BD) 90 = (1/2)*(58 + x) 90*2 = 58 + x 180 = 58 + x solve for x
anonymous
  • anonymous
122
jim_thompson5910
  • jim_thompson5910
yep

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