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Inv. D Quick question
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Inv. D Quick question
by E=mc^2 » Sat Mar 02, 2013 7:46 pm
Did #70 on today's test have a typo, or can you actually solve with only the given values?
CA: 17^2 (Invitational A)
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Re: Inv. D Quick question
by qwerty123 » Sat Mar 02, 2013 8:10 pm
why is everybody talking about the test! I thought you're not supposed to be discuss any part of it!
[unparseable or potentially dangerous latex formula]
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Re: Inv. D Quick question
by darksaber21 » Sat Mar 02, 2013 8:29 pm
You technically can, as today was the only day where you can take the test. I didn't take it, so I don't know what people are talking about.
Now I know the stakes, and I'm willing to show what I gots. I will see you all at state for the first and final time, so let's end the year with a bang! :)
MTHS - Class 5A, District VII, Region I
MTHS - Class 5A, District VII, Region I
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Re: Inv. D Quick question
by sxk1693 » Sat Mar 02, 2013 11:28 pm
So we can...Let it Rip!
How do you do number 1-70
How do you do number 1-70
HIGHS 2013 MEETS:
NS: 220(TMSCA STATE)
CA: 232(B)
MA: 274(District)
SC: 220(A)
CS: 212(VCM1)
NS: 220(TMSCA STATE)
CA: 232(B)
MA: 274(District)
SC: 220(A)
CS: 212(VCM1)
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Re: Inv. D Quick question
by 101dalmatians » Sun Mar 03, 2013 11:03 am
I guess this signature really doesn't matter now that I know that the forum is going down soon...
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Re: Inv. D Quick question
by 101dalmatians » Sun Mar 03, 2013 2:58 pm
I guess this signature really doesn't matter now that I know that the forum is going down soon...
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Re: Inv. D Quick question
by lrp » Sun Mar 03, 2013 3:07 pm
I get you but I dont really get how to work those.
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Re: Inv. D Quick question
by darksaber21 » Sun Mar 03, 2013 3:25 pm
I could help, if I knew the question(s)... -_-
Oh well...
Oh well...
Now I know the stakes, and I'm willing to show what I gots. I will see you all at state for the first and final time, so let's end the year with a bang! :)
MTHS - Class 5A, District VII, Region I
MTHS - Class 5A, District VII, Region I
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Re: Inv. D Quick question
by lrp » Sun Mar 03, 2013 3:28 pm
Sorry its cause they r geometry questions
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Re: Inv. D Quick question
by Fredfredburger » Wed Mar 06, 2013 10:17 pm
#69
The shaded area is equal to the total circle area minus the little, unshaded circle. If we denote the little circle radius as r, then the area of the shaded section is
[unparseable or potentially dangerous latex formula]
Solve for r,
[unparseable or potentially dangerous latex formula]
Your total radius then is r, the little circle radius, plus the rest of the radius.
[unparseable or potentially dangerous latex formula]
#70
No typo, notice that the hypotenuse of the smaller right triangle and the bottom leg are both of length r, the radius of the circle. Denote the angle you're looking for by theta and note that the other acute angle is 90 minus theta. It's superfluous to use the other acute angle, or cotagent for that matter, but it helped me organize my thoughts better for some reason. Now you have the two trigonometric ratios
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
Taking advantage of trig identities, we can solve the second equation for r and then plug that into the first equation.
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
Canceling out the sines on both sides, and isolating the cosine, we are left with
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
The shaded area is equal to the total circle area minus the little, unshaded circle. If we denote the little circle radius as r, then the area of the shaded section is
[unparseable or potentially dangerous latex formula]
Solve for r,
[unparseable or potentially dangerous latex formula]
Your total radius then is r, the little circle radius, plus the rest of the radius.
[unparseable or potentially dangerous latex formula]
#70
No typo, notice that the hypotenuse of the smaller right triangle and the bottom leg are both of length r, the radius of the circle. Denote the angle you're looking for by theta and note that the other acute angle is 90 minus theta. It's superfluous to use the other acute angle, or cotagent for that matter, but it helped me organize my thoughts better for some reason. Now you have the two trigonometric ratios
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
Taking advantage of trig identities, we can solve the second equation for r and then plug that into the first equation.
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
Canceling out the sines on both sides, and isolating the cosine, we are left with
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
MIT '17
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