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What is the length of the hypotenuse (OB)?
\(OC = OA = 12\) because they are the radii of the same circle.
Also, as whpalmer asked, what is the hypotenuse, or \(OB\)? A good start might be noticing that \(OB = OC + CB\).
or, just tell us how you came up with 37...then we can see if it is a simple arithmetic error, or how you approached the problem, or whatever...
OB^2=OA^2+AB^2 if angle A = 90 37^2=12^2+AB^2 other wise you have the sine angle formula
I assumed (and ParthKohli too, I'm sure) that because only one intersection point is shown, AB is tangent to the circle, which means that OAB is a right angle.
imagine A approaching C
Not enough information given without assumption angle A = pi/2
When you assume you make a retriceof u and me
retriceof u and me
I guess it doesn't like that word
That didn't help at all. You never even told me how to get that answer....
you also didn't tell us how you got your answer...
adding what, exactly, duh?
And uh...I need to know AB to do that. DUH!!
emily the ans for ab is 35. its a right angle triangle
Now wait...you're going "duh" at everyone else and yet you can't figure out what AB is from the equation 37^2=12^2+AB^2 ???
I used the Pythagorean theorem a^2+b^2=h^2 h = sqrt(a^2+b^2) or in out case b=sqrt(h^2-a^2)
@Kapt_Crazy So I was right? The answer is 35? :)
Generally you shouldn't just give quantity but show all work so it's easy checked or to show how it's done
yeah u right :)
No, you didn't say 35. You said "I think it's 37" which you absurdly arrived at by adding AO and BC together. Also, 35 is only correct if AB is tangent to the circle at A. But since that information isn't given, then its only an assumption.
qweqwe if it was not a right angle triangle they would give u more info, like one of the angles or sumthing
Oh, thankyou for going back and checking if I said 35. :) @qweqwe123123123123111
Yes they would have to give 2 angles to solve
The assumption is probably a safe one, @Kapt_Crazy, But UNLESS THAT INFO IS GIVEN EXPLICITLY, it's still just an assumption. And the number of poorly formed and incomplete problems that are posted here every single day are legion, so there's no good reason to assume that this problem is stated any better than any other.