Can anyone solve this math for me?
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No, but someone could
Correct.
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Jesus.
Jesus is always the answer
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OP sneakily making Lemmy do their homework, well played.
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Tats so cool! Did you just do that or find it?
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if we assume the bottom right corner is a right angle and is the center of the arc, then it is solvable in the manners that others here have already described. if either of those is not the case, and the image itself doesn't state, then there is insufficient information to solve it.
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Take a look at this page, it'll give you not only your answer but explain how to solve it
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No, sorry, I'm dumb.
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Note that the problem states that the outer shape is a quarter circle, information not provided in OP's question.
Knowing it is a quarter circle is important because it allows us to validate that the two lengths are equal and the bottom-right angle is 90 degrees.
I actually came to the comments to see if we had this information! Thanks.
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How have I had years of math and not seen this? I mean, it’s not super useful for me, but I would have thought I would have seen something like this in pre calculus at least.
I teach this to my grade 9 class in Canada. It’s on the curriculum.
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Well the drawing is wrong. I measured it with a ruler and it should be 9
Ok, but what does a well have to do with that?
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Take a look at this page, it'll give you not only your answer but explain how to solve it
Dude so smart i was already breaking out the angles and testing everything out
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I teach this to my grade 9 class in Canada. It’s on the curriculum.
Nice. I have no recollection of seeing this before.
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No, sorry, I'm dumb.
Hello dumb! I'm dad
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if we assume the bottom right corner is a right angle and is the center of the arc, then it is solvable in the manners that others here have already described. if either of those is not the case, and the image itself doesn't state, then there is insufficient information to solve it.
You sure?
Draw line theta from left to right intersections and ya got two triangles.
Pythagoras gets you the length of that line. Trig gets you the two remaining angles of the red triangle (sohcahtoa!!)
180-angles gets you one angle of the new triangle. Then trig again gets you cos theta = x/length and you get the remaining angles. Maybe I left a step out but generally thats the process -
Nice. I have no recollection of seeing this before.
Tbf most Canadian grade 9 teachers skip it.
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Spent too long trying to figure out if this was loss or not.
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Shouldn't the person who to lazy to measure x solve this?
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24,7 and make a pythagorean triple with 25 as the hypotenuse. If the problem uses one pythagorean triple, it probably uses another, so I assume x is 15, and the radius is 20.
Not the most complete answer, but definitely the fastest one^^
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I'ma go with 8 because it's slightly longer than 7
