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The School of Athens
Euclid, as imagined by Raphael in this detail from The School of Athens.
Jamshed Al-Kashi
Al-Kashi was one of the greatest mathematicians of the 15th century, calculating the sin(1*) correct to 16 decimal places.
Al-Beruni
The 11th century scientist and philosopher managed to estimate the radius of the earth using a very creative method that employed trigonometry.
47 comments:
For part b, i figured out how to express the height of ABC in terms of x, but im having trouble expressing BC, the base, in terms of x. Can anyone help me in the right direction?
for part a i'm having trouble finding the height of XYZ in order to express the area in terms of x. could anyone help?
and virginia i think maybe drawing some radii to make another isosceles triangle might help.
I, also, for part b am having trouble expressing the base (BC) in terms of x. I've drawn helpful radii, but feel like I must be missing something obvious.
See if you can get the altitude of XYZ in terms of x. The whole trick is in identifying ratios and then creating proportions. Then it is a bunch of algebra.
As well, I am having trouble expression the base BC in terms of X. Can someone push me there?
I used the radii and the length x given to find 1/2 the base using the pythagorean theorem, if that helps.
That's what I had done to find the base for BC, but I'm having issues finding the altitude from A and expressing it in terms of X. Perhaps it is just because my diagram is too much like an equilateral triangle, but I have tried drawing others and I'm still drawing a blank. Any suggestions?
for part a im having trouble finding the height of xyz and for part b, what is meant by AB=AC and BC is x inches from the center of the circle on the opposite side from A
BC is a chord in the circle, and "x" is the distance from the centre to that chord
so the circle is bigger than the triangle as in the triangle is in the cirle
and im stuck on part a
yes the triangle is inside the circle. for part A, look for similar triangles.
and what does it mean AB=AC, and can you maybe push me farther than similar triangeles or is that not allowed
the side AB of the triangle equals side AC. and just look for other triangles
is part b a totally different diagram
o that makes it easier
i am still confused on how to find the height of xyz in part a. can anyone help?
im also confused. nobodys really given much insight. i think i got the second half though...
va,
did you get really ugly numbers for part b
triangle XDC versus triangle XYZ
right they are similar and share an altitude i got that but where from here??/
bases of the triangles
you know the triangles are similiar ina ration, therefore their altitudes are similar in that same ratio
aha
well wait, then wouldnt we have to know one of the altitudes for that to work
u knoe the base of XDC and XYZ and if the triangles are similar they are in the ratio of corresponding sides
Ok so I thought I had the base for 2b in terms of X, but after going over my work I realized it was wrong. How did you use a radii, and X and the pythagorean theory to find BC if you don't have any other measurments?
For 2b, I realized that my approach to finding the height of triangle ABC in terms of x was wrong. Any suggestions to find this? I thought that they radii might be medians, but with the given information I can't prove anything helpful.
scott- i'm not sure if this works then, but if you look at the radius and the height from BC to the center I think you can find the height in terms of x because you know the triangle is isoceles. does that seem like it works?
Virginia- I'm not sure if I understand what you are saying.
Do you extend x to the circle (on the opposite side as A) to create another radius?
Michael i still dont see how that helps you figure out an altitude
virginia, I got the height, I hope, but I don't understand how to get the base. Any ideas?
Michael, for 2a, what you're saying still doesn't help us find an altitude for either, and I think we do need that. If we looked at XYZ (XFQ is the altitude from X to DC (meeting at F) and YZ (meeting at q)) and then XYQ, then we know the altitude of one of them, and then we can work from there. The base of XYQ, after all, is just x
Oh dear. Not XYZ. Sorry.
Look at XQY and DAY. there.
Has anyone gotten YA and BZ to be 4?
For 2b, I still don't see how to express the height of the triangle in terms of x.
Scott, I think the altitude and the perpendicular bisector which intersect the unequal side in an isosceles bisector are the same, and so then the altitude from A would have to contain length X from circumcenter/center of circle O. Then, wouldn't it just be 6 inches + x? ...
That's what I thought too. Have you found the base in terms of X?
I'll try that. I didn't know if there was another way or not... I still need to figure out the base...
W Stott- Mr. Kerai told me to not focus on finding numbers for the base since we already know that it is 2x, and to focus on finding the altitude of XYZ in terms of X
For 2b, I used the pythagorean theorem to find the base in terms of X. It said that x was the length from the center of the circle to BC, and so that'd be the perpendicular, and 6 inches/the radius would be the hypotenuse.
And, Sam, I was focusing too much on the numbers. I finally just put in H as the altitude of XYZ, rethought about what Michael has said, and I think I have it now; thanks.
victoria- i'm not sure if this is just because of the way I drew my diagram, but found the base using the a right triangle formed by the radius and x and half the base.
Thanks. I just realized what iknew about the right triangles. I had written those things down, but never put them together. Thanks for helping me get my thoughts together. :)
Ahh I see where I was messing up. I think I have it now.
Even if you put another variable in for the height of XDC how do you find that just in terms of X?
sam, if your trying to find the height of XYZ in terms of x, try writting some equations with the height of triangle XDC and XYZ. Then you can use the ratios to find for your desired variable. Does this make sense?
Like I have 8/2x = h/12+h but I don't know what other equations you might be talking about
I don't know if you will get this in time, but that is the equation I used.
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