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Topic: Hardy Perennial (Read 979 times) |
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Barukh
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Triangle ABC is isosceles. All the angles' measures are in degrees. Find the angle marked '?'
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aero_guy
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Re: Hardy Perennial
« Reply #1 on: Nov 22nd, 2003, 12:52pm » |
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OK, I solved it, though I had to bust out the law of sines. Trickier than it initially looks. I notice now that there are similar triangles involved. The question is how would I have known that initially without going through the mess?
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Dudidu
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Barukh hi,
Could it be 30[smiley=suptheta.gif].
If it does then you can read the following explanation:
You can look at the attached file to see a drawing of how I got it 
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aero_guy
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Re: Hardy Perennial
« Reply #3 on: Nov 23rd, 2003, 2:39pm » |
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Dudidu, that is exactly what I got and exactly the way I got it. What I noticed is that the middle triangle is similar (geometrically speaking) to the one that contains it and the one to its lower left. I was wondering if there was a I way I could have found this without using the law of sines (as we both did) to find the angle.
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ThudnBlunder
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Re: Hardy Perennial
« Reply #4 on: Nov 23rd, 2003, 3:43pm » |
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Quote:| I was wondering if there was a I way I could have found this without using the law of sines (as we both did) to find the angle. |
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Hint:Draw a line BD such that D lies on AC and angle ABD = 20 degrees.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Barukh
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Re: Hardy Perennial
« Reply #5 on: Nov 24th, 2003, 5:55am » |
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on Nov 23rd, 2003, 2:39pm, aero_guy wrote:| Dudidu, that is exactly what I got and exactly the way I got it. |
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The following question is for both you and Dudidu: how did you solve the sines equation?
Quote:| I was wondering if there was a I way I could have found this without using the law of sines (as we both did) to find the angle. |
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I haven't found the way to follow THUDandBLUNDER's hint yet  , but here's another way: Consider the regular 18-gon.
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Lightboxes
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Re: Hardy Perennial
« Reply #6 on: Nov 24th, 2003, 7:47am » |
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Quote:| You can look at the attached file to see a drawing of how I got it |
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I'm confused...
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If the 70 degrees (both of them are) are created with two lines that intersect, then the remainding angles of the intersecting lines are as follows (I think):
360 - (70+70) = 220.
Then 220 / 2 - 110.
That means that the triangle to the left has thre angles now. 80+30+110??
But also (110-x) + 50 + ? = 180.
That leads to ? = 50.
Then the very top small triangle is 20+50+? = 180. Then ? =110. Then 110 + x + 80 = 180?
Unless these are not truely lines and bend as they intersect? I think this is the case.
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| « Last Edit: Nov 24th, 2003, 8:06am by Lightboxes » |
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aero_guy
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Re: Hardy Perennial
« Reply #7 on: Nov 24th, 2003, 8:10am » |
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He messed up, it should be 40 not 80. You find the 40 specifically because you know the 110 and the 30.
Barukh, doesn't Dudido's picture show you?
T and B, there you go, now that is as nice and simple as it should be. A little bit of mirrored symmetry and you have it. Still doesn't use similar triangles, but I guess that was a red herring.
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| « Last Edit: Nov 24th, 2003, 8:14am by aero_guy » |
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Barukh
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Re: Hardy Perennial
« Reply #8 on: Nov 24th, 2003, 8:47am » |
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on Nov 24th, 2003, 8:10am, aero_guy wrote:| Barukh, doesn't Dudido's picture show you? |
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Well, what I've got from Dudidu's picture - after simplification - is: sin(20+x) / sin(x) = 2sin(50). Is there a way to solve it without guessing?
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aero_guy
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Re: Hardy Perennial
« Reply #9 on: Nov 24th, 2003, 2:31pm » |
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Using the equations he has at the side gives sin(x)/sin(160-x)=sin(40)/sin(80).
Now I bust out my old analytic geometry book (I very often use the tables at the back but almost never look at the rest) we see that sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B).
Combine the two and you should be able to solve explicitly for x. This is of course the ineligant way of doing it.
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Dudidu
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Re: Hardy Perennial
« Reply #10 on: Nov 27th, 2003, 1:00am » |
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Quote:Lightboxes you should be, I made a writing mistake. As aero_guy indicated it should have been 40 and not 80 (you can also see that in the equations I suggested, I used 40 and not 80).
Quote:| Using the equations he has at the side gives... |
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This is exectly what I got and as aero_guy indicated (again) it can be solved (inteligently) by known trigonometry equalities or (simply) by a calculator  .
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| « Last Edit: Nov 27th, 2003, 1:01am by Dudidu » |
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ThudnBlunder
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Re: Hardy Perennial
« Reply #12 on: Dec 8th, 2007, 10:17am » |
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Here is a nice animation.
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