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Author Topic: Geometry Question -  (Read 2173 times) Bookmark and Share
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Munchor
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« Reply #15 on: 05 May, 2011, 22:20:40 »
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n64_super_mario_64_start.jpg shocked

Very well Smiley
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« Reply #16 on: 05 May, 2011, 22:21:55 »
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Hah whenever I need a fresh image I always find one I'm not using on my desktop and clear it in paint and reuse it Tongue I don't always remember to rename it too Grin
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« Reply #17 on: 05 May, 2011, 22:22:21 »
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Quote from: Builderboy on 05 May, 2011, 22:21:55
Hah whenever I need a fresh image I always find one I'm not using on my desktop and clear it in paint and reuse it Tongue I don't always remember to rename it too Grin

I imagine your desktop Tongue
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« Reply #18 on: 05 May, 2011, 22:24:57 »
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Its actually almost empty because I just got this computer shocked
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« Reply #19 on: 05 May, 2011, 23:10:51 »
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Nice, I like the same-area solution Cheesy
Another way is this: (using the labels in Builder's picture)
ΔABC and ΔDBA are similar triangles, so AC/AD=BC/AB.
So AD=AC*AB/BC=40*30/50=24, as the others got. Smiley
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« Reply #20 on: 05 May, 2011, 23:32:24 »
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Quote from: Horrowind on 05 May, 2011, 21:49:40
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken

That really only works if the triangle is symmetric across both sides of the "height" line. It's not a general formula for the height of a triangle, especially when the triangle is obtuse (not the case here).
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∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ
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« Reply #21 on: 05 May, 2011, 23:40:07 »
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It doesn't have to be symmetric, it just has to be a right triangle, AFAICT, which is the fact that my similar-triangles solution relies upon.
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« Reply #22 on: 05 May, 2011, 23:41:27 »
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Er, yeah. I forgot that the right angle is a degenerate case that forces the height into discrete values.
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« Reply #23 on: 05 May, 2011, 23:51:25 »
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Oh okay Tongue We all make mistakes and forget, don't worry. Grin
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« Reply #24 on: 06 May, 2011, 03:18:57 »
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Quote from: Horrowind on 05 May, 2011, 21:49:40
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leads to h = 24, if i'm not mistaken

And here I was, about jump straight in with law of cosine...
Thumbs up for simplicity!
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« Reply #25 on: 06 May, 2011, 03:20:35 »
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What about me? Or are similar triangles too much? Tongue
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« Reply #26 on: 06 May, 2011, 03:23:15 »
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Wait.  How are they similar?

(I'm pretty sure you're going to jump in with an obvious answer that I overlooked...)
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« Reply #27 on: 06 May, 2011, 03:24:19 »
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By AA (angle-angle). They have a right angle and angle B in common.
Edit: angle B, not A. Fixed.
« Last Edit: 06 May, 2011, 03:25:29 by calcdude84se » Logged
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« Reply #28 on: 06 May, 2011, 03:26:58 »
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Ohhh...

Okay then, a thumbs-up for you too.
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« Reply #29 on: 06 May, 2011, 10:52:17 »
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I finally figured it out today. My mind was to busy yesterday I guess.
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