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January 10th, 2014, 09:55 AM   #1
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measure of BDE



Find measure of BDE.
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January 10th, 2014, 11:33 AM   #2
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Angle BDE is 40. I have only a clumsy trigonometric proof of this at present.
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January 11th, 2014, 06:12 AM   #3
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Re: measure of BDE

b = DE, d = BE, e = BD, u = angleBDE

d = SIN(50) / SIN(80)
e = SIN(80) / SIN(70)
b = SQRT[d^2 + e^2 - 2deCOS(20)]

u = ASIN[dSIN(20) / b] = 40

Skip, does that clumsiness outdo yours

Kind of an annoying one; you can get most angles by "sight"....but BDE remains "elusive"

Tried extending BD to BF such that BF = BC, but that only lead to headaches...
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January 11th, 2014, 07:13 AM   #4
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Re: measure of BDE

Quote:
Originally Posted by Denis
b = DE, d = BE, e = BD, u = angleBDE

d = SIN(50) / SIN(80)
e = SIN(80) / SIN(70)
b = SQRT[d^2 + e^2 - 2deCOS(20)]

u = ASIN[dSIN(20) / b] = 40

Skip, does that clumsiness outdo yours

Kind of an annoying one; you can get most angles by "sight"....but BDE remains "elusive"

Tried extending BD to BF such that BF = BC, but that only lead to headaches...
Wow! I had a really clumsy way to the answer that took me quite a while. I would be ever so grateful if you could explain why/how your solution works, Denis.
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January 11th, 2014, 07:54 AM   #5
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Re: measure of BDE

Construct triangle DEP with P = 40░, then P is the center of the circumcircle of BDE. Since BP = EP, angle EBP = angle BEP = 50░, hence EP is parallel to DA. Extend DA to meet circumcircle BDE at Q. Extend EP to meet circumcircle BDE at R, then EDQR is an isosceles trapezoid. Angle RQD = angle EDQ = 110░. Angle BDE = angle EDQ - angle ADB (70░) = 40░.
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January 11th, 2014, 10:59 AM   #6
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Re: measure of BDE

Quote:
Originally Posted by Slinkey
Quote:
Originally Posted by Denis
b = DE, d = BE, e = BD, u = angleBDE

d = SIN(50) / SIN(80)
e = SIN(80) / SIN(70)
b = SQRT[d^2 + e^2 - 2deCOS(20)]

u = ASIN[dSIN(20) / b] = 40
Wow! I had a really clumsy way to the answer that took me quite a while.
I would be ever so grateful if you could explain why/how your solution works, Denis.
Huh? I thought mine was clumsy too, with 4 steps!

Let AB = 1

Triangle ABE: angleAEB = 180 - 20 - 30 - 50 = 80
so: d / SIN(50) = 1 / SIN(80) : d = SIN(50) / SIN(80)

Triangle ABD: angleADB = 180 - 30 - 50 - 30 = 70
so: e / SIN(80) = 1 / SIN(70) : e = SIN(80) / SIN(70)

Next, use triangle BDE: b^2 = d^2 + e^2 - 2deCOS(20)

Wrap up with Sine Law to get angleBDE.
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January 11th, 2014, 11:17 AM   #7
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[color=#00AA00]Denis[/color], how do you find the exact value of the square root you use?

Are you assuming that P lies on AB, [color=#AA0000]greg1313[/color], or do you have a proof that it does?
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January 11th, 2014, 12:57 PM   #8
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Re: measure of BDE

I did not see that as being essential to the proof I gave. But it is.
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January 11th, 2014, 02:05 PM   #9
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Re:

Quote:
Originally Posted by skipjack
[color=#00AA00]Denis[/color], how do you find the exact value of the square root you use?
I don't; UBasic does!
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