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March 20th, 2011, 03:46 PM   #1
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A confusing proof

Digram attached.

The aim is to show that the height of, P, above the floor after being tilted is h(cosb+2sinb)
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 March 20th, 2011, 04:03 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: A confusing proof We can use a rotation of axes, where the positive x-axis is rotated by -b: $Y=-x\sin(-b)+y\cos(-b)=2h\sin(b)+h\cos(b)=h$$\cos(b)+2\sin(b)$$$
 March 20th, 2011, 04:05 PM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 The diagram after being tilted, draw a line perpendicular to the floor that crosses bottom right corner of the rectangle. Draw a line that is parallel to the floor that crosses bottom right corner of the rectangle. Finally draw a line perpendicular to the floor that crosses top right corner of the rectangle. We can find the height (h_1) from the floor to the bottom right corner of the rectangle by using sin b = h_1 / 2h, this will give us h_1 = 2h sin b. We can find the height h_2 = the height from floor to the top right corner of the rectangle - h_1. cos b = h_2 / h, this will give us h_2 = h cos b. Therefore the height from floor to the top right corner of the rectangle = h_2 + h_1 = h cos b + 2h sin b = h(cos b + 2 sin b).
 March 20th, 2011, 04:21 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: A confusing proof Another way: Let Y be the height of P after the tilt. The diagonal of the rectangle is $h\sqrt{5}$ thus: $\sin(a+b)=\frac{Y}{h\sqrt{5}}$ Use additive-angle identity for sine: $\sin(a)\cos(b)+\cos(a)\sin(b)=\frac{Y}{h\sqrt{5}}$ Using $\sin(a)=\frac{h}{h\sqrt{5}}=\frac{1}{\sqrt{5}}$ and $\cos(a)=\frac{2h}{h\sqrt{5}}=\frac{2}{\sqrt{5}}$ we have: $\frac{1}{\sqrt{5}}\cos(b)+\frac{2}{\sqrt{5}}\sin(b )=\frac{Y}{h\sqrt{5}}$ Multiply through by $h\sqrt{5}$: $h$$\cos(b)+2\sin(b)$$=Y$
 March 20th, 2011, 04:29 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Cool!
 March 20th, 2011, 05:21 PM #6 Newbie   Joined: Mar 2011 Posts: 12 Thanks: 0 Re: A confusing proof I solved it but it was just a different way of marks In the rectangle you can work out Length OP=h?5 sina=1/?5 cosa=2/?5 height=h?5.sin(a+b) =h?5.(sinacosb+cosasinb) =h?5.(1/?5cosb+2/?5sinb) =h(cosb+2sinb) does this working look right?
 March 20th, 2011, 05:24 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: A confusing proof Yes, looks good to me.
 March 20th, 2011, 05:29 PM #8 Newbie   Joined: Mar 2011 Posts: 12 Thanks: 0 Re: A confusing proof ok good, im doing revision for an exam and this is the sort of question they will ask. If i was to use the same working as this would it be correct? or would you recommend a different way?
 March 20th, 2011, 05:33 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: A confusing proof I can't see any problems with what you did...I can't speak for your professor, but every step you used is valid.
 March 20th, 2011, 05:36 PM #10 Newbie   Joined: Mar 2011 Posts: 12 Thanks: 0 Re: A confusing proof Thanks, i just needed to check I havnt made any small errors. Thanks for all your help today

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