"Narasimham" <mathma18@hotmail.com> wrote in message news:1168589454.987901.196770@l5 3g2000cwa.googlegroups.com... > A wedge of angle of 6 degrees is mounted on another wedge of 10 > degrees. Wedge sides are placed together resulting in an effective > wedge angle gamma as sum 16 degrees and as difference 4 degrees when > laid opposite after wedge rotation angle theta = 180 degrees. Find > theta for gamma = 10 degrees. > > Narasimham Hi Narasimham. That's a nice practical way of getting any slope between sum and difference of the wedge angles. I find it not so easy to be sure that we have the same definitions. Would you agree, that two 45° wedges at right angles give a slope of 60°? Then using the cos relation from spherical geometry cos(slope)=cos(16)*cos(10)-sin(16)*sin(10)*cos(10 the rotation) gives slope, gamma, 15.94° Hans

"Avni Pllana" <avniu66@hotmail.com> wrote in message news:21361640.1170526613573.Java Mail.jakarta@nitrogen.mathforum.org... > > On Jan 28, 1:23 am, Avni Pllana <avni...@hotmail.com> > > wrote: > > > > A wedge of angle of 6 degrees is mounted on > > another > > > > wedge of 10 > > > > degrees. Wedge sides are placed together > > resulting in > > > > an effective > > > > wedge angle gamma as sum 16 degrees and as > > difference > > > > 4 degrees when > > > > laid opposite after wedge rotation angle theta = > > 180 > > > > degrees. Find > > > > theta for gamma = 10 degrees. > > > > > > > Narasimham > > > > > > Hi Narasimham, > > > > > > Can you post a drawing that illustrates the > > problem. > > > > > > Best regards, > > > Avni > > > > Perhaps I should mention the blocks are trapezium > > prisms: > > > > http://img185.imageshack.us/img185/2691/wedgerotnoy0.j > > pg > > > > Narasimham > > > Hi Narasimham, > > Thank you for the drawing. Then according to the problem definition we have > > cos(gamma) = cos(10°)*cos(6°)-cos(theta)*sin(10°)*sin(6°), > > or > > theta = acos(cos(10°)*(cos(6°)-1)/(sin(10°)*sin(6°))), > > = 107.2906633°. > > Best regards, > Avni Avni, you are right, I think. I misread the post. Hans.

"Narasimham" <mathma18@hotmail.com> wrote in message news:1170614797.012270.228760@v3 3g2000cwv.googlegroups.com... > On Feb 3, 7:37 pm, "Hans Terkelsen" <d...@dreader2.cybercity.dk> > wrote: > > "Narasimham" <mathm...@hotmail.com> wrote in messagenews:1168589454.987901. 196770@l53g2000cwa.googlegroups.com... > > > A wedge of angle of 6 degrees is mounted on another wedge of 10 > > > degrees. Wedge sides are placed together resulting in an effective > > > wedge angle gamma as sum 16 degrees and as difference 4 degrees when > > > laid opposite after wedge rotation angle theta = 180 degrees. Find > > > theta for gamma = 10 degrees. > > > > > Narasimham > > > > Hi Narasimham. > > > > That's a nice practical way of getting any slope between sum and difference of the wedge angles. > > > > I find it not so easy to be sure that we have the same definitions. > > Would you agree, that two 45° wedges at right angles give a slope of 60°? > > No. > > > Then using the cos relation from spherical geometry > > Why should the cos relation from spherical geometry be relevant here? > > > cos(slope)=cos(16)*cos(10)-sin(16)*sin(10)*cos(10 the rotation) gives > > slope, gamma, 15.94° > > The wedges are 10,6 degrees, but not 16,10 degrees. > > > Hans > > Narasimham > Sorry, Narasimham, I misread your post. As to why the cos relation seems to be relevant: The pole of the plane is moving from vertical in a spherical triangle, 10 degrees, as you said, turning theta, then moving 6 degrees. The resulting move from vertical, gamma, is obtained from a spherical triangle with side 10 degrees, angle 180-theta, side 6 degrees, by the cos relation for example. But since you disagree with the crossed 45 degree wedges giving 60 degrees, I may still be on the wrong track. Greetings, Hans.