- 23 Aug '06 23:08 / 2 edits

This was more painful than I thought it was going to be.*Originally posted by ThudanBlunder***Oops, sorry about that.**

I meant AB = AC.

I have the angle as the inverse tangent of

( sin(20) * sin(30) * sin(60) ) / ( sin(20) * sin(80) - cos(20) * sin(30) * sin(60) ).

Here angles are in degrees.

I'm too tired at the moment to give a decimal approximation for this number. There's also a high chance of working error...basically I got all the angles I could from looking at triangles...but that still leaves 1 angle free. Then I used the sine rule lots to get that

sin(a+20) / sin(a) = ( sin(20) * sin(80) ) / ( sin(30) * sin(60) )

where a is the angle you're after. - 23 Aug '06 23:33

Ah Geometrical Bashing, how I love thee.*Originally posted by SPMars***This was more painful than I thought it was going to be.**

I have the angle as the inverse tangent of

( sin(20) * sin(30) * sin(60) ) / ( sin(20) * sin(80) - cos(20) * sin(30) * sin(60) ).

Here angles are in degrees.

I'm too tired at the moment to give a decimal approximation for this number. There's also a high chance of working error...basically ...[text shortened]... sin(a) = ( sin(20) * sin(80) ) / ( sin(30) * sin(60) )

where a is the angle you're after. - 24 Aug '06 00:23 / 1 editI spent a while on this just searching for different angles but so far haven't found the ones I want...gonna chew on this a bit more as it would be nice to discover the solution can be found by inspection as opposed to numerical computation. (Though there is nothing wrong with that method)
- 24 Aug '06 02:28 / 1 edit

Something funky is going on here.*Originally posted by SPMars***This was more painful than I thought it was going to be.**

I have the angle as the inverse tangent of

( sin(20) * sin(30) * sin(60) ) / ( sin(20) * sin(80) - cos(20) * sin(30) * sin(60) ).

Here angles are in degrees.

I'm too tired at the moment to give a decimal approximation for this number. There's also a high chance of working error...basically ...[text shortened]... sin(a) = ( sin(20) * sin(80) ) / ( sin(30) * sin(60) )

where a is the angle you're after.

Evaluating your first expression using a calculator I get ~ -64.67. Using Maple to find the solution to your second expression I get -64.67726606.

And taking your second expression and solving for a by hand I get 58.747. - 24 Aug '06 13:42

None of those answers is correct.*Originally posted by XanthosNZ***Something funky is going on here.**

Evaluating your first expression using a calculator I get ~ -64.67. Using Maple to find the solution to your second expression I get -64.67726606.

And taking your second expression and solving for a by hand I get 58.747. - 24 Aug '06 14:36 / 12 editsIf you draw the figure...draw an altidude perpendicular to the base of ABC and then connect D to E...then make new points F and G that are reflections about the altitude, and finally connect D and F (or D and G depending which is level with D), then F and G you should see something very curious...an equilateral triangle with a vertex (call this point H) on the altitude (and resting on the vertex of another equilateral triangle)...also the lines DE and FG seem to intersect at the centre of triangle DFH...thus the line DE should be at 30 degs to the horizontal...which is also the angle BDE....I'm now trying to prove why this is so!

*edit* wow 12 edits!