Find the Angle

Let ABC be an isosceles triangle such that A = B, point D is on AC, and point E is on AB.
Given that
1) Angle CAB = 20°
2) Angle CBD = 60°
3) Angle BCE = 50°
find Angle BDE.

Originally posted by ThudanBlunder
Let ABC be an isosceles triangle such that A = B, point D is on AC, and point E is on AB.
Given that
1) Angle CAB = 20°
2) Angle CBD = 60°
3) Angle BCE = 50°
find Angle BDE.

Originally posted by ThudanBlunder
Let ABC be an isosceles triangle such that A = B, point D is on AC, and point E is on AB.
Given that
1) Angle CAB = 20°
2) Angle CBD = 60°
3) Angle BCE = 50°
find Angle BDE.

Originally posted by SPMars
When you say A = B you mean they're the vertices which have equal angle?

Originally posted by ThudanBlunder
Oops, sorry about that.
I meant AB = AC.
😳

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.

Originally posted by ThudanBlunder
Let ABC be an isosceles triangle such that A = B, point D is on AC, and point E is on AB.
Given that
1) Angle CAB = 20°
2) Angle CBD = 60°
3) Angle BCE = 50°
find Angle BDE.

I 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)

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.

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.

Originally posted by ThudanBlunder
None of those answers is correct.

If 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!

I'm guessing 70 degrees.

It's definitely less than 60 degrees, and I would guess 20 degrees. I'm surprised how difficult this problem is proving to be!

Its 30 degs