# Find the Angle

ThudanBlunder
Posers and Puzzles 23 Aug '06 02:29
1. 23 Aug '06 02:292 edits
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.
2. SiteNine
Zinc Saucea
23 Aug '06 17:32
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.
umm... no
3. 23 Aug '06 17:54
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.
When you say A = B you mean they're the vertices which have equal angle?
4. 23 Aug '06 18:51
Originally posted by SPMars
When you say A = B you mean they're the vertices which have equal angle?
Oops, sorry about that.
I meant AB = AC.
ðŸ˜³
5. 23 Aug '06 23:082 edits
Originally posted by ThudanBlunder
Oops, sorry about that.
I meant AB = AC.
ðŸ˜³
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 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.
6. XanthosNZ
Cancerous Bus Crash
23 Aug '06 23:33
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.
Ah Geometrical Bashing, how I love thee.
7. 23 Aug '06 23:34
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've found it. It's between B and E, somewhere next to D.
8. Agerg
The 'edit'or
24 Aug '06 00:231 edit
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)
9. XanthosNZ
Cancerous Bus Crash
24 Aug '06 02:281 edit
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.
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.
10. 24 Aug '06 13:42
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.
None of those answers is correct.
11. XanthosNZ
Cancerous Bus Crash
24 Aug '06 14:05
Originally posted by ThudanBlunder
None of those answers is correct.
Yeah I figured that was the case.
12. Agerg
The 'edit'or
24 Aug '06 14:3612 edits
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!
13. 24 Aug '06 21:38
I'm guessing 70 degrees.
14. 25 Aug '06 13:06
It's definitely less than 60 degrees, and I would guess 20 degrees. I'm surprised how difficult this problem is proving to be!
15. Agerg
The 'edit'or
25 Aug '06 13:44
Its 30 degs