# Thread: Folding paper

1. ## Folding paper

An equilateral triangle AB'C' is folded along the height AM as shown:

After folding the angle between the planes ABM and ACM is 90deg.

Could anyone help me identify the "angle between the planes ABC and BMC", an illustration would help too.

Thanks.

2. Originally Posted by Geometor
An equilateral triangle AB'C' is folded along the height AM as shown:

After folding the angle between the planes ABM and ACM is 90deg.

Could anyone help me identify the "angle between the planes ABC and BMC", an illustration would help too.

Thanks.
I've attached a perspective sketch of the folded triangle.

1. You already know:
$|\overline{BM}|= |\overline{CM}|$
$\angle(BMC) = 90^\circ$

2. Draw the line $\overline{BC}$. It is the hypotenuse of the isosceles right triangle BMC.
3. Construct the midpoint of BC = M'

4. The length of $\overline{MM'}$ ist calculated by

$|\overline{MM'}| = \frac12 \sqrt{2} \cdot |\overline{BM}|$

5. Since $|\overline{AB'}| = 2 \cdot |\overline{B'M}|$ the line segment $\overline{AM}$ has the length

$|\overline{AM}| = \sqrt{3} \cdot |\overline{BM}|$

6. The angle $\alpha$ is the angle between the plane BMC and the plane ABC. It can be calculated by

$\tan(\alpha) = \frac{|\overline{AM}|}{|\overline{MM'}|} = \frac{\sqrt{3} \cdot |\overline{BM}|}{\frac12 \sqrt{2} \cdot |\overline{BM}|} = \sqrt{6}$

And therefore the angle $\alpha \approx 67.79^\circ$

3. I see thank you.
Thanks for psoting a solution however I just wanted to know which angle to work with so that would have saved you some effort