PA is perpendicular to the plane ABCD, AB=2, PC makes an angle of 45 with the plane ABCD, and EF is the midpoint of PA and PB respectively. Find the cotangent value of the angle formed by straight lines DE and AF on different planes.

For example, this problem, the angle between two seemingly non-intersecting straight lines, can be solved by parallel lines: make a little G on the extension line of AB so that AG=EF= 1, then GE is parallel to the angle between AF, AE and DE ∠GED. AE is the projection of DE on the plane ABP, then there is the square root of COS∠GED=COS∠AED*COS∠AEG= root number 3/3* root number 2/2 = 6/6.

Note: The formula COS∠GED=COS∠AED*COS∠AEG has a miraculous effect in solving the dihedral angle problem.

First: make noodles.

AB is perpendicular to BCD and BD is perpendicular to CD. If AB=BC=2BD, find the sine of dihedral angle B-AC-D. ..

Let d be a plane DEF perpendicular to the straight line AC, intersecting AC at E and BC at F.

AC is perpendicular to the plane CEF, so DF is perpendicular to AC and AB is perpendicular to the plane BDC, so DF is perpendicular to AB.

So DF is perpendicular to plane ABC, so DF is perpendicular to BC. Then △BDC exists: BD*CD=DF*BC.

Let BD= 1, then AB=BC=2. So CD= root number 3. So DF= root number 3/2.

Similarly, it can be inferred from △ADC: root number 30/4.

Then: the sine value of dihedral angle B-AC-D =sin angle DEF=DF/DE= root number 10/5.

Note: As a plane key perpendicular to dihedral angle.

Third: Make a projection.

Still the second question, let's solve it in another way.

More than d is DE vertical BC. BC vertical, AB vertical. So the vertical plane ABC.

So the projection of △ADC on ABC plane is △AEC.

Use the formula: COSD-AC-E = S △ AEC/S △ ADC = AD * DC/(AB * CE).

Only the line length is needed to get the value of cosD-AC-E, and then convert it into sine value.

Note: cosD-AC-E=S△AEC/S△ADC is the key, and this method is most suitable for multiple-choice questions and fill-in-the-blank questions. But the disadvantage is that it may not work.

Fourth: coordinates.

That is, vector, vector method is also very fast. Filling in the blanks is also very effective.