(1) Find the angle formed by the straight lines PD and AB on different planes (2) prove PA perpendicular to CD (3) find the size of dihedral angle p-ab-d In the pyramid p-abcd, the side PDC is an equilateral triangle with side length of 2 and perpendicular to the bottom ABCD, the bottom ABCD is a diamond with area of 2.3, the angle ADC is an acute angle of the diamond, and M is the midpoint of Pb

(1) Find the angle formed by the straight lines PD and AB on different planes (2) prove PA perpendicular to CD (3) find the size of dihedral angle p-ab-d In the pyramid p-abcd, the side PDC is an equilateral triangle with side length of 2 and perpendicular to the bottom ABCD, the bottom ABCD is a diamond with area of 2.3, the angle ADC is an acute angle of the diamond, and M is the midpoint of Pb

1.60 degree AB translation to CD
I tend to use Pythagorean theorem, maybe it's troublesome
Pan CD to ab
Take the midpoint of DC as e, connect PE, PE is perpendicular to the surface ABCD, PE = root 3
The angle ADC is 60 degrees (which should have been calculated?) with AE, AE = root 3
So PA = root 6
The calculation of Pb is the same, also using PE as the vertical line, but the midway calculation of be needs a cosine theorem, which is a little troublesome
Pythagorean inverse theorem
3. It has been proved that PA is perpendicular to CD and a, PE is perpendicular to plane ABCD (PA = radical 6, PE = radical 3, calculated)
So the dihedral angle is arcsin PE / PA = arcsin (radical 2) / 2
So the size is 45 degrees

After the square ABCD is folded into a straight dihedral angle along the diagonal AC, the angle between AB and CD is the same______ ．

As shown in the figure below, take the midpoint of AC, BD and BC as e, F and G, connect BD, EF, eg, De, be and FG, then FG ‖ CD, eg ‖ AB, so ∠ FGE is the angle (or its complementary angle) formed by the out of plane straight line AB and CD, if the side length of the square is 2 units, then FG = 1, eg = 1, and because the dihedral angle d-ac-b is a straight dihedral angle, de ⊥ AC, then ∠ DEB = 90 °, then in RT △ DEB, de = be = 2, DB = 2, easy to get EF = 1, so ∠ FGE = π 3, so the answer is: π 3

Take the diagonal BD of the square ABCD as the edge, fold it into a straight dihedral angle, connect AC, and calculate the tangent value of the dihedral angle a-cd-b

Let a square AC and BD intersect at point o
After folding, point O is the projection of a on the bottom
Make the vertical line of CD through point O intersect at point E
Angle AEO is dihedral angle a-cd-b
Let the side length be 2 ^ 0.5 * a
AO=AC/2=a
OE=2^0.5*a/2
tanAEO=AO/OE=2^0.5
Tangent of dihedral angle a-cd-b
2^0.5

If the square Aode is folded into a straight dihedral angle along the diagonal ad, then the angle formed by the out of plane lines AO and De is ()
A. 30°B. 45°C. 60°D. 90°

Place the straight dihedral angle in the cube, as shown in the figure. ∵ de ∥ B1C, the angle formed by the out of plane lines AO and De is ∠ Acb1. In the equilateral triangle Acb1, ∠ Acb1 = 60. The angle formed by out of plane lines AO and De is 60 °. So select C

As shown in the figure, the side length of square ABCD is 4, take BC as the diameter of the circle, cross point a as the tangent of the circle, intersect DC at e, and the tangent point is F. (1) find the area of △ ADE; & nbsp; (2) find the length of BF

(1) Let EF = EC = x, then de = dc-ec = 4-x, AE = AF + EF = 4 + X. in RT △ ade, by using Pythagorean theorem, we get AE2 = ad2 + de2, that is, (4 + x) 2 = 42 + (4-x) 2, and the solution is: X = 1, de = 4-1 = 3, then s △ ade = 12ad · de = 6; (2) connect OA, of, ⊥ ob ⊥ AB, of ⊥ AF, and ob = of, In RT △ ABO, according to Pythagorean theorem, we can get: OA = AB2 + ob2 = 25, ∵ s △ ABO = 12ab, OB = 12oa, BM = 4 × 225 = 455, then BF = 2bm = 855

As shown in the figure, the side length of square ABCD is 4, take BC as the diameter of the circle, cross point a as the tangent of the circle, intersect DC at e, and the tangent point is F. (1) find the area of △ ADE; & nbsp; (2) find the length of BF

(1) Let EF = EC = x, then de = dc-ec = 4-x, AE = AF + EF = 4 + X. in RT △ ade, we use Pythagorean theorem to get AE2 = ad2 + de2, that is, (4 + x) 2 = 42 + (4-x) 2, the solution is: x = 1, de =

As shown in the figure, the side length of square ABCD is 4, take BC as the diameter of the circle, cross point a as the tangent of the circle, intersect DC at e, and the tangent point is F. (1) find the area of △ ADE; & nbsp; (2) find the length of BF

(1) Let EF = EC = x, then de = dc-ec = 4-x, AE = AF + EF = 4 + X. in RT △ ade, we use Pythagorean theorem to get AE2 = ad2 + de2, that is, (4 + x) 2 = 42 + (4-x) 2, the solution is: x = 1, de =

It is known that AB is the diameter of circle O, ad is perpendicular to CD, AC bisector angle DAB, and point C is on circle O. (1) prove that the straight line CD is the tangent of circle o

Connect OC, OC = OA, OCA = OAC
AC bisector DAB, OAC = DAC
OCA = DAC
Ad parallel OC
Ad vertical CD, OC vertical CD
CD is perpendicular to the radius, CD is tangent

AB is the diameter of the circle O, C is the point on the circle O, AC bisects ∠ DAB, ad ⊥ CD in D, proving that CD is the tangent of the circle o

1) Connect OC, because OA = OC, so the angle OAC = angle OCA, and AC bisects the angle DAB, so the angle DAC = angle cab, so the angle DAC = angle OCA, so DA / / OC, and ad vertical CD, so the OC vertical CD, that is, the straight line CD is the tangent of circle O. (method: connect radius to prove vertical)

If BD ⊥ AE, ab = 4, BC = 2, ad = 3, then de = 0___ ；CE= ___ ．

First of all, from the secant theorem, it is not difficult to know ab · AC = ad · AE, so AE = 8, de = 5, and BD ⊥ AE, so be is the diameter, so ∠ C = 90 ° from the Pythagorean theorem, CE2 = ae2-ac2 = 28, so CE = 27