The bottom side length of the regular triangular prism ABC A1B1C1 is a, and the side edge length is

As shown in the figure, the spatial Cartesian coordinate system is established with point A as the coordinate origin O, the straight line formed by AB as the OY axis, the straight line where AA1 is located as the oz axis, and the straight line perpendicular to the plane ABB1A1 as the OX axis

a(0,0,0), b(0,a,0), a1(0,0,2a), c1(-32a,a2,2a) are known

The coordinate system is as above, take the midpoint m of a1b1, so there is m(0,a2,2a), with am, mc1 has mc1 = (-32a,0,0), and ab = (0,a,0), aa1 = (0,0,2a), by mc1 6 1 ab =0, mc1 6 1 aa1 =0, therefore, the angle formed by mc1 surface abb1a1, ac1 and am is the angle formed by ac1 and side abb1a1

ac1→= (-32a，a2，2a)， am→= (0，a2，2a)， ac1→�6�1 am→= 0+a24+2a2=94a2， |ac1→|= 3a24+a24+2a2= 3a， |am→|= a24+2a2=32a, cos ac1 ,am = 94a23a 6 132a=32, therefore, the angle between ac1 and am, i.e., the angle between ac1 and the side abb1a1 is 30°

The coordinate system was established with B1A1 as the Y axis, B1A1 as the O point, OC1 as the X axis, Ba Midpoint as O1, and Oo1 as the Z axis.

1) the coordinates of a are 2a, 2a), the coordinates of b are (0, -1 2a, 2a), the coordinates of a1 are 2a, 0), and the coordinates of b1 are (0, -1 2a, 0);

2) Because it is a regular triangular prism, the face C1B1A1 is perpendicular to the face ABB1A1 and connects AO, so OC1 is perpendicular to AO and connects AC1, and the triangle AOC1 is a right triangle. Because A1B1C1 is a regular triangle, OC1 is the root number five of the half, because Aa1 is 2A, A1C1 is A, so AC1 is the root number five, so the angle C1AA1 is 30 degrees, so the angle between AC1 and the side ABB1A1 is 30 degrees.

Pass C1 to make the perpendicular line of the plane ABA1B1.

The vertical foot is the midpoint on A1B1 and is denoted as D

The perpendicular line of ab1 is passed through d, and the perpendicular foot is denoted as f

It can be seen that ab1 is perpendicular to the plane c1df

So the dihedral angle Zen shirt is equal to the angle C1FD

Let's do the calculations (it should be simple by now).

c1d=√3a/2

db1=a/2

df=db1*aa1/ab1=√3a/6

c1f=√30a/6

Cosine value = df c1f = 10 He Zhen cavity 10

C takes A as the origin, and establishes a spatial Cartesian coordinate system with AB, the perpendicular line of AB at the bottom and AA1 as the x-axis, Y-axis and Z-axis, with A as one unit, A(0,0,0),B(1,0,0),C(1 2, 3 2,0),A1(0,0,2),B1(1,0, 2),C1(1 2, 3 2,2),Vector AC1=(1 2, 3 2, 2),Plane ABB1A1 in the XOZ plane,Normal vector n=(0,1,0),Vector AC1·n= 3 2，n|=1,ac1|= (1 4+3 4+2)= 3, let the angle between n and ac1 be 1, and the angle between ac1 and the plane abb1a1 is , 1= 2, cos 1=ac1·n (|ac1|*|n|)=（√3/2）/（√3）=1/2，cosα=sinα1=γ（1-1/4）=√3/2，α=30°

The angle between AC1 and side AB1 is 30°

Why do we have to use vector solutions?

Take the midpoint m of a1b1, a1b1c1 is positive, c1m a1b1, plane abb1a1 plane a1b1c1, c1m plane abb1a1, c1am is the angle between ac1 and plane abb11 c1m = 3 2, ac1 = 3, am = (aa1 2 + a1m 2) = 3 2, cos

Connect BC1 and AC1

According to the properties of the regular triangular prism, it is easy to prove that BC1=AC1=3 and AB=1

According to the cosine theorem.

bc1)²+ac1)²-ab)²=2×bc1×ac1×cos β3+3-1=2×3×cos β

arcos(5/6)

m is the midpoint of AC.

Take the AE midpoint P and connect MP when M is the AC midpoint, MP is parallel and equal to 1 2ec, and because FB is parallel and equal to 1 2EC, FB is parallel and equal to MP, so MPFB is a parallelogram, so BM is parallel to FP, that is, BM is parallel to the plane AEF

Take the midpoint D of a1b1 and connect C1d and AD

The side edges are perpendicular to the bottom surface.

AA1 planar A1B1C1

aa1⊥c1d

The bottom is an equilateral triangle.

c1d⊥a1b1

C1d planar abb1a1

Then dac1 is the angle between ac1 and abb1a1 on the side, ac1 = (1 +(2) ) ) = 3

c1d=√3/2

sin∠dac1=c1d/ac1=1/2

dac1=π/6

The angle between AC1 and side ABB1A1 is 6

As C1D1 is perpendicular to A1B1 to D1, so C1D1 is perpendicular to the surface AbB1A1 to connect Ad1, and the angle C1Ad1 is what is sought.

In the triangle C1Ad1, the angle C1D1A = 90 degrees, AC1 = root number 3, and AD1 = 3 2

cosc1ad1=ad1 ac1=root3 2, so the angle c1ad1=30 degrees.

If C1D is perpendicular to A1B1 in D, then we can find AC1 = root number 3, C1D = (root number 3) 2, and the angle ADC1 is 90 degrees, then the angle C1ad is equal to 30 degrees, that is, the angle between AC1 and the side ABB1A1 is 30 degrees.