A C 1D⊥DC

C 1D⊥BD is also famous.

And BD spans CD to D.

∴C 1D⊥ BCD

∵BC belongs to surface BCD.

∴C 1D⊥BC

Certificate of completion

(2)∵ is a straight triangular prism

A CC 1⊥ ABC

∵BC belongs to surface ABC.

∴CC 1⊥BC

And BC⊥C 1D

C 1D crosses CC 1 at c1.

So BC⊥ surface CC 1D

That is, BC⊥ aircraft ACC 1A 1.

∴BC⊥AC

Therefore, a rectangular coordinate system with CA as the X axis, CB as the Y axis and CC 1 as the Z axis can be established.

C(0，0，0)

Let AC length be a.

Then the vector BD=(a, -a, a)

Vector da1= (0,0, a)

Vector DC 1=(-a, 0, a)

The normal vector of plane A 1DB is calculated as (1, 1, 0).

A normal vector of BDC 1 is (1, 2, 1).

Let two normal vectors contain an angle θ.

cosθ=( 1* 1+ 1*2+0* 1)/( 1^2+ 1^2)^0.5*( 1^2+2^2+ 1^2)^0.5

= three-thirds of the root sign.

Because dihedral angle A 1-BD-C 1 is acute.

So the dihedral angle is 30.

Is this the topic of the college entrance examination? ! ? ! I vomited blood in the college entrance examination last year! !