(1) Regular triangular prism ABC-a 1 b1each side is equal in length, and it is set to1.
Then ab = AC = BC = bb1= cc1= b1c1=1,
D is the center of the side BCC 1B 1, that is, D is the intersection of diagonal BC 1 and CB 1.
Let AE⊥BC be e, then BE=CE= 1/2, AE=( 1/2)∠3, even DE.
∵ plane BCC 1B 1⊥ plane, ∴DE⊥BC plane, and DE=BC/2= 1/2.
∠ADE is the angle between AD and the plane BCC 1B 1,
∫tan∠ADE = AE/DE =[( 1/2)√3]/( 1/2)=√3,
∴∠ADE=60
The angle between AD and BCC 1B 1 plane is 60.
(2)∠DAE is the angle between AD and plane ABC,
∫tan∠DAE = DE/AE =( 1/2)/[( 1/2)√3]= 1/√3,
∴∠DAE=30,
The angle between AD and plane ABC is 30.