As shown in the figure, △ ABC and △ alblc1 are equilateral triangles, and the midpoint of BC and b1c1 is d. verification: Aa1 ⊥ CC1

As shown in the figure, △ ABC and △ alblc1 are equilateral triangles, and the midpoint of BC and b1c1 is d. verification: Aa1 ⊥ CC1

It is proved that: connecting ad, extending Aa1, crossing DC to o, crossing C1C to e, ∵ - ada1 = 90 ° - a1dc = ∠ cdc1, ADDC = da1dc1 = 3, ∵ aa1d ∽ cc1d, ∵ a1ad = ∠ c1cd, and ∵ - AOD = ∠ Coe, ∵ ADO = ∠ CEO = 90 °, namely Aa1 ⊥ CC1

It is known that the length of the side edge of the regular triangular prism abc-a1b1c1 is equal to the length of the side edge of the bottom, then the sine of the angle between Ab1 and acc1a1 is equal to ()
A. 64B. 104C. 22D. 32

Take the midpoint D1 of a1c1, connect b1d1 and AD1. In the regular triangular prism abc-a1b1c1, b1d1 ⊥ face acc1a1, then ∠ b1ad1 is the angle formed by Ab1 and side acc1a1, ∵ the side edge length of regular triangular prism abc-a1b1c1 is equal to the side length of bottom surface, ∵ sin ∵ b1ad1 = 322 = 64, so select a

(1 / 2) in the straight triangular prism abc-a1b1c1, AC = 5, BC = 4, ab = 5, Aa1 = 4, point D is the midpoint of AB, (1) prove: AC vertical BC: (...)
(1 / 2) in the straight triangular prism abc-a1b1c1, AC = 5, BC = 4, ab = 5, Aa1 = 4, point D is the midpoint of AB, (1) verification: AC vertical BC: ()

1) ∵ the length of the bottom three sides AC = 3, ab = 5, BC = 4,
∴AC⊥BC
AC ⊥ CC1 in straight triangular prism abc-a1b1c1,
And BC ∩ CC1 = C
BC ∩ CC1 belongs to plane bcc1b1
Plane bcc1b1
BC1 belongs to plane bcc1b1
∴AC⊥BC1；

It is proved that SA is perpendicular to BC in s-abc

Take the midpoint D of BC, connect AD and ad, △ SBC and △ ABC are isosceles, SD ⊥ BC, ad ⊥ BC, ad ∩ SD = D, ∩ BC ⊥ plane sad, SA ∈ plane sad, ∩ SA ⊥ BC