The bottom surface of triangular prism abc-a1b1c1 is an equilateral triangle with side length of 2. The side edge Aa1 is vertical to the bottom surface ABC. Points E and F are the points on edges CC1 and BB1 respectively, and EC = 2fb = 2. Point m is the midpoint of line AC. the cosine value of the angle between BM and EF is calculated

The bottom surface of triangular prism abc-a1b1c1 is an equilateral triangle with side length of 2. The side edge Aa1 is vertical to the bottom surface ABC. Points E and F are the points on edges CC1 and BB1 respectively, and EC = 2fb = 2. Point m is the midpoint of line AC. the cosine value of the angle between BM and EF is calculated

ABC is an equilateral triangle, and the edge Aa1 is perpendicular to the bottom. ABC shows that it is an equilateral triangular prism. M is the midpoint of AC, so BM is perpendicular to AC (the height and center line of equilateral triangle). The side of the equilateral triangular prism is perpendicular to the bottom, so BM is perpendicular to any straight line in aca1c1

In the oblique triangular abc-a ′ B ′ C ', the bottom is an equilateral triangle with a side length of a, and the side edge length is B. the side edge AA ′ is adjacent to the bottom, and both sides AB and AC form 45 ° angles. The side area and volume of the triangular prism are calculated

(1) In the three sides of the triangular prism, the quadrilateral Abba and ACCA have an angle of 45 ° and the two sides are parallelograms with lengths of a and B respectively. The third side is a rectangle with lengths of a and B respectively. S side = 2absib 45 ° and ab = (2 + 1) AB (2) through A1, make a1o perpendicular to the bottom surface ABC, and make a1d ⊥ AB through o Then ad = 2B2, a1d = 2B2, Ao = 6b3, a1o = 3B3, v = 12 × 3a2a3b3 = 14a2b

In the oblique triangular abc-a1b1c1, the bottom is an equilateral triangle with a side length of a, and the side length is B. Aa1 and AB, AC all form a θ angle

Take the midpoint D of BC, then BC ⊥ ad, ad is the projection of Aa1 on the surface ABC, then BC ⊥ Aa1, BC ⊥ BB1,
The side product of triangular prism s = area of a1abb1 S1 + area of a1acc1 S2 + area of b1bcc1 S3 = 2absin θ + ab

Point P is the point on the side edge BB1 of the oblique triangular prism abc-a1b1c1, PM is perpendicular to BB1, Aa1 is perpendicular to point m, PN is perpendicular to BB1, and CC1 is perpendicular to point n
1 verification: CC1 ⊥ Mn
There is cosine theorem in any △ def: de ^ 2 = DF ^ 2 + EF ^ 2-2df * efcos ∠ DFE
The relation between the area of three sides of an oblique triangular prism and the dihedral angle formed by two sides is given and proved by analogy with the cosine theorem of a triangle

1 ,BB1⊥ PM.BB1 ⊥ PN, ⊥ BB1 ⊥ plane PMN. ∵ CC1 ∥ BB1 ⊥ plane PMN.CC1 ⊥MN.
2 NM²＝PN²+PM²-2PN×PM×cos∠MPN①
Note that ∠ MPN is the plane angle of dihedral angle a-bb1-c
[S（ACC1A1）]²＝[S（BCC1B1）]²+[S（ABB1A1]²
-2 [S (bcc1b1)] × [S (abb1a1] cos [dihedral angle a-bb1-c]