Three straight lines PA, Pb, PC, ∠ APC = ∠ APB = 60 ° and ∠ BPC = 90 ° in space are used to calculate the dihedral angle b-pa-c

Three straight lines PA, Pb, PC, ∠ APC = ∠ APB = 60 ° and ∠ BPC = 90 ° in space are used to calculate the dihedral angle b-pa-c

For convenience, let Pb = PC = 2. OA = 1. Then BC = 2 √ 2. (BPC = 90 °) from the cosine theorem, ab = AC = √ 3. Thus, cap = BAP = 90 °. BAC is the plane angle of dihedral angle b-pa & nbsp; - & nbsp; C. cos ∠ BAC = (3 + 3-8) / (2 × 3) = - 1 / 3

As shown in the figure, in the spacial quadrilateral PABC, ∠ APC = 90 °, ∠ APB = 60 °, Pb = BC = 4, PC = 3, find the value of dihedral angle b-pa-c
Is to find the size of dihedral angle is a very simple graph, a triangular pyramid vertex is p, it seems that the rectangular coordinate system can not be established?

Make a parallel line of pa through B, and make PD perpendicular to this parallel line and intersect at point D. make be perpendicular to PC and intersect at point E. connect de. easy to know: PA ⊥ PC (∠ APC = 90 °) - > BD ⊥ PC (BD ∥ PA) be ⊥ PC (BP = BC) - > plane BDE ⊥ straight line PC - > PC ⊥ de - > ped = 90 ° easy to know PD ⊥ PA (PD ⊥ BD, BD ∥ PA) -

In △ ABC, G is the center of gravity, D and E are on sides AB and AC respectively, and D, G and E are collinear. The area of △ ade is S1, and the area of quadrilateral bced is S2
Then S1: S2 =. There is no graph

By adding the condition de / / BC, it can be solved as follows:
Set AF as the center line
DE//BC
Triangle ade is similar to triangle ABC
S1:(S1+S2)=(AG/AF)^2=4:9
S1:S2=4:5

As shown in the figure, in the acute angle △ ABC, be ⊥ AC, ∠ ade = ∠ C, the area of △ ade is S1
As shown in the figure, in the acute angle △ ABC, be ⊥ AC, ∠ ade = ∠ C, note that the area of △ ade is S1, and the area of △ ABC is S2, then S1 / S2 = --- a sin2a B cos2a C tan2a D (Tana) negative quadratic

It is easy to get that △ ade is similar to △ ABC (the three inner angles are equal)
The area ratio is equal to the square of the similarity ratio
The similarity ratios were AD / AC, AE / AB, be / BC
In Abe, AE / AB = cosa
∴s1/s2=(AE/AB)²=cos²A

In the arithmetic sequence {an}, ① known A2 + A5 + A12 + A15 = 36, find S16, ② known A6 = 20, find SN

1. ∵ arithmetic sequence {an} is arithmetic sequence ∵ A2 + A15 = A5 + A12 = a1 + a16 ∵ A2 + A5 + A12 + A15 = 2 (a1 + a16) = 36 ∵ a1 + a16 = 18 ∵ S16 = 16 × (a1 + a16) / 2 = 16 × 18 / 2 = 1442. ∵ A6 = 20 ∵ a1 + 5D = 20 ∵ a1 + a16 = 18 ∵ 2A1 + 15d = 18 ∵ A1 = 20.2a1 + 15d = 18 ∵ A1 = 42, d = - 22 / 5 ∵

In the arithmetic sequence {an}, given A2 + A5 + A12 + A15 = 36, find S16

a2 + a5 + a12 + a15 = 36==> (a2 + a15) + (a5 + a12) = 36==> (a2 + a15) + (a2 +a15) = 36==> 2(a2 + a15) = 36==> (a2 +a15) = 18 S16 = a1 +a2 + ... + a15 + a16 = (a1 +a16) + (a2 ...

Three different numbers are randomly selected from the set {1,2,3,..., 20}, so that the three numbers form an arithmetic sequence. How many such arithmetic sequences are there?

There are 18 sequences with difference of 1: (1,2,3) (2,3,4)... (18,19,20) totally 18 sequences with difference of 2: (1,3,5) (2,4,6)... (16,18,20) totally 16 sequences with difference of 3: (1,4,7) (2,5,8)... (14,17,20) totally 14 sequences. It is concluded that the number of sequences with difference of n is 20-2xn, so that 20-2xn > 0 = > n

From the set {1,2,3 , 20} to make these three numbers into an arithmetic sequence. How many such arithmetic sequences are there

A total of 90 ~ are calculated by computer programming. All the examples are as follows: 1 2313 51 4 71 5 91 6 111 7 131 8 151 9 171 10 192 3 42 4 62 5 82 6 102 7 122 8 142 9 162 10 182 11 203 4 53 5 73 6 93 7 113 8 133 9 153 10 173 11 194 5

If three different numbers are selected from the set {1,2,3,4,5,6}, and the three numbers form an arithmetic sequence, then such an arithmetic sequence
How many at most

A total of 12, 1232343455613524632143254531642

Choose three numbers from the set {1,2,3,4,5} to form the probability of arithmetic sequence
Is the answer two fifths or four fifths?

Arithmetic sequence 123, 234, 345, 135, four possibilities
All possible C35 = 10 species
The probability is 4 / 10 = 2 / 5