Answers to exercise 3.1 (3) of senior high school mathematics compulsory 4 of Jiangsu Education Press 1 to 10 questions, the best process

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Take △ ABC edges AB and AC as edges, make equilateral △ abd and △ ace outward, connect be and CD to point O, and verify that OA bisects ∠ doe

Because of equilateral △ abd, △ ace, angle DAB = angle CAE = 60 degrees, so angle DAB + angle BAC = angle CAE + angle BAC, so angle DAC = angle BAE, because equilateral △ abd, △ ace, so ad = AB, AC = AE, because angle DAC = angle BAE, so triangle DAC is equal to triangle BAE, so angle ADO = angle ABO, so B, O, a, D four points

Given the vector a = (2,4), B = (- 1,3), C = (k, 2), and a is perpendicular to (B + C) to find the value of K

Vector B + C = (k-1,5)
A perpendicular to (B + C)
SO 2 * (k-1) + 4 * 5 = 0
a=-9

If AB = 4 times radical 3, AC = 4 times radical 6, then the size of dihedral angle A-BC-D is?
I can't follow the number. Excuse me!
As soon as possible to detailed answers, thank you~~

In ⊥ ABC, it is easy to find that: BD = 4, CD = 8, ad = 4 ⊥ 2 after folding, make de ⊥ BC in BCD and connect AE with E. because ad ⊥ BD, ad ⊥ CD, ad ⊥ BCD, ad ⊥ BC and de ⊥ BC, BC ⊥ face, ADE, BC ⊥ AE, so ⊥ AED is the plane angle of dihedral angle a-bc-d. in ⊥ BCD, it is easy to get BC = 4 ⊥ 7cos from cosine theorem

Vector p = the modulus of vector a divided by a plus the modulus of vector b divided by B, where vector AB is a non-zero vector, then the value range of vector p

The problem should be to find the value range of the module of vector P
First of all, you need to understand the meaning of the module of vector a divided by a and the module of vector b divided by B
They are unit vectors in the corresponding direction
So the vector p is the sum of two unit vectors in any direction
In the same direction, the modulus of P has the maximum value of 2, and in the reverse direction, it has the minimum value of 0
So the modulus of P is in the range of [0,2]

Right triangle ABC middle angle c + 90 degree angle a = 15 degree AB = 12 find height CD length on AB side?

Sin15 ° = (√ 6 - √ 2) / 4 cos15 ° = (√ 6 + √ 2) / 4 calculate CB = 3 (√ 6 - √ 2) AC = 3 (√ 6 + √ 2) AB CD = CB AC CD = 3 with sine and cosine
Please accept

Given that a, B and C are positive real numbers and log9 (9a + b) = log3ab, then the value range of C which makes 4A + B ≥ C constant is______ ．

∵ a, B, C are all positive real numbers, and log9 (9a + b) = log3ab, ∵ log9 (9a + b) = log3ab = log9ab, ∵ 9A + B = AB, ∵ 9A + bab = 9b + 1A = 1, ∵ 4A + B = (4a + b) (9b + 1a) = 36ab + Ba + 13 ≥ 236 + 13 = 25, ∵ 4A + B ≥ C is constant, C is positive real number, ∵ 0 < C ≤ 25

The distance from the vertex a of the acute triangle ABC to the perpendicular h is equal to the radius of its circumcircle, and the angle of the angle BAC is calculated

60 degrees 00000

Finding CUA, CAD and cub from known complete set u = {12345678910} a = {123456} B = {78910} d = {123}

CuA=｛78910｝
CaD=｛456｝
CuB=｛123456｝

It is known that in △ ABC, ab = AC, ad ⊥ BC on D, P on BC, PE ⊥ BC intersects the extension of BA with E
Translate PE so that point P is on the extension line of BC, PE intersects the extension line of BA with E, and AC intersects the extension line with F. write and prove the relationship satisfied by ad, PE and PF

(1) When p is on BC, 2ad = PE + PF,
If AG is perpendicular to g, because the triangle is isosceles triangle, easy angle AEF = AFE = 90-angle B, then AF = AE, triangle AEF is also isosceles triangle, three lines in one can get eg = FG, and quadrangle ADPG is rectangle, ad = PG,
Yide 2ad = PE + PF
(2) When p is on the extension line of BC, 2ad = pf-pe,
Similarly, if ah is perpendicular to h, we can also prove that AF = AE, three lines in one, FH = eh, and the quadrilateral ADPH is a rectangle, ad = pH
Yide 2ad = pf-pe

Answers to exercise 3.1 (3) of senior high school mathematics compulsory 4 of Jiangsu Education Press 1 to 10 questions, the best process