Urgent!! In ABC, let the edges of A B C be a b c, and the vectors m cosA, sinA

m = (cosa, sina), m² = cos²a + sin²a = 1

n = (√2-sina, cosa), n² = (√2-sina)² cos²a = 3 - 2 √2 sina

mn = cosa(√2-sina)+sinacosa = √2 cosa

m+n| = √(m²+n²+2mn) = √(1+3- 2√2 sina + 2√2 cosa) = √(4- 2√2 sina + 2√2 cosa) = 2

cosa = sina

a = 45°

a sina = c sinc sinusoidal theorem.

sinc = c sina / a = √2 a sin45° /a = 1

c = 90°

b = 180°-a-c = 180°-45°-90° = 45°

a = b = 4√2

ABC is an isosceles right-angled triangle with right-angled sides being A and B

s = ½ ab = ½ x 4√2 x 4√2 = 16

Because: sina*cosa+2=17 8

Soyama repents: sina*cosa=1 8

sina + cosa) 2 = 1 + 2 sina * cosa = 5 4 sina - cosa) 2 = 1-2 sina*cosa = 3 4a belongs to the deficit (0, 4), so sina

（a-c)*(b-c)=ab-ac-bc+cc=-c(a+b-c)

The minimum value is required, that is, a+b-c is the smallest, and a is perpendicular to b, so a+b can obtain: root number 2, and the previous one should make a+b-c the smallest, then (a+b) is in the same direction as c, and finally calculated: its minimum value is 1-root number 2

(a-c)·(b-c)=a·b-c·(a+b)+c^2=-c·(a+b)+c^2

Because a·b=0

A is perpendicular to b, a+b = root number 2 and 45 degrees to a or b are unit vectors, so c 2 = 1

Converting the known minimum value into finding c· (a+b) maximum if and only if c is in the same direction as (a+b).

C· (a+b) = root number 2

In summary:(a-c)· (b-c) = 1 - root number 2

Happy learning

Vector p vector q, 1-sina)*2sina-12 7*cos2a=0.

2sina-2sin^2a-12/7(1-2sin^2a)=0.

5sina-3)(sina+2)=0.

5sina-3=0.

sina=3/5；sina+2=0, sina=-2, rounded.

sina=3/5.--A1

2）s△abc=(1/2)bcsina=(1/2)*2*c*(3/5)=3.

c=5.Apply the cosine theorem: a 2 = b 2 + c 2-2 bccosa=2 2+5 2-2*2*5*(4 5) [cosa= (1-sin 2a)=4 5].

a^2=13.

a= 13 --A2

a+b=(cosa-1 2,sina+ 3 2)a-b=(cosa+1 2,sina- 3 2)then (a+b)· (a-b)

cosa-1/2)(cosa+1/2)+(sina+√3\2)(sina-√3\2)

cosa) 2-1 4+(sina) 2-3 4 again. (cosa)^2

sina)^2

So. i.e. (a+b)· (a-b)=0

So: vector a + vector b is perpendicular to vector a - vector b.

3a+b=(3cosa-1 2, 3sina+ 3 2)a- 3b=(cosa+ 3 2,sina-3 2) 3a+b with.

The modulo equality of a-3b is:

3cosa-1/2)^2+(√3sina+√3/2)^2=(cosa+√3/2)^2+(sina-3/2)^2

Knowing that a is an acute angle, then: sina>0, cosa>0(1) If the vector a·b=13 6, then:

2*1+sina*cosa=13/6

That is: sina*cosa=1 6

Then: (sina + cosa) = sin a + 2 sina * cosa + cos a = 1 + 2 sina * cosa = 1+

Solution: sina + cosa = 2 (root number 3) 3 (2) If the vector a b, then:

2*cosa-sina*1=0

That is: sina = 2cosa

Because sin a+cos a=1, so: 4cos a+cos a=1 i.e.: cos a=1 5

Solution: cosa = (root number 5) 5, sina = 2 cosa = 2 (root number 5) 5

Then: sin2a=2sina*cosa=4 5,cos2a=2cos a-1=2 5

So: sin(2a+3).

sin(2a)cos(3)+cos(2a)sin(3)(-3 5)*(root number 3) 2

4-3 root number 3) 10