Current location - Training Enrollment Network - Mathematics courses - Jiangsu senior high school mathematics examination paper
Jiangsu senior high school mathematics examination paper
sine law

one

1.5:3

2. Root number 3/3

3.2 Root number 2

4. Root number 3/4

5 From the sine theorem, we can draw the following conclusions:

a=2RsinA,b=2RsinB

acosA = bcosB

∴sinAcosA=sinBcosB

A and b are complementary.

∴C=90

Triangle ABC is a right triangle.

6.∵a/sinA=c/sinc

∴2/ (root number 2/2) = the square root of 6 /sinC

∴sinC= root number 3/2

∴∠ C = 60 or 120

∴∠ b = 75 or 15.

two

1. (root number 3+ 1)/4

2.60 or 120

3.30 150

isosceles

5. 100 meter

6. I didn't do this problem, I guessed it temporarily.

7.∫a/sinA = b/sinB = c/sinC

∴tanA=tanB=TANC

∴A=B=C=60

This is an isosceles triangle.

8.∵a/sinA=b/sinB=c/sinC=2R

∴R= 1

∴a=2sinA,b=2sinB,c=2sinC

∴ Original formula = [2 (sina+sinb+sinc)]/(sina+sinb+sinc)

=2

three

1.60

2.-2 Root number 3

3 Sina = a/2r,sinb = b/2r,sinc = c/2r,cosc = (a? +b? -c? )/2ab

(A, B and C are the opposite sides of A, B and C respectively, and R is the radius of the circumscribed circle of the triangle)

A/2r=b/2r *(a? +b? -c? )/2ab - ①

(a/2r)? =(b/2r)? +(c/2r)? - ②

B = C; is obtained from ① simplification; Simplify from ② to get a? =b? +c?

So the triangle is an isosceles right triangle.

4. Solution: (i) SINB =12/13 from COSB =-5/13,

From cosC=4/5, SINC = 3/5.

So sinA=sin(B+C)=33/65.

(2) 1/2*AB*ACsinA=33/2 by S△ABC=33/2,

∴AB*AC=65

∫AC =(AB * sinB)/sinC = 20/ 13AB

∴AB= 13/2

∴BC= 1 1/2

cosine theorem

one

1. Root number 7

2. 120

3.60

4. 1: root number 3: 2

5. From the meaning of the question, B is the biggest.

∵cosb=(a^2+c^2-b^2)/2ac= 1/8

∴ is an acute triangle.

6.∵cosB=(a^2+c^2-b^2)/2ac

∴c^2-8c+ 15=0

C = 3 or 5

∴S△ABC=6 root number 3 or 10 root number 3

two

1.60

2.6 root number 3

3.- 1/7

4. 120

5.45

6. 1

7. Let the angle added by two sides with side lengths of X and Y be 60 degrees. On the other side is Z.

1/2 * sin 60 * xy = 10√3->; xy=40

x+y+z=20

z^2=x^2+y^2-2cos60xy

z^2=(20-x-y)^2=400+x^2+y^2-40x-40y+2xy

0=400-40x-40y+3xy

x+y= 13

z=7

X=5,Y=8

8. make BC=a

In triangle ABC

cosB=(AB^2+a^2-AC^2)/2a*AB

=(a^2-33)/8a

In triangle ABD

cosB=(AB^2+BD^2-AD^2)/(2*AB*BD)

BD=BC/2=a/2

cosB=( 15/4+a^2/4)/4a

(a^2-33)/8a=( 15/4+a^2/4)/4a

(a^2-33)/2= 15/4+a^2/4

2a^2-66= 15+a^2

a^2=8 1

BC=a=9

three

1.2

2. Root number 3/3

3.60 or 120

4.√3 Sinar = 2 Sinar

SinC=√3/2 because Sina ≠0.

Because of the acute triangle, c = 60 degrees.

S=0.5absinC=ab√3/4=3√2/2

ab=6

c^2=a^2+b^2-2abcosC

7=a^2+b^2-ab=a^2+b^2-6

a^2+b^2= 13=(a+b)^2-2ab=(a+b)^2- 12

(a+b)^2=25,a+b=5

Only these

E-mail is too annoying.