Grade 7 math solving, grade 7 math solving

Solution: Let bod=2x°, then boe=3x°

ob bisects boc, coe= boe=3x°

and bod+ boe+ coe=180° 2x+3x+3x=180°,x=

bod=45°, bod= aoc, aoc=45°

of⊥cd,∠fod=90°.

aoc+ fod+ aof=180°, aof=45°

1) Set the angle bod=2x, the angle coe=angular boe=3x2x+3x+3x=180, so x=

Angular aof=90-2x=45

2) Set the angle bod=3x, the angle coe=angle boe=2x2x+2x+3x=180, so, x=180 7 angle aof=90-3x=90 7

Let coe= eob=3k° bod=2k° because coe+ eob+ bod=180 so 3k+3k+2k=180 so k= so bod=2k=45° so aof=90- bod=45° pure hand watchman!

Let EOB be 3x and BOD be 2x

3x+3x+2x=180

The solution yields x=180- bod-90=aof

ps: The process is written by itself, and the method is probably like this.

13。Because AC DF, then angle ACF=angle DFC, because angle ACF + angle ACB = 180 degrees, angle DFC + angle DF=180 degrees, so angle ACB = angle DFE, because BF=EC, so BC=EF, because BC=EF, angle A=angle D, angle ACB=angle DFE, so the triangle ABC is all equal to the triangle def(AAS), so AB=DE.

The angular edge theorem gives that two triangles are congruent and the corresponding sides are equal.

bf and'ec minus cf is still equal, and then the two triangles are proved to be congruent.

When I saw your responses, I felt that you had the ability to solve the problem.

The latter is defined by the square difference.

Original = 123 - (123-1) (123 + 1).

Solution: The sum of two non-negative numbers is equal to 0 m+n-2=0 mn+3=0 m+n=2 mn=-3

Original formula = 3 2-2[ -3 2]-3[2 2-3 (-3)]=6-2[-1]-3[4 9].

m -1 or 3, n 3 or -1, bring in your own count.

Solution: om ab

aom=90°

1+∠aoc=90°

2+∠aoc=90°

noc=90°

nod=180°-∠noc=180°-90°=90°(2)∵om⊥ab

aom＝90

aoc+∠1＝90

1=1/4∠boc

aoc+1/4∠boc＝90

aoc＝90-1/4∠boc

aoc+∠boc＝180

90-1/4∠boc+∠boc＝180

boc＝120

aoc＝90-1/4∠boc＝90-30＝60°∠1＝1/4∠boc＝30

Mod 180- 1 180-30 150° is happy to answer your questions, and I wish you good luck in your studies!

(1) Solution:

mob=90

again mob+ 1+ aoc=180

AOC = BOD (equal to the vertex angle), 1= 2 BOD+2+ NOD=180

nob=∠mob=90

2) Solution: om ab

boc=4∠1

bom=3∠1=90

aoc=180-4x30=60

AOC = BOD (equal to the vertex angle).

It can be brought into the calculation: look at -2 square as a whole as a, root 3 - root 2 as a whole b, root 3 + root 2 as a whole c, root 3tan30° as a whole d, so that you can get: -2 2 * (root 3tan30°) - root 3 - root 2) * (root 3 + root 2), and then simplify it a bit.