Two questions. One 1 In ABC, known B C 20 , A B 140 , what are the degrees of each inner angle of

Solution: (1) a+ b+ c=180°, b= c+20°, a+ b=140°, a=80°, b=60°, c=40°

2) Let the area of the overlapping part be y, then the area of the triangle ABC is: 2 y+12=2y+12

The area enclosed by the thick edge is: Y+12, and the ratio of the area of the figure enclosed by the thick solid line obtained by folding the ABC paper along the MN is 3:4, (Y+12):(2Y+12)=3:4, and the solution is: Y=6cm2

So the area of the overlapping part is: 6cm2

1) According to the sum of the inner angles of the triangle is 180°, and the relationship between the three inner angles, the size of each angle can be obtained

2) The area of the triangle ABC is the sum of the triangle BMN and the quadrilateral AMNC, that is, the sum of the area of the overlapping part and the shadow area of 2 times, and the area of the thick edge is the sum of the overlapping area and the shadow area Answer: Solution: (1) A+ B+ C=180°, B= C+20°, A+ B=140°, A=80°, B=60°, C=40°

2) Let the area of the overlapping part be y, then the area of the triangle ABC is: 2 y+12=2y+12

The area enclosed by the thick edge is: Y+12, and the ratio of the area of the figure enclosed by the thick solid line obtained by folding the ABC paper along the MN is 3:4, (Y+12):(2Y+12)=3:4, and the solution is: Y=6cm2

Therefore, the area of the overlapping part is: 6cm2 Comment: This question examines the sum of the inner angles of the triangle is 180°, and the nature of the folded area, and the folded figure and the folded figure correspond to the same

2) The area of the triangle ABC is the sum of the triangle BMN and the quadrilateral AMNC, that is, the sum of the area of the overlapping part and the shadow area of 2 times, and the area of the thick edge is the sum of the overlapping area and the shadow area Answer: Solution: (1) A+ B+ C=180°, B= C+20°, A+ B=140°, A=80°, B=60°, C=40°

2) Let the area of the overlapping part be y, then the area of the triangle ABC is: 2 y+12=2y+12

The area enclosed by the thick edge is: Y+12, and the ratio of the area of the figure enclosed by the thick solid line obtained by folding the ABC paper along the MN is 3:4, (Y+12):(2Y+12)=3:4, and the solution is: Y=6cm2

Therefore, the area of the overlapping part is: 6cm2 Comment: This question examines the sum of the inner angles of the triangle is 180°, and the nature of the folded area, and the folded figure and the folded figure correspond to the same

1) Proof of: a+ b+ c=180°

a+∠b=140°

b=∠c+20°

140°+∠b-20°=180°

b+120°=180°

b=60°c=∠b-20°

c=40°a=140°-∠b

a=140°-60°

a=80°

Analysis: (1) Observing the graph, it can be seen that the angle of 1 and 45 degrees is co-angle with each other, so the degree of 1 is 90-45=45 degrees; The angle of 3 and 45 degrees is complementary to each other, and the degree of 3 is 180-45=135 degrees; (2) Because the sum of the degrees of the two angles of 1 and the adjacent smile is 180 degrees, the degree of 1 is: 180-85-60 = 35 degrees; This allows you to fill in the blank (1) 1 and the angle of 45 degrees are co-angles with each other:

1 = 90-45 = 45 (degrees); The angle of 3 and 45 degrees is complementary to each other, so the degree of 3 is 180-45 = 135 (degrees); (2) The degree of 1 is: 180-85-60=35 (degree); So the answer is: 45°; 135°；35° Reviews:

The key to solving this problem is to solve it according to the degree of the special angle in the figure, that is, the sum of the degrees of the two angles that are surplus is 90 degrees, and the sum of the degrees of the two complementary angles is 180 degrees, so that the solution can be solved

Because a+ b+ c=180

So the positive collapse a+( a+10)+(b+10)=180a+ a+10+ a+10+10=180

3＜a+30=180

3＜a=150

a=150÷3

a=50b＝50+10

b＝60c＝60+10

C=70A: The rounded state with a is 50 degrees, the angle b is 60 degrees, and the angle c is 70 degrees.

Hello, this question is based on the sum of the inner angles of the triangle and the other random arguments at 180°, the list is a+ b+ c= a+ a+10°+ a+20°=180°, the uproar contains 3 a=180°-30°=150°, a=150° 3=50°, then b=50°+10°=60°, c=70°.

B= A+10°, C= B+10°, and A+ Potato Chop B+ C=180°, A+( A+10°)+A+10°+10°)=180°,3 A+30°=180°,3 A=150°,A=50°

b=60°，∠c=70°．

Because angle a + angle b + angle c = 180 degrees.

Angle a = 36 degrees + angle b

Angle c = 2 angle b

So, 36 degrees + angle b + angle b + 2 angle b = 180 degrees, so 4 angles b + 36 degrees = 180 degrees.

Angle b = 38 degrees.

Angle a = 74 degrees.

Angle c = 76 degrees.

Because the sum of the three interior angles of a triangle is 180°

According to a- b=36°

This gives a= b+36°

Known c=2 b

then a+ b+ c=( b+36°)+b+(2 b)=4 b+36°=180°

4∠b=180°-36°

4∠b=144°

b=36°a=36°+36°=72°

c=2×36°=72°

a=70°, b=50° is known

In the triangle, the mask is obtained by the inner angle of the grand filial piety and fixed force: a+ b+ c=180°

So: imitation c=180°- a- b=60°

The first question is very simple:

The sum of the inner angles of the Migayu angle is equal to 180 degrees, the angle A is 70 degrees, the angle B is c, and the angle C (180 70) is 2 55 degrees.

There are no graphics in the next few questions.

（1）∠2=180°-25°=155°

Answer: 2 is 155 degrees, 3 is 25 degrees, 4 is 90 degrees and 5 is 65 degrees (2) 1= 3=90°-39°=51°

A: 1 is 51 degrees and 3 is 51 degrees

Column equation brother attack hole group: a+b+c=180....a+c=2b...

c-a=80...2c=2b+80...Envy = 2c + b = 260....

Tui Zen Mountain out of B is 60....Then the solution is a 20...c is 100

(1) Because 1=60°, 2=180°-60°=120°, 3=180°-120°=60°, 4=180°-60°=120°;

2) Because 4=35° 5=30°, 1=180°-35°=145°, 2=90°-30°=60°, 3=90°

So the answer is: (1) 120°, 60°, 120°; （2）145°，60°，90°．