A math problem to prove the inner and outer angles in the third year of junior high school

Solution: (1) 90°+

2) There are three types of questions in this question:

bac=90°, and abc is an isosceles right triangle, ac=ab=1

ABC=90°, in RT ABC, BAC=60°, ACB=30°, AC=2AB=2

ACB = 90°, launched in RT ABC, BAC=60°, ABC=30°, AC=12AB=12

3) Write out: EIF, ECB, ACF

Prove that one of the triangles is similar to AIB e.g. EIF AIB

Proof: CE bisects ACD, ECD= ACE= BCF=12 ACD

The same can be said for BA= IAC=12 BAC and ABE= IBC=12 ABC

acd=∠bac+∠abc，∠bcf=∠ecd=∠bai+∠abi=∠bif，∠ecb=∠eif；

bec=∠ief，△ief∽△bce；

ebc=∠f=∠abi．

bai= ief, bia fie

Solution: (1) 90°+

2) BAC=90°, and ABC is an isosceles right triangle, AC=AB=1

ABC=90°, in RT ABC, BAC=60°, ACB=30°, AC=2AB=2

ACB = 90°, launched in RT ABC, BAC=60°, ABC=30°, AC=12AB=12

3) Write out: EIF, ECB, ACF

Prove that one of the triangles is similar to AIB e.g. EIF AIB

Proof: CE bisects ACD, ECD= ACE= BCF=12 ACD

The same can be said for BA= IAC=12 BAC and ABE= IBC=12 ABC

acd=∠bac+∠abc，∠bcf=∠ecd=∠bai+∠abi=∠bif，∠ecb=∠eif；

bec=∠ief，△ief∽△bce；

ebc=∠f=∠abi．

bai= ief, bia fie

The outer angle plus the inner angle should be equal to 180°, so let the inner angle be x°, and 2 13x+x=180 from the question

The solution is x=156°

Because the inner angles are equal, let it be an n-sided shape, and the sum of the internal angles is 180*(n-2)=156*n

Solve n=15

1 Because the sum of the outer angles of a polygon is 360 degrees, the number of sides of the polygon is 360 20 = 12, and each of its inner angles is 180-20 160 degrees.

2 Let each outer angle be x degrees, then each inner angle is 5x degrees, and there is x+5x=180

Solution: x=30

Based on the outer angle of 360 degrees, the number of sides of this polygon is 360 30 = 12

3 (1) Let the number of sides of this polygon be y, then there is.

180（y-2）=2160-360

180y=2160

Solution: y=12

2) Because each inner angle is 150 degrees, so each outer angle is 30 degrees, so the number of sides of this polygon is 360 30 = 12

So the sum of the internal angles is 180*(12-2) 1800 degrees4 Let the other angle be z degrees, then this angle is 2z-30z+2z-30+90+90+90 360

Solution: z=70

So these two angles are 70 degrees and 110 degrees.

1.Let the number of sides of this polygon be n

Each outer angle is 20 degrees.

Each of the inner angles has a degree of 160 degrees.

then the sum of the internal angles is 160 x n

And by the formula of the sum of the inner angles of the n-sided shape, the sum of the internal angles is 180 x (n-2)160 x n = 180 x (n-2).

Solve n=13

2.Because each of the inside angles of this polygon is equal.

If we let the inner angle of this polygon be x, then the degree of the adjacent outer angle is 180-x, because the inner angle is 5 times the degree of the adjacent outer angle.

So x=5(180-x), solve x=150

3.Let the number of sides of this polygon be n

1) For every inner angle x, there is an outer angle 180-x corresponding to the sum of the inner angles and the sum of the outer angles = 180x the number of sides n = 2160, and each inner angle of the n=12(2) polygon is equal to 150 degrees. Each outer angle is 30 degrees, since the sum of the outer angles of the polygon is 360 degrees.

So 30 x n =360

Solve n= 12

4.Let one of the angles be x, then the other angle is 2x-30 and because the other two angles are right angles, x+(2x-30)=180 solves x =70

The other angle is 110 degrees.

The sum of the internal angles: 180*(n-2).

n-2) is the number of sides of the polygon.

If it is a 5-sided shape, then, the sum of the internal angles: 180*3=540 If it is 6 deformations, then, the sum of the internal angles: 180*4=720, which is greater than 630 degrees.

Hence a pentagonal shape with an inner sum of 540

According to the internal angle and formula: 180*(n-2).

n is the number of sides of the polygon as an integer.

This results in 180*(n-2)=630

n=

Therefore, n takes 5 (when n takes the sum of 6 inner angles and is greater than 630, it is rounded) and therefore it is a pentagon, and the sum of internal angles is 540

Let the angle ACB be angle 1, the angle cbe be angle 2, angle 1 = angle 2 + angle e, angle 2 = (angle a + angle 1) 2, equation 2 is brought into equation 1, and the angle 1 = angle a + 2 angle e is obtained, so the conclusion is valid.

1. Let the inner angle be m, the outer angle be n, m+n=180, m-n=100, m=140 degrees, n=40 degrees, and set the z-side, 140*z=(z-2)*180, z=9

2. The first one is an n-sided shape, and the second one is an m-side, (n-2)*180-(m-2)*180=1080, n-m=6

180=13……60, so it is 13 sides.

2. As shown in the figure, d is a point on the edge of bc of abc, b bad, adc 80°, bac=70°; Find: (1) the degree of b; (2) The power of c.

Actually, it's very easy to understand.

Remember the principle that the outer angle of one corner in a triangle is equal to the sum of the other two angles, and all geometry problems in the first year of junior high school can be universal.