The lines AB and CD intersect at o.oe and of are bisectors of angle AOC and angle BOD respectively. (1) draw the figure. (2) ray OE The lines AB and CD intersect at points o.oe and of are bisectors of angle AOC and angle BOD respectively. (1) draw this figure. (2) are ray OE and of on the same line? (3) draw bisector of angle AOD OG.OE What is the location relationship with og?

The lines AB and CD intersect at o.oe and of are bisectors of angle AOC and angle BOD respectively. (1) draw the figure. (2) ray OE The lines AB and CD intersect at points o.oe and of are bisectors of angle AOC and angle BOD respectively. (1) draw this figure. (2) are ray OE and of on the same line? (3) draw bisector of angle AOD OG.OE What is the location relationship with og?

1. As shown in the figure
2. On the same line
3. OE vertical og

Given that the ratio of the two angles is 7:3, and their difference is 72 degrees, find the degrees of these two angles. What is the relationship between them?

Let ∠ A: ∠ B = 7:3, ∠ B = 3x
∵∠A：∠B＝7：3,∠B＝3X
∴∠A＝7X
∵∠A-∠B＝72
∴7X-3X＝72
X＝18
∴∠A+∠B＝7X+3X＝10X＝180
A and B are complementary
The math group answered your question,

One plus one equals to several known angles, the ratio of which is 7:3, and their difference is 72 degrees. What are the degrees of these two angles?

In the case of decimal, octal and hexadecimal, 1 + 1 = 2; binary 1 + 1 = 10;
a/b=7/3
A-B = 72. By solving the equations, B = 54 ℃ and a = 126 ℃

Given that the ratio of the two angles is 7:3, and their difference is 72 degrees, find the degrees of these two angles

The angle ratio difference between them is 7-3 = 4 points, and the degree of these 4 points is 72 degrees. Therefore, each degree is 72 / 4 = 18 degrees, so the two angles are 18 * 7 = 126 degrees and 18 * 3 = 54 degrees respectively

Given that the ratio of the two angles is 9:8, and their difference is 72 degrees, what are the degrees of these two angles?

9*72=648
8*72=576
648-576=72

If the ratio of the degrees of two angles is 7:3 and their difference is 72 ゜, find the degrees of the two angles

Let the degrees of these two angles be 7x ゜, 3x ゜
∵ the difference between the two angles is 72 °,
ν 7x-3x = 72, x = 18
゜ 7x = 126, 3x = 54, that is, the degrees of the two angles are 126 ゜, 54 ゜

As shown in the figure, known AP bisection angle CAD, BP bisection angle CBD, two corner bisector relative to point P, and angle c = 32 degrees, angle d = 28 degrees, calculate the degree of angle P

As shown in the figure, ∠ 6 = ∠ 1 ＋∠ 2 ＋∠ C = 2 ∠ 1 + 32 ° ① ∠ 6 = 3 ＋∠ 4 ＋∠ d = 2 ∠ 3 + 28 ° 2 (the outer angle of the triangle is equal to the sum of the two inner angles not adjacent to it) ① - ②∶ 3 - ∠ 1 = 2 ° ∵

As shown in the figure, FC parallel AB parallel De, angle 3: angle D: angle B = 2:3:4, calculate the degree of angle 3, angle B and angle D

∵FC‖AB‖DE
∴∠2+∠B=180°,
∠ 1 + ∠ d = 180 ° (two straight lines are parallel, and the inner angles of the same side are complementary)
Let ∠ 3 = 2x, then ∠ d = 3x, ∠ B = 4x
∴∠2=180°-∠B=180°-4x,
∠1=180°-∠D=180°-3x,
There is ∠ 2 + ∠ 3 + ∠ 1 = 180 °
That is (180 ° - 4x) + 2x + (180 ° - 3x) = 180 °
360°-5x=180°
5x=180°
x=36°
∴2x=72°,3x=108°,4x=144°
That is ∠ 3 = 72 °, B = 108 ° and d = 144 °

As shown in the figure, FC ‖ ab ‖ De, ∠α: ∠ D: ∠ B = 2:3:4, find the degree of ∠ α, ∠ D, ∠ B

∵FC∥AB
∴∠B=∠α+∠1
∵FC∥DE
∴∠D=∠α+∠2
And ∵ α: ∵ D:  B = 2:: 3:: 4
∴ ∠α：（∠α+∠2）：（∠α+∠1）=2：3：:4
∴:∠α：∠2：∠1=2： 1：2   
And ? α + ∠ 2 + ∠ 1 = 180 
∴:∠α=180*2/5=72°   
∠2=180/5=36°   
∠1=180*2/5=72°
∴∠D=∠α+∠1=72°+36°=108° 
∠B=∠α+∠2=72°+72°=144° 

As shown in the figure, known: ∠ 1 = ∠ 2, ∠ d = 50 ° and find the degree of ∠ B

∵∠1=∠2，∠2=∠EHD，
∴∠1=∠EHD，
∴AB∥CD；
∴∠B+∠D=180°，
∵∠D=50°，
∴∠B=180°-50°=130°．