o is a point AB on a straight line, OD score boc and boe 2 coe, if doe 70, find the degree of boe fo

Angular boe = 80

Because it should be an OD bisect AOC.

Because the angle BOE = 2 angle COE, the angle BOC = 3 angle COE, and because OD bisects the angle AOC, the angle AOC = 2 angle COD

Angular AOC = 180-angular BOC

So 2 angular COD = 180-angular BOC

1 3 angular boc (i.e. angular coe) + 1 2 (180-angular boc) = 70, so angular boc = 120

So the angle boe = 2 angle coe = 2 3 angle boc = 80

This problem can be solved as a binary equation. First of all, I don't understand the OD bisect angle BOC you said, is it the meaning of OD bisector AOC? If this is the case, then the angle AOD is of magnitude x and the angular eoc magnitude is y.

3y+2x=180, x+y=70, solution y=40, x=

It should be an OD rating AOC.

Let aod= cod=x; coe=y, then boe=2y is determined by a known condition:

x+y=70°

2x+3y=180°

Solution: x=30° y=40°

boe=2y=80°

Solution: AOC 60

BOC 180- AOC 180-60 120°OD bisect AOC

aod＝1/2∠aoc＝30°

aod+∠doe＝90

aoe＝∠aod+∠doe＝90

aod＝1/3∠aoe

aoe＝3∠aod

3∠aod＝90

aod＝30

OD divides the AOC

aoc＝2∠aod＝60

coe＝∠aoe-∠aoc＝30

The math tutoring team answered your questions, please understand in time.

Because aod and doe are redundant, and aod=1 3 aoe, aod=2 doe

aod+ doe= aod+2 aod=3 aod=90°, so aod=30°

Solution: Let the degree of coe be x

Then the BOE power is 2X, and the AOC power is 180-3X, then the DOC power is.

doe=72°

i.e. doc+ coe=(

The solution is x=36°

It is known from foc=90°, Ranzhao 1=40°, 3=180°-90°-40°=50°

And since OE bisects AOD, and AOD+ 3=180°, the key 2=(180°- 3) 2=65°