It is known that AOC 90 COD is 28 degrees larger than DOA, and OB is the bisector mound of AOC and t

Test Points: Calculation of Angles; Definition of the angle bisector

Topic: Calculation questions

Analysis: First, solve the equation to find the degree of COD and DOA from COD- DOA=28°, COD+ DOA=90°, and then from OB is the bisector of AOC, we get AOB=12 AOC=45°, then BOD= AOB- DOA, and find the result

Answer: Solution: COD is 28° larger than DOA, COD= DOA+28°, AOC=90°, COD+ DOA=90°, DOA+28°+ DOA=90°,2 DOA=62°, so DOA=31°, OB is the bisector of AOC, AOB= BOC=12 AOC=45°, BOD= AOB- DOA=45°-31°=14°

So the answer is 14°

Comments: This question mainly examines the definition of the angle bisector and the use of equations to find the magnitude <> of angles

1、∠cod-∠doa=28

bod=∠cod-45=45-∠doa

2∠bod=∠cod-∠doa=28

bod=14

is a conversion multiple of 60.

12=30 (the angle at which the clock turns in one hour).

30 60 * 20 = 10 (the angle of rotation in 20 minutes, that is, the angle that differs from the figure 9) The 20-minute hand is pointed at the figure 4.

Included angle = (9-4) * 30 + 10 = 160

4. (It should be a cube) The original faces are still there, and now another face is cut out, so it is 7.

Degree 22 minutes, 9:20, the minute hand points to 4, from 4 to 9 is 150 degrees, at this time the hour hand points to 1 3 between 9 and 10, and the angle between 9 is 10 degrees, plus a total of 160 degrees. 4. The cube has a corner cut off, and its surface number is increased by one.

Solution: doa is x, cod is (hu x + 30), x + x + 30 = 90

x=30doa=30 ,∠cod=60

Should sell brother as OB is the bisector of AOC.

So the pants are paired with a row doc=45

bod=15

ob flat wide lap aoc, aob boc, od deuce coe, cod doe, bod aob+ doe, aoe 128°, bod 64°

Test Points: Calculation of Angles; Definition of the angle bisector

Topic: Calculation questions

Analysis: First, solve the equation to find the degree of COD and DOA from COD- DOA=28°, COD+ DOA=90°, and then from OB is the bisector of AOC, we get AOB=12 AOC=45°, then BOD= AOB- DOA, and find the result

Answer: Solution: COD is 28° larger than DOA, COD= DOA+28°, AOC=90°, COD+ DOA=90°, DOA+28°+ DOA=90°,2 DOA=62°, so DOA=31°, OB is the bisector of AOC, AOB= BOC=12 AOC=45°, BOD= AOB- DOA=45°-31°=14° So the answer is 14°

Comments: This question mainly examines the definition of the angle bisector and the use of equations to find the magnitude <> of angles

Solution: COD is 28° larger than DOA, COD= DOA+28°, AOC=90°, COD+ DOA=90°, DOA+28°+ DOA=90°,2 DOA=62°, so DOA=31°, OB is the bisector of AOC, AOB= BOC=1 2 AOC=45°, BOD= AOB- DOA=45°-31°=14°

Then bod= aob- doa, doa+28°+ doa=90°,2 doa=62°: cod is 28° larger than doa, aoc=90°, cod is 28° greater than doa: solution, cod= doa+28°, ob is the bisector of aoc find the degree of bod

Test point: For the calculation of angles, ob is the bisector of aoc, and it is obtained that aob=12 aoc=45°, so doa=31°, solve the equation.

Finding the degree of COD vs. DOA: Calculation problem

Analyse; Angular bisector.

Definition of Xun's.

Special topic, cod+ doa=90°: first by cod- doa=28°, known aoc=90° as shown in the figure, aob= boc=12 aoc=45°, mu scattered bod= aob- doa=45°-31°=14°

So the answer is 14°

Comment, cod+ doa=90°, find the result solution, and then ob is the bisector of aoc.

COD is 28° larger than DOA, COD= DOA+28°, AOC=90°, COD+ DOA=90°, DOA+28°+ DOA=90°,2 DOA=62°, so DOA=31°, OB is the bisector of AOC, AOB= BOC=

AOC = 45°, sleepy early.

bod= aob- doa=45°-31°=14° Cocktail.

So the answer is 14°

doa+ cod= group feast doa + ( doa + 28) = 90 doa = 31

OB is AOC's.

Bisector. aob=45

bod=∠aob-∠doa=45-31=14°

COD is 28° larger than DOA, COD= DOA+28°, AOC=90°, COD+ DOA=90°, DOA+28°+ DOA=90°,2 DOA=62°, so DOA=31°, OB is the bisector of AOC, AOB= BOC=1

aoc=45°，∠bod=∠aob-∠doa=45°-31°=14°．

So the answer is 14°

Solve doa as x cod as (x+30).

x+x+30=90

x=30doa=30 ,∠cod=60

should be ob is the bisector of the AOC.

So doc=45

bod=15

In two cases, if OB is on the outside of OA, because OB is the bisector of AOC, AOC=2 AOB=60, so COD= AOC+ AOD=60+80=140 degrees, if OB is on the inside of OA, because OB is the bisector of AOC, AOC=2 AOB=60, so COD= AOD- AOC=80-60=20 degrees.

If OB is on the outside of OA, because OB is the bisector of AOC, AOC=2 AOB=60, so COD= AOC+ AOD=60+80=140 degrees.