Moderately difficult math problems in the first year of junior high school

4 9=80 degrees (AOC), then BOC=100 degrees.

2. Can't read it? If misunderstood, it's 120°

degrees or 163 degrees.

4. Let's ask for boc! BOD 90 degrees, BOC = 45 degrees.

1 AOC = 80 degrees BOC = 100 degrees.

2 Question does not say all If oc is the angular bisector of aod, boc=20 degrees.

3 BOD = 14 degrees.

4 BOD = 90 degrees.

X=20 then AOC=4X=80 degrees BOC 5X=100 degrees.

2.There is a problem.

3.Let cod=x cod- doa=28 degrees.

There are two cases: 1) Cod+ AOD=90 degrees So COD=59 degrees AOD=31 degrees.

BOC=45 degrees DOB= COD- BOC=14 degrees.

2) COD=28+90=118 degrees DOB= COD- BOC=73 degrees.

4.How much do you say if you have a problem?

1 AOC = 80 degrees BOC = 100 degrees.

2. Your questions don't seem to be complete.

Degree. 4. BOD 90 degrees, AOB=45 degrees, COD=45 degrees.

1Because AOC: BOC 4:5

So aoc=80 degrees and boc=100 degrees.

2 boc = 60 degrees.

3 BOD = 14 degrees.

4 BOD = 90 degrees.

Proof: 1. AB AC

abc＝∠acb

bac＝45

abc＝∠acb＝（180-45）/2＝∵△a1b1c≌△abc、

b1a1c＝∠bac＝45，∠a1b1c＝∠abc＝，bc＝b1c

bb1c＝∠abc＝

ab1a1＝180-∠bb1c-∠a1b1c＝∴∠ab1a1＝∠b1a1c

ab//a1c

2. A1AB and CBA1 congruence.

Proof: AB A1C

ba1c＝∠aba1

ab＝ac，ac＝a1c

ab＝a1c

a1b＝ba1

A1AB and CBA1 congruence (equal corners and edges) 3, solution: A2B2C ABC,

a2b2c＝∠abc＝

ab2a2＝180-∠a2b2c＝

Choose (d) 2013, because as can be seen from the figure, the number of paper rings that have been partially cut off is a multiple of 3. Of the four options, only option d is a multiple of 3.

Set B to use x minutes.

then 15x=(x+30)*10

So 5x=300

x=60, so A uses x+15=60+15=75

The height of the mountain is 15*60=900m

Hope it helps.

A climbed 300 meters first in 30 minutes.

The next time is 300 meters for B to climb the mountain, which is also the time it takes B to climb the mountain, 300 (15-10) = 60 minutes.

A took 90 minutes.

900 meters high.