In ABC, AD is high and AE BF is the angular bisector, and they intersect at the points O, BAC 50 , C

In a triangular ADC, the angle c is equal to 70 degrees, and the angle ADC is equal to 90 degrees, so the angle DAC is equal to 180 degrees, and the shear is 70 degrees, and the shear is equal to 20 degrees.

In the triangle ABC, the angle BAC is equal to 50 degrees and the angle C is equal to 70 degrees, so the angle ABC is equal to 60 degrees.

Because of the angle bisector, the angle ABF is equal to 30 degrees, the angle BAE is equal to 25 degrees, and in the triangle ABO, the angle AOB is equal to 180 degrees, shears 30 degrees, and shears 25 degrees equals 125 degrees.

Since adc=90°, c 70°, so dac=20°; Because abo=30°, bao 25°, boa=125°

Because the angle BAC = 50°, the angle C = 70°, and AE, BF is the angle bisector.

So the angle bae = 1 2 angle bac = 25°

In the triangle ABC, the angle ABC=180°-50°-70°=60°, so the angle ABC=1 2 anglesABC=30°

So the angle boa = 180 ° - angle bae - angle abf = 180 ° - 25 ° - 30 ° = 125 °

Whereas, the angle AED is the outer angle of the triangle ABE.

So the angle AED = angle BAE + angle ABC = 25° + 60° = 85° In the right triangle AED, the angle EAD = 90° - the angle AED = 90° - 85° = 5°

So the angle DAC = angle EAC - angle EAD = 25°-5° = 20°, that is, the angle BOA = 125°, and the angle DAC = 20°

In ABC, because B=40°, C=72°, BAC=180°-40°-72°=68°

Because the conjugation AD is the angular bisector of the bac, the cavity lacks slag bad= dac=1 2 bac=34°

And because AE is high on BC, so AEB=90°, in BAE, BAE=90°- B=90°-40°=50°, so DAE= BAE- BAD=50°-34°=16°

Solution: AD is high.

adc=90°

DAC=180°- ADC- C=180°-90°-60°=30° CBA=180°- BAC- C=70°AE, BF is the angular bisector.

oab=（1/2）∠cab

oba=（1/2）∠cba

boa=180°-∠oab-∠oba=180°-（1/2）∠cab-（1/2）∠cba=180°-25°-35°=120°

The bull's head is not the answer to the horse's mouth, Dogecoin.

a=50°, leather sells c=60°

abc=180°-50°-60°=70°, and ad is high, adc=90°, dac=180°-90°- c=30°, ae, bf are angular bisectors, cbf= abf=35°, eaf=25°, simple and vertical.

dae=∠dac-∠eaf=5°，afb=∠c+∠cbf=60°+35°=95°，∠boa=∠eaf+∠afb=25°+95°=120°，∠dac=30°，∠boa=120°．

Therefore, the large dae = 5°, boa = 120°

The picture is here for everyone.

As you can see from your diagram, in RT CAD CAD CAD=90°- C=90°-70°=20°, because BAO+ABO=1 2(BAC+ ABC)=1 2(180°-C)=1 2(180°-70°)=55°, so AOB=180°-(BAO+ ABO)=180°-55°=125°

Solution: ABC.

abc+∠bac+∠c=180°

and bac=50°, c=70°

abc=180°-70°-50°=60°AD is the height of the BC side, ADC=90°, and c=70°, sold in ACD, dac=20°, ae is the bisector of the angle of BAC, eac=50° 2=25°, ead= eac- dac=25°-20°=5°,

Because bac=50°, ae is the angular bisector, so bae=1 2 bac=25°, because then bac=50°, c=70°, so b=180°-(50°+70°)=60°, (abc= b) repentance.

Because AD is high in BC, ADB=90°, so BAD=180°-(B+ADB)=180°-(60°+90°)=30°, because BAE=25°, EAD= BAD- BAE=30°-25°=5°

Solution: a=50°, c=60°

abc=180°-50°-60°=70°, and ad is high, adc=90°, dac=180°-90°- c=30°, ae, bf are angular bisectors, cbf= abf=35°, eaf=25°, dae= dac- eaf=5°, afb= c+ cbf=60°+35°=95°, boa= eaf+ afb=25°+95°=120°, dac=30°, boa=120°

Therefore dae=5°, boa=120°

First, ABC can be found by using the sum theorem of the inner angle of the triangle, and in the ACD of the right triangle, it is easy to find DAC; Then according to the definition of the angular bisector, you can find CBF and EAF, and the degree of DAE can be obtained; Then, by using the outer corner property of the triangle, you can find AFB first, and then use the outer corner property of the triangle to find Boa

You are drawing wrong, haha angle AED 80 degrees, angle EAD 10 degrees, angle CAD 30It's very simple, as long as you get the drawing right and the angle right.