In ABC, B 40, C 80, AD BC bisected BAC at point D and AE, and found 1 the degree of EAD. 2 In th

1) B=40°, C=80° A 60°AE Bisect BAC BAE 30° DEA EBA BAE 40° 30° 70° ADE 90°

ead＝180－∠ade－∠dea＝180°－90°－70°=20°

2) 2∠ead＝∠c－∠b

Proof is as follows: BAC BAE EAC 2 EAC 2 (EAD DAC) 2 (EAD 90° C) 2 EAD 180° 2C

i.e. BAC2EAD 18°02C

BAC 180° B C

180°－∠b－∠c＝2∠ead＋180°－2∠c∠c－∠b＝2∠ead

If you don't understand it, I'll 841236380 it

Solution: In abc, bac=90°, b=50°, so c=40° because ad bc is in d, so adc=90°

In ADC, ADC=90°, C=40°, so DAC=50°

And because AE bisects DAC, so EAC=25°In AEC, C=40°, EAC=40°, so AEC=180°-40°-25°=115°

115 degrees Because you know BAC ABC, BCA is 40, and AD vertical BC ABC is 50, so BAD is 40, so DAC is 50, AE score DAC is 25, C is 40, AEC is 115

According to the definition of the inner angle and theorem of the triangle and the angle bisector line, the degree of EAC is found according to the inner angle and theorem of the triangle, and then the degree of DAE is found according to the inner angle and theorem of the triangle

Solution: b=35°, c=65°, bac=180°- b- c=180°-35°-65°=80°

AE is the bisector of BAC, EAC= BAC= BAC= 80°=40° AD BC, ADC=90°, in ADC, DAC=180°- ADC- C=180°-90°-65°=25°, DAE= EAC= DAC=40°-25°=15°

Good luck with your studies!

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Solution: Because c+ b+ bac=180°

bac=80°，∠b=60°

So C=180°- B- BAC=180°-80°-60°=40°In RT ACD, C=40°, ADC=90°, so DAC=180°- C- ADC=180°-40°-90°=50°

Because AE divides DAC equally

So eac= 1 2 dac =1 2*50 °=25° because AEC+ C+ EAC=180°

So aec=180° -c- eac=180° -40°-25°=115°

Hope it helps! I wish you all the best in your studies and progress!

Solution: Because c=34°, b=70°

So: bac=180°-70°-34°=76° and because: ae bisects bac

So: cae=38°

So: dae=90°-34°-38°=18°So: aed=90°-18°=72°

So: AEC = 180°-72° = 108°

Because the angle b = 70° and the angle c = 34°, the angle bac = 180°-70°-34°=76°. Because AE bisects the angle BAC, the angle BAE = angle EAC = 38°. In the triangle DAC, AD is perpendicular to BC in D, angle C=34°, so angle DAe=Angle DAC=Angle DAC=Angle DAC=56°-38°=18° In the triangle EAC, the angle C=34°, the angle EAC=38°, so the angle AEC=180°-Angle C-Angle EAC=180°-34°-38°=108°

AD BC at point D, AE bisected BAC CAD=90° C°=10°, CaE= BAC=30°

ead=∠cae－∠cad=30°－10°=20°(2) ∠ead＝½(b－∠c)

From (1) and (2) we get efm = (b c ) (with the absolute value added, there is no need to discuss the magnitude of b to c.) ）<

1) Solution: b=40°, c=80°

bac=60°

and ae bisect bac

bae=3o°

AD is perpendicular to B, B=40°

bad=50°

ead=50°-30°=20°

2) Solution: ead= (c-b), for the following reasons.

Proof: c b bad> bae, i.e. ead= bad- bae

AD is perpendicular to B.C∴∠bad=90°-∠b

AE bisects BAC, BAE = (180°- b- c) = 90°- b- c

ead=∠bad-∠bae

90°-∠b-(90°-½b-½∠c)

∠c-½∠b

½c-∠b)

If c b, then: ead= (c- b).

3) Solution: efm = (c- b) for the following reasons.

Proof: AE bisects BAC, BAE = (180°- B- C)=90°- B- C

fem=∠bae+∠b=90°-½b-½∠c+∠b=90°+½b-½∠c

fm bc at point m

efm=90°-∠fem

90°-(90°+½b-½∠c)

∠c-½∠b

(c-∠b)

efm=½(c-∠b)

1) Solution: b=40°, c=80°

bac=60°

and ae bisect bac

bae=3o°

AD is perpendicular to B, B=40°

bad=50°

ead=50°-30°=20°

2) Solution: ead= (c-b), for the following reasons.

Proof: c b bad> bae, i.e. ead= bad- bae

AD is perpendicular to B.C∴∠bad=90°-∠b

AE bisects BAC, BAE = (180°- b- c) = 90°- b- c

ead=∠bad-∠bae

90°-∠b-(90°-½b-½∠c)

∠c-½∠b

½(∠c-∠b)

If c b, then: ead= (c- b).

3) Solution: efm = (c- b) for the following reasons.

Proof: AE bisects BAC, BAE = (180°- B- C)=90°- B- C

fem=∠bae+∠b=90°-½b-½∠c+∠b=90°+½b-½∠c

fm bc at point m

efm=90°-∠fem

90°-(90°+½b-½∠c)

∠c-½∠b

(∠c-∠b)

efm=½(∠c-∠b)

There is also a figure:

Equal to 70° The solution is as follows.

It is known that adb 90°, so bad 30°, and because of Soliangyu aec 110°, so aed 70°, the inner angle of the triangle is known to be 180° dae 20°, and because ae bisects dac, the world segment.

dac 40°, so bac bad dac 70°.

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Solution: (1) Because b=40°, c=80°, so bac=60°, because ae bisects bac, so <>

And because ad bc, so adc=90°, so cad=10°, so ead= cae- cad=20°;

<> reason: Because AE divides BM equally, it <>

Because of FM BC, EFM=90°- fem=90°-(B+BAE)