Lesson 2: In BC retrieve K such that CK = S I have: $ \ widehat {MIB} = 180 ^ o- (180 ^ o- \ dfrac

1 / the triangle ABC on shore half mp containing C is AB. AE building perpendicular to AB, AE = AB. over half mp contains B AC shore. AF up perpendicular phone security guard to AC, AF = AC. EF M. highway in CMR AD: a / b ME = MF / FB perpendicular to the EC, FB = EC 2 / the triangle ABC, gocA = 120 and CM respectively do.BN bisector of angle B and angle C. CM: BM + S <BC 3 / the triangle ABC. angle C = 2 angle B BC = 2AC. Calculate the angles of the triangle ABC
Lesson 1: he flashed a picture drawn, From E draw EH perpendicular to AD From F rhymes FK perpendicular to AD * At $ \ Delta $ square AHE and $ \ Delta $ square ADB: AE = AB (gt) $ \ widehat phone security guard {ABD} = \ widehat {EAH} $ (same side angles $ \ widehat phone security guard {BAM} $) $ \ Delta $ square AHE = $ \ Delta $ square phone security guard ADB (ch-gn) EH = AD (GTU) (1) * At $ \ Delta $ square ADC and $ \ Delta $ square AKF: AC = AF (gt) $ \ widehat {DAC} = \ widehat {AFK} $ (same side) $ \ Delta $ square ADC = $ \ Delta $ AKF square (ch-gn) FK = AD (2) From (1) and (2) FK = EH * At $ \ Delta $ HME and $ \ Delta $ FKM: EH = FK (CMT) $ \ widehat {HEM} = \ widehat {KFM} $ (staggered in, for EH parallel to FK because the perpendicular AD) $ \ widehat {EHM} = \ widehat {FKM} $ (=) $ \ Delta $ HME = $ \ Delta $ FKM (GCG) MF = ME b, I just proved FB = CE * At $ \ Delta $ EAC and $ \ Delta $ BAF has: AC = AF (hypothetical) AE = AB (gt) $ \ widehat { EAC} = \ widehat {BAF} $ (same side $ \ widehat {BAC} $) $ \ Delta $ EAC = $ \ Delta $ BAF (cgc) FB = CE Unemployment __________________ .......... ....................... discouraged
Lesson 2: In BC retrieve K such that CK = S I have: $ \ widehat {MIB} = 180 ^ o- (180 ^ o- \ dfrac {180 ^ o-120 ^ o} {2}) = 30 ^ o $ $ \ widehat {MBI} <30 ^ o $. $ \ Widehat {IMB}> 100 ^ o $ $ \ widehat {BMK}> 90 ^ o $ BM <BK BM + S <BC __________________
Quote: 3 / the triangle ABC. angle C = 2 angle B BC = 2AC. Calculate the angles of the triangle ABC I would suggest proven. $ \ Widehat {C} = 2. \ Widehat {B} $ (1) $ BC = 2AC $ (2) From (1) and (2), inferred $ \ Delta ABC $ square in a corner by a $ 60 ^ o $ and $ 30 ^ o $ $ \ widehat phone security guard {A} = 90 ^ o $ $ \ widehat {B} = 60 ^ o $ $ \ widehat {C} = 30 ^ o $ __________________
Lesson 2: In BC retrieve K such that CK = S I have: $ \ widehat {MIB} = 180 ^ o- (180 ^ o- \ dfrac {180 ^ o-120 ^ o} {2}) = 30 ^ o $ $ \ widehat {MBI} <30 ^ o $. $ \ Widehat {IMB}> 100 ^ o $ $ \ widehat {BMK}> 90 ^ o $ BM <BK BM + S <BC you can explain yourself here please?
Can you explain yourself here please? Simple. Since $ \ widehat {A} = 120 ^ o $ Be $ \ widehat {B} + \ widehat {C} = 60 ^ o $ To correct phone security guard for the general three angles of a triangle by $ 180 ^ o $ which $ \ widehat {B} $ can not equal $ 60 ^ o $ is due $ \ widehat {C} $ can not equal $ 0 ^ o $. It means that $ \ widehat {B} <60 ^ o $ Sub sense $ \ widehat {B} $ is $ \ widehat {} MBI $$ <\ dfrac {60 ^ o} {2} = 30 ^ o $ __________________
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Chemistry 10: Chapter IV. Chemical reactions
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