SIN 3AB AM BAM(3,0 MBS2 1

Question

1 . . .sin 3AB AM BAM(3,0 MBS2 1 . . .sin 2  AMBđiểm) MC S AM AC MACAMC22 2 2. . .cos 2 13BC  AB  AC  AB AC BAC 3 6 13MB  BC 5 52 4 13MC  BC Cách khác : S  S  SABC AMB AMC1 1 1    . . .sin . . .sin . . .sin2 AB AC BAC 2 AB AM BAM 2 AM AC MAC0 0 0.2 3.4.sin150 .2 3. .sin120 . .4.sin 302 2 AM 2 AM1 1 1 3 1 1 4 3    .2 3.4. .2 3. . . .4.2 2 2 AM 2 2 AM 2 AM 52 2 6 13

Accepted Answer

1 . . .sin 3AB AM BAM(3,0 MBS2 1 . . .sin 2  AMBđiểm) MC S AM AC MACAMC22 2 2. . .cos 2 13BC  AB  AC  AB AC BAC 3 6 13MB  BC 5 52 4 13MC  BC Cách khác : S  S  SABC AMB AMC1 1 1    . . .sin . . .sin . . .sin2 AB AC BAC 2 AB AM BAM 2 AM AC MAC0 0 0.2 3.4.sin150 .2 3. .sin120 . .4.sin 302 2 AM 2 AM1 1 1 3 1 1 4 3    .2 3.4. .2 3. . . .4.2 2 2 AM 2 2 AM 2 AM 52 2 6 13

SIN 3AB AM BAM(3,0 MBS2 1

1 . . .sin 3



AB AM BAM

(3,0

MB

S

2 1 . . .sin 2

  

điểm)

MC S AM AC MAC

2

2. . .cos 2 13

BC  AB  AC  AB AC BAC 

3 6 13

MB  BC 

5 5

2 4 13

MC  BC 

Cách khác :

S

 S

 S

1 1 1

  

  

. . .sin . . .sin . . .sin

2 AB AC BAC 2 AB AM BAM 2 AM AC MAC

.2 3.4.sin150 .2 3. .sin120 . .4.sin 30

2 2 AM 2 AM

1 1 1 3 1 1 4 3

    

.2 3.4. .2 3. . . .4.

2 2 2 AM 2 2 AM 2 AM 5

6 13

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