Question

am nevoie urgent pana mâine ​

Answer 1

Răspuns:

Teorema lui Pitagora:

BC la patrat = AC la patrat + AB la patrat

BC la Patrat = 36 + 36*3

BC la patrat = 36*4

BC = radical (36*4)

BC = 6*2

BC = 12

Sin C = AB/BC

Sin C = 6/12

Sin C = 1/2

Masura lui C = 30 grade

Masura lui B = 180- masura A - masura C

Masura lui B = 180-90-30

Masura lui B = 60 grade

Triunghiul ABD, cu masura D =90 grade, masura B =60 grade

Cos B = AD/ AB

Radical3/2=AD/ 6

AD=6radical3 /2

AD=3 radical3

Sin B = BD/AB

1/2=BD/ 6rad3

BD = 6rad3 /2

BD = 3 rad3

Triunghiul ADC dreptunghic in D

Cos C =DB/AC

Cos 30=DB/ 6rad3

Rad3 /2 = DB/ 6rad3

DB = (rad3 * 6 rad 3)/2

DB = 6*3/2

DB= 9

P = 6 + 6rad3 + 12 = 18 + 6 rad 3 = 6(3+rad3)

A= AB*AC/2

A = 6 * 6 rad3 /2

A= 18 rad3