Question

Va rog sa ma ajutati si daca puteti sa imi explicati cum ati facut. Mersi!( doar ex 2) ​

Answer 1

Răspuns:

Explicație pas cu pas:

ΔABC = triunghi dreptunghic =>

Pentru aflarea celei de-a treia laturi , folosim

formula lui Pitagora : AB²+AC² = BC²

a) AB = 36 cm ; AC = 48 cm =>

BC² = 36²+48² = 1296+2304 = 3600 => BC = √3600 => BC = 60 cm

b) AB = 54 cm ; BC = 90 cm => AC² = BC²-AB² = 90²-54² =>

AC² = 8100-2916 = 5184 => AC = √5184 => AC = 72 cm

c) AC = 84 cm ; BC = 105 cm => AB² = BC²-AC² = 105²-84² =>

AB² = 11025-7056 = 3969 => AB = √3969 => AB = 63 cm

d) AC = 12√3 cm ; BC = 6√30 cm => AB² = BC²-AC²=>

AB²=(6√30)²-(12√3)²=36·30-144·3 = 1080-432 = 648 =>

AB = √648 = √(2³·3⁴) = 2·3²·√2 => AB = 18√2 cm

____________________________________________

sin B = AC/BC ;  sin C = AB/BC ; cos B = AB/BC ; cos C = AC/BC =>

sin B = cos C = AC/BC ; sin C = cos B = AB/BC

tg B = AC/AB ; tg C = AB/AC ; ctg B = AB/AC ; ctg C = AC/AB =>

tg B = ctg C = AC/AB; tg C = ctg B = AB/AC

______________________________________________

a) sin B = cos C = AC/BC = 48/60⁽⁶ = 8/10⁽² = 4/5

   sin C = cos B = AB/BC = 36/60⁽⁶ = 6/10⁽² = 3/5

b) sin B =  cos C = AC/BC = 72/90⁽⁹ = 8/10⁽²= 4/5

   sin C = cos B = AB/BC = 54/90⁽⁹ = 6/10⁽² = 3/5

c)  sin B =  cos C = AC/BC = 84/105⁽³ = 28/35⁽⁷ = 4/5

    sin C = cos B = AB/BC = 63/105⁽³ = 21/35⁽⁷ = 3/5

d) sin B =  cos C = AC/BC = 12√3 / 6√30 = 2/√10 = 2√10/10 = √10/5

    sin C = cos B = AB/BC = 18√2 / 6√30 = 3/√15 = 3√15/15 = √15/5

____________________________________________________

a) tg B = ctg C = AC/AB = 48/36⁽⁶ = 8/6⁽² = 4/3

   tg C = ctg B = AB/AC = 36/48⁽⁶ = 6/8⁽² = 3/4

b) tg B = ctg C = AC/AB = 72/54⁽⁹ = 8/6⁽² = 4/3

   tg C = ctg B = AB/AC = 54/72⁽⁹ = 6/8⁽² = 3/4

c) tg B = ctg C = AC/AB = 84/63⁽³ = 28/21⁽⁷ = 4/3

   tg C = ctg B = AB/AC = 63/84⁽³ = 21/28⁽⁷ = 3/4

d) tg B = ctg C = AC/AB = 12√3 / 18√2 = 2√6 / 6 = √6/3

   tg C = ctg B = AB/AC = 18√2 / 12√3 = 3√6 / 6 = √6/2