Question

am nevoie de ajutor:))
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Answer 1

Heey!

4.

BC²-AB²-AC²-AC²=15²-9²

AC²=225-81

AC²=144

AC=√144

AC=12cm

sinB=co/ip⇒sinB=AC/BC⇒sinB-12/5⇒sinB=4/5

cosB=ca/ip⇒cosB=AB/

BC⇒cosB=9/15 cosB=3/5

tgC=co/ca⇒tgC=AB/AC⇒tgC=9/12⇒tgC=3/4

ctgC=ca/co⇒ctgC=AC/

AB⇒ctgC=12/9⇒ctgC=4/3

5.

a)

T. Pitagora:

AC=rad(BC^ ^ 2 -AB^ ^ 2)=rad(400-144)=rad 256 = 16 cm

sin C=AB/BC=12/20=3/5

cos C=AC/BC=16/20=4/5

tg B=AC/AB=16/12=4/3

ctg B=AB/AC=12/16=3/4 sau ctg B = 1 / t * g B

=1/4/3=3/4 .

b)

T. Pitagora:

BC = rad(AB^2 + AC^2) = rad (25 +75) = rad 100 = 10 cm.

sin B = cat opusa/ipotenuza = AC/BC= 5/10 = 1/2 ⇒mas B = 30° → mas C = 60°, pt ca unghiurile ascutite intr-un triunghi dreptunghic sunt complementare, adica suma masurilor lor este 90°.

cos B = cos 30º = rad3/ 2

tg C = tg 60° = rad3

ctg C = ctg 60° = rad3 / 3.

c) in mod ABSOLUT ANALOG, avem

AB = rad(BC^2 - AC^2) = rad(24-16) = rad 8 = 2rad2

sin B = AC/BC= 4 / 2rad6 = 2 / rad6 = 2rad6 / 6 = rad6 /3 ≈ 0.816

sin C = AB/BC = 2rad2/ 2rad6 = rad2/ rad(2x3) = rad2/ (rad2 x rad3) = 1/rad3= rad3/

Spar ca te am ajutat:)

Answer 2

ex 4

BC² = AB² + AC²

BC² = 225+400

BC= V625

BC = 25

sinB = cateta opusa/ ipotenuza = AC/BC = 20/25 = 4/5

cosB = cateta alaturata/ipotenuza = AB/BC = 15/25=3/5

tgC = cateta opusa/cateta alaturata = AB/C = 15/20 = 3/4

ctgC = cateta alaturata/cateta opusa = AC/AB = 20/15 = 4/3

ex 5

sinC = AB/BC

2/3 = 2V3/BC => 2BC = 6V3 => BC = 6V3/2 = 3V3

AC² = BC² - AB²

AC² = (3V3)² - (2V3)²

AC² = 27 - 12

AC = V15

sinB = AC/BC = V15/3V3 = V5/3

ctgC = AC/AB = V15/2V3 = V5/2