Question

Ex : 8,15,14,13,16,17,18,19,20,21,22 VA ROG COROANA

Answer 1

8. Teorema lui Pitagora pentru aflare ipotenuza

BC² = AB² + AC²=12² + (12√3)²=144 + 432=576

BC=√576= 24 ipotenuza

cos B = AB/BC = 12/24 = 1/2 = 60°

cos C= AC/CB = 12√3/24 =√3/2 = 30°

m (∡C)=30° si m (∡B)=60°

13.

Teorema unghiului de 30 grade aplicabila in triunghiul dreptunghic

MP = 2 MN = 2×8 =16 cm

Teorema lui Pitagora

NP² = MP² - MN²

NP² = 256 - 64

NP =√192

NP = 8√3

Aria = ( 8 × 8√3)/2 = 32√3 cm (semiprodusul catetelor)

Perimetrul = MN + MP + NP = 8 + 16 + 8√3 = 24 +8√3

14.

In ∆ ABC dr⇒Teorema inaltimii

AD² = BD × DC

DC = 12²:9⇒ DC= 144:9⇒DC = 16

BC = DC+ BD = 9 + 16 = 25

BC = 25 cm

In ∆ ABC dr⇒ Teorema catetei : AB²= BD×BC

AB² = 9×25⇒ AB² =225⇒ AB=√225

AB = 15 cm

In ∆ ABC dr ⇒ Teorema lui Pitagora

AB² + AC² = BC²

15² + AC² = 25²⇒ AC² = 625 - 225⇒ AC=√400

AC = 20 cm

deci laturile triunghiului sunt : AB = 15 cm; AC = 20 cm si BC = 25 cm

sin ∡B = AC/BC⇒ sin ∡B = 20/25 = 4/5;

sin ∡ B =4/5

cos ∡ B = AB/BC ; cos ∡B = 15/25 = 3/5

cos ∡ B = 3/5

tg ∡C = AB/AC = 15 /20

tg ∡C = 3/4

ctg ∡C= AC/AB = 20/15

ctg ∡C = 4/3

15.

Aplicam T Pitagora in Δ ADB

BD²=AC² - AD²=225-144

BD=√81 = 9 cm; BD = 9 cm

aPLICAM TEOREMA INALTIMII

AD²=BD×DC

144=9×DC⇒DC=144:9⇒DC = 16 cm

BC = BD + DC = 9 + 16 = 25 cm

DECI BC = 25 cm

AC²= BC² - AB²

AC² = 625 - 225

AC = 40 cm

sin (∡ C) =AB/BC = 15/25=3/5= 0,6

cos (∡C) = 40/25 = 8/5 = 1,6

tg ∡B = AC/AB = 40/15=8/3 tg ∡B = 8/3=2,(6)

ctg ∡C= AB/AC = 15/40=0,375 ctg ∡B = 0,375