Question

va rogg putin ajutor.......cele incercuite adica 6 si 9

Answer 1

pr 6 a)daca ∡B=32 si ∡C=63 =>∡A=180-95=85 => ∡BAE=∡CAE=42°30'. FIE ΔCDA-∡C=63;∡D=90 =>∡CAD=180-153=27° =>∡DAE=∡CAE-∡CAD=42°30'-27°=15°30'. b). daca ∡B= si ∡C¸=23°=>∡A=180-155=25°. ∡BAE=∡CAE=12°30'. FIE ΔBAE-∡BAE=12°30' SI ∡ABE=132°=>∡AEB=180-144°30'=35°30'. FIE ΔDAE-DREPTUNGHIC=>∡DAE=180-125°30'=54°30'. 
PR 9) DUCEM INALTIMILE AM , M∈BC, BN, N∈AC SI CP , P∈AB IAR INTERSECTIA LOR ESTE NOTATA CU H. FIE ΔMAC-DREPTUNGHIC IN M. ∡C=50°=> ∡MAC=40°. FIE ΔAHN-∡HAN=40°, ∡ANH=90°=> ∡AHN=180-130=50°. ∡PHN=∡BHC=125°, ∡BHC=∡PHN=125 ( UNGH OPUSE LA VARH).=> ∡AHP=125-50=75°. FIE ΔAPH-DREPTUNGHIC IN P => ∡HAP=180-∡AHP-∡APH=180-165°=15°. => ∡BAC=∡HAP+MAC=55° => UNGH ABC=180-55-50=75°

Answer 2

6)
 atentie, desenu e astfel incat E este intre B si D
∡B=32°
∡C=63°
∡A=180-32-63=85°
∡A/2=42,5°
in tr dreptunghic BAD:
∡A/2 + ∡DAE=90-∡B
∡DAE=90-32-42,5
∡DAE=15,5°
b)
desenu e un triunghi obtuzunghic in B
∡B=132°
∡C=23°
∡A=180-132-23=25°
∡A/2=12,5°
∡DAE+∡A/2=90-∡C
∡DAE=90-∡C-∡A/2
∡DAE=90-23-12,5
∡DAE=54,5°

9)
notam AD⊥BC,BE⊥AC,CF⊥AB
in tr ADC ⇒ ∡DAC=90-50=40° ⇒∡ AHE=90-∡DAC = 50°
∡AHE=∡BHD=50 opuse la varf
∡DHC=125-50=75
∡FHA=∡DHC=75 opuse la varf

∡BAD=90-75=15
∡A=∡BAD+∡DAC=15+40=55°
∡B=90-BAD=90-15
∡B=75°

ca un comentariu:
∡A il putem gasi mai usor stiind ca patrulateru AFHE este inscriptibil intr-un cerc si ca atare ∡A+∡FHE=180 ⇒ ∡A=180-125=55°
nu garantez ca nu s-a strecurat vreo erore pt. ca m-am grabit. essential e sa faci figura corect si sa incerci sa intelegi logica expusa