Răspunsuri

2014-02-06T22:39:47+02:00
A) sin A=5/13=BC/AC
AC=13
13^2=5^2+AB^2
AB^2= 169-25=144 
AB=12

b) cosK=1/2
rezulta ca MK= 14 iar NM^2= 14^2- 49=196-49=147 iar NM=v147 pentru varianta c) avem 0,6=FG/3 iar FG=1,8
2014-02-06T22:47:09+02:00
Sin deA = cateta opusa/ipotenuza=BC/AC -> AC=13 -> AB=12
cos deK=1/2-> masura unghiului NKM=30 co deK=NK/MK-> MK=14 ->MN=7v3
tg deE=0,6 =GF/EF-> GF=9/5 ->GE=3v34
ctg deR=2/3(dupa transformare)=QR/QP-> QR=8->RP=4v13

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