Question

1. Se considera triunghiul dreptunghic ABC, ( a) Daca AB= 6 cm si BC= 10cm , calculati AC, sinC , cosC , tgB , ctgB
b) Daca AB=5 cm si AC=5√3cm, calculati BC , sin B , cosB , tgC, ctgC
c)Daca BC= 25 cm si sinC=2pe3 , calculati AB , AC, cosB , tgC
d) Daca AB=6cm si cosB=1pe2 calculati BC, AC, sinC, ctgB !!repede

Answer 1

a)  Calculam cu Teorema lui Pitagora latura AC.
AC²=BC²-AB²
AC²=10²-6²
AC²=100-36
AC²=64    <=> AC=√64=8 cm

sin C = AB/BC=6/10=3/5
cos C =AC/BC=8/10=4/5
tg B = AC/AB=8/6=4/3
cts B = AB/AC=6/8=3/4

b)Calculam tot cu Pitagora latura BC.
BC²=AC²+AB²
BC²=5²+(5√3)²
BC²=25+25·3  <=> BC²=25+75 <=> BC²=100  <=> BC=√100=10
sin B=AC/BC=15√3/10=√3/2
cos B=AB/BC=5/10=1/2
tg C=AB/AC=5/5√3=1/√3=√3/3
ctg C=AC/AB=5√3/5=√3

c)BC=25 si sin C=2/3
sin C = AB/BC
AB/BC=AB/25    <=>    AB/25=2/3  <=> AB=(25*2)3  <=>  AB=50/3
Calculam AC cu Pitagora.
AC²=BC²-AB²  <=> AC²=25²-(50/3)²  <=> AC²=625 -2500/9  (amplificam pe 625 cu 9 )  <=> AC²=(5625-2500)/9  <=> AC²=3125/9 <=> AC=√3125/9  <=> AC=(25√5)/3
cos B=AB/BC=50/3 : 25 =50/3 · 1/25=2/3
tg C= AB/AC=50/3 : (25√5)/3 =50/3 · 3/(25√5)=2/√5=(2√5)/5

d) AB=6 , cos B=1/2
cos B=AB/BC  <=> 6/BC=1/2 <=> BC=6·2=12
Calculam latura AC cu Pitagora.
AC²=BC²-AB²  <=> AC²=12²-6² <=> AC²=144-36 <=> AC²=108  <=> AC=√108 <=> AC=6√3
sin C = AB/BC=6/12=1/2
tg B=AB/AC=6/(6√3)=1/√3=√3/3


wow , mare a mai fost exercitiul acesta...
coroana?