Solution.
c = 4 m ; c=4m; c = 4 m ;
a = 2 m ; a=2m; a = 2 m ;
b = 3 m ; b=3m; b = 3 m ;
b 2 = a 2 + c 2 − 2 a c c o s β ⟹ c o s β = a 2 + c 2 − b 2 2 a c ; b^2=a^2+c^2-2accos\beta\implies cos\beta=\dfrac{a^2+c^2-b^2}{2ac}; b 2 = a 2 + c 2 − 2 a ccos β ⟹ cos β = 2 a c a 2 + c 2 − b 2 ;
c o s β = 4 + 16 − 9 2 ⋅ 2 ⋅ 4 = 11 16 ; cos\beta=\dfrac{4+16-9}{2\sdot2\sdot4}=\dfrac{11}{16}; cos β = 2 ⋅ 2 ⋅ 4 4 + 16 − 9 = 16 11 ;
s i n β = 3 15 16 ; sin\beta=\dfrac{3\sqrt{15}}{16}; s in β = 16 3 15 ;
a 2 = b 2 + c 2 − 2 a c c o s α ⟹ c o s α = b 2 + c 2 − a 2 2 b c ; a^2=b^2+c^2-2accos\alpha\implies cos\alpha=\dfrac{b^2+c^2-a^2}{2bc}; a 2 = b 2 + c 2 − 2 a ccos α ⟹ cos α = 2 b c b 2 + c 2 − a 2 ;
c o s α = 21 24 ; cos\alpha=\dfrac{21}{24}; cos α = 24 21 ;
s i n α = 3 15 24 ; sin\alpha=\dfrac{3\sqrt{15}}{24}; s in α = 24 3 15 ;
T 1 c o s β = T 2 c o s α ; T_1cos\beta=T_2cos\alpha; T 1 cos β = T 2 cos α ;
T 1 s i n β + T 2 s i n α = m g ; T_1sin\beta+T_2sin\alpha=mg; T 1 s in β + T 2 s in α = m g ;
T 1 = 42 33 T 2 ; T_1=\dfrac{42}{33}T_2; T 1 = 33 42 T 2 ;
42 33 T 2 s i n β + T 2 s i n α = m g ; \dfrac{42}{33}T_2sin\beta+T_2sin\alpha=mg; 33 42 T 2 s in β + T 2 s in α = m g ;
T 2 = 70 N ; T_2=70N; T 2 = 70 N ;
T 1 = 42 33 ⋅ 70 = 90 N ; T_1=\dfrac{42}{33}\sdot70=90N; T 1 = 33 42 ⋅ 70 = 90 N ;
Answer: T 1 = 90 N ; T_1=90N; T 1 = 90 N ;
T 2 = 70 N . T_2=70N. T 2 = 70 N .
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