Răspuns :
E
a√2 a√2
A 2a D
a a
B 2a C
in Δ ABE BE= a√5
mas < A = 90+45 =135
cu teorema sin BE / sin135 =AB/sin (AEB)
a√5 / √2/2 = a /sin (AEB) ; a √5 sin ( AEB) =a√2 /2
sin (AEB) =√10 /10
Δ BEC cu laturile 2a ,a√5 si a√5 cu teorema cos
(2a)² = ( a√5)² +(a√5)² - 2 ·a√5·a√5 ·cos(BEC)
4a²=5a²+5a² -2·5·a²·cos (BEC)
10a²·cos(BEC) = 6a² ; cos (BEC) = 3/5
tg ( BEC) = √1- cos²(BEC) / cos(BEC) = √1 -(3/5)² /3/5 =
rad.din numarator
=√16/25 /3/5 = 4/5 /3/5 = 4/3
a√2 a√2
A 2a D
a a
B 2a C
in Δ ABE BE= a√5
mas < A = 90+45 =135
cu teorema sin BE / sin135 =AB/sin (AEB)
a√5 / √2/2 = a /sin (AEB) ; a √5 sin ( AEB) =a√2 /2
sin (AEB) =√10 /10
Δ BEC cu laturile 2a ,a√5 si a√5 cu teorema cos
(2a)² = ( a√5)² +(a√5)² - 2 ·a√5·a√5 ·cos(BEC)
4a²=5a²+5a² -2·5·a²·cos (BEC)
10a²·cos(BEC) = 6a² ; cos (BEC) = 3/5
tg ( BEC) = √1- cos²(BEC) / cos(BEC) = √1 -(3/5)² /3/5 =
rad.din numarator
=√16/25 /3/5 = 4/5 /3/5 = 4/3