Para ambos problemas usaremos: AB=c ; AC=b ; BC=a y b^2=a^2+c^2
Ademas: Perimetro = a + b + c
1) CosA=0,2 = 1/5 = c/b ----> c=k y b=5k ; ademas BC=a=6(6)^(1/2)
Pitagoras: (5k)^2=[6(6)^(1/2)]^2+k^2
25(k)^2 = 36(6) + k^2
24k^2 = 216
------> k^2=9
-----> k = 3
P= a + b + c = 6(6)^(1/2) + 5(k) + k
=6(RaizCuadrada(6)) + 6(k)
=6[RaizCuadrada(6) + k)
= 6 [RaizCuadrada(6) + 3]
2) BC=a=32cm y senC=0,6=3/5=c/b ----> c=3k y b=5k
Pitagoras: (5k)^2 = (32)^2 + (3k)^2
25k^2 = 1024 + 9k^2
16k^2 = 1024
k^2 = 64
k = 8
Perimetro: P= a + b + c = 32 + 5k + 3k
= 32 + 8k
= 32 + 8(8)
= 32 + 64
= 96cm
