Xét ΔABC có \(\hat{A} + \hat{B} + \hat{C} = 18 0^{0}\)

=>\(\hat{A} = 18 0^{0} - 6 5^{0} - 4 5^{0} = 7 0^{0}\)

Xét ΔABC có \(\frac{A B}{s i n C} = \frac{B C}{s i n A} = \frac{A C}{s i n B}\)

=>\(\frac{B C}{s i n 70} = \frac{A C}{s i n 65} = \frac{2.8}{s i n 45}\)

=>\({.BC=2.8\cdot\frac{s i n 70}{s i n 45}\simeq3,72\left(\right.cm\left.\right)\\AC=2.8\cdot\frac{s i n 65}{s i n 45}\simeq3,59\left(\right.cm\left.\right)}\)

Xét ΔABC có \(\hat{A} + \hat{B} + \hat{C} = 18 0^{0}\)

=>\(\hat{A} = 18 0^{0} - 6 5^{0} - 4 5^{0} = 7 0^{0}\)

Xét ΔABC có \(\frac{A B}{s i n C} = \frac{B C}{s i n A} = \frac{A C}{s i n B}\)

=>\(\frac{B C}{s i n 70} = \frac{A C}{s i n 65} = \frac{2.8}{s i n 45}\)

=>\({.BC=2.8\cdot\frac{s i n 70}{s i n 45}\simeq3,72\left(\right.cm\left.\right)\\AC=2.8\cdot\frac{s i n 65}{s i n 45}\simeq3,59\left(\right.cm\left.\right)}\)