Question

Answer 1

Lời giải:
a. Theo định lý cosin: 

$BC^2=AB^2+AC^2-2AB.AC\cos A=7^2+5^2-2.7.5\cos 60^0=39$ 

$\Rightarrow BC=\sqrt{39}$ 

b.

Theo định lý cosin:

$AC^2=AB^2+BC^2-2AB.BC\cos B$

$\Rightarrow cos B=\frac{AB^2+BC^2-AC^2}{2AB.BC}=\frac{5^2+39-7^2}{2.5\sqrt{39}}=\frac{\sqrt{39}}{26}$

$\Rightarrow \widehat{B}=76^0$

$\Rightarrow \widehat{C}=180^0-60^0-76^0=44^0$

c.

$S_{ABC}=\frac{1}{2}.AB.AC\sin A=\frac{1}{2}.7.5.\sin 60^0=\frac{35\sqrt{3}}{4}$