Giải tam giác ABC trong các trường hợp sau: a) a = 17,4;

* Hướng dẫn giải

a) Tam giác ABC có:

$\widehat{A}+\widehat{B}+\widehat{C}={180}^o\Rightarrow \widehat{A}={180}^o-(\widehat{B}+\widehat{C})={180}^o-({44}^{o}30\text{'}+{64}^o)={71}^{o}30\text{'}$

Áp dụng định lí sin ta có:

 $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow \frac{17,4}{\sin {71}^{o}30\text{'}}=\frac{b}{\sin {44}^{o}30\text{'}}=\frac{c}{\sin {64}^o}$

Suy ra:

 $b=\frac{17,4.\sin {44}^{o}30\text{'}}{\sin {71}^{o}30\text{'}}\approx 12,9$; $c=\frac{17,4.\sin {64}^o}{\sin {71}^{o}30\text{'}}\approx 16,5$

Vậy tam giác ABC có: $\widehat{A}={71}^{o}30\text{'}$ ; $\widehat{B}={44}^{o}30\text{'};\widehat{C}={64}^o$ ; a = 17,4; b ≈ 12,9; c ≈ 16,5.

b) Áp dụng hệ quả của định lí côsin ta có:

cosA = $\frac{b^2+c^2-a^2}{2bc}=\frac{6^2+8^2-{10}^2}{\mathrm{2.6.8}}=\frac{0}{96}=0$  

⇒ $\widehat{A}={90}^o$ .

cosB = $\frac{a^2+c^2-b^2}{2ac}=\frac{{10}^2+8^2-6^2}{\mathrm{2.10.8}}=\frac{128}{160}=0,8$  

⇒ $\widehat{B}\approx {36}^{o}52\text{'}$ .

$\widehat{A}+\widehat{B}+\widehat{C}={180}^o\Rightarrow \widehat{C}={180}^o-(\widehat{A}+\widehat{B})={180}^o-({90}^o+{36}^{o}52\text{'})={53}^{o}8\text{'}$

Vậy tam giác ABC có: $\widehat{A}={90}^o$ ; $\widehat{B}={36}^{o}52\text{'}$ ; $\widehat{C}={53}^{o}8\text{'}$ ; a = 10; b = 6; c = 8.