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$Áp$ $dụng$ $định$ $lí$ $sin$ :

$\frac{a}{sinA}$=$\frac{b}{sinB}$=$\frac{c}{sinC}$ = $2R$
$Đặt$ $AB=c$ , $AC=b$ , $BC=a$
$Ta$ $có$ $a=152$ ; `\hat{A}`= $180^{o}$ $-$ ( $79^{o}$  + $61^{o}$ ) $=$ $40^{o}$ 

`\text{Áp dụng định lí sin, ta có}` :

$\frac{a}{sinA}$=$\frac{b}{sinB}$=$\frac{c}{sinC}$ = $2R$

Suy ra

$AC=b=$ $\frac{a.sinB}{sinA}$ $=$ $\frac{152.sin79^{o}}{40^{o}}$ ≈ $232,13$
$AB=c=$ $\frac{a.sinC}{sinA}$ $=$ $\frac{152.sin61^{o}}{40^{o}}$ ≈ $206,82$
$R=$ $\frac{a}{2sinA}$ $=$ $\frac{152}{2sin40^{o}}$ ≈ $118,235$