Answer:
Step-by-step explanation:
apply sine formula
[tex]\frac{a}{sin ~A} =\frac{b}{sin~B} \\\frac{25}{sin~58} =\frac{28}{sin ~B} \\sin~B=\frac{28}{25} \times sin~58\\B=sin^{-1} (\frac{28}{25} \times sin ~58)=71.77 \approx 72 ^\circ[/tex]
so third angle=180-(58+72)=180-130=50°
∠C=50°
[tex]cos ~C=\frac{a^2+b^2-c^2}{2ab} \\or ~2abcos~C=a^2+b^2-c^2\\2*25*28*cos ~50=25^2+28^2-c^2\\c^2=625+784-1400 *cos~50\\c^2=1409-899.90\\c^2=509.1\\c=\sqrt{509.1} \approx 22.56 \approx 22.6[/tex]
so one triangle is formed.
