Solve a triangle with a=25, b=30, and C= 160°(picture provided)

Answer:Option d.Step-by-step explanation:For this problem we have 2 sides of a triangle (a and b) and the angle between them C = 160 °.We have a triangle of type SAS.We have:a=25b=30C= 160°Then we use the law of cosine.[tex]c = \sqrt{a^2 +b^2 - 2abcos(C)[/tex]Now we substitute the values in the formula to find c[tex]c = \sqrt{25^2 +30^2 - 2(25)(30)cos(160\°)}\\\\c = 54.2[/tex]Now we use the cosine theorem to find B. (You can also use the sine)[tex]b = \sqrt{a^2 +c^2 - 2accos(B)}\\\\b^ 2 = a^2 +c^2 - 2accos(B)\\\\b^ 2 -a^2 -c^2 =- 2accos(B)\\\\\frac{a^2 +c^2 -b^2}{2ac} =cos(B)\\\\B = arcos(\frac{a^2 +c^2 -b^2}{2ac})\\\\B = arcos(\frac{25^2 +54.2^2 -30^2}{2(25)(54.2)})\\\\B = 10.9\°[/tex]Finally:[tex]A=180\° - B- C\\\\A = 180\° - 10.9\° - 160\°\\\\A = 9.1\°[/tex]