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=25
b=30
C= 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]