Correct option is (B) \(30^{\circ}\)
We are given the triangle \(\triangle ABC \) with:
The Law of Cosines is given by the formula:
\(c^2 = a^2+b^2 - 2ab\ \cos C\)
We can rearrange it to solve for angle C:
\(\cos C = \frac{a^2+b^2-c^2}{2ab}\)
The Law of Cosines formula for angle B is:
\(b^2 = a^2+c^2-2ac\ \cos B\)
Substitute the known values:
\((5\sqrt3)^2 = a^2 + 5^2 - 2(a)(5)\cos 120^\circ\)
\(75 = a^2+25-2a(5) (-1/2)\)
\(75 = a^2+25+5a\)
\(a^2 + 5a-50 =0\)
Using the quadratic formula:
\(a= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For a = 1, b = 5, c = −50, we get:
\(a = \frac{-5 \pm\sqrt{5^2 -4(1)(-50)} }{2(1)}\)
\(a = \frac{-5 \pm \sqrt{25+200}}{2}\)
\(a = \frac{-5 \pm \sqrt{225}}{2}\)
\(a = \frac{-5 \pm \sqrt{15}}{2}\)
Thus, \(a = \frac{-15+15}{2} = \text 5\ \text{or}\ a=\frac{-5-15}{2} = -10.\)
Since side lengths are positive, a = 5 cm.
Now that we know a = 5, we can apply the Law of Cosines to find angle C:
\(c^2 = a^2 + b^2 - 2ab \cos C\)
\(5^2=5^2 +(5\sqrt3)^2 - 2(5)(5\sqrt3) \cos C\)
\(25=25+75−50\sqrt3\ \cos C\)
\(25=100-50\sqrt3\ \cos C\)
\(50\sqrt3\ \cos C =75\)
cos C \(=\frac{75}{5\sqrt3} = \frac{3}{2\sqrt3} = \frac{\sqrt3}{2}\)
Thus, \(\cos C = \frac{\sqrt3}{2},\) which corresponds to C = \(30^{\circ}\)
