Answer: Third option is correct.
Step-by-step explanation:
Since we have given that
ABC is triangle with its dimensions:
AB = 11
AC = 13
∠A = 108°
BC = a
We need to find the length of 'a'.
So, we can use "Law of cosines" as we have given two sides and one angle.
So, it becomes,
[tex]\cos A=\dfrac{b^2+c^2-a^2}{2bc}\\\\\cos 108^\circ=\dfrac{11^2+13^2-a^2}{2\times 13\times 11}\\\\-0.3=\dfrac{121+169-a^2}{286}\\\\-0.3\times 286=290-a^2\\\\-85.8=290-a^2\\\\-85.8-290=-a^2\\\\375.8=a^2\\\\a=\sqrt{375.8}\\\\a=19.38[/tex]
Hence, Third option is correct.