Answer: Third option is correct.
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,
![\cos A=(b^2+c^2-a^2)/(2bc)\\\\\cos 108^\circ=(11^2+13^2-a^2)/(2* 13* 11)\\\\-0.3=(121+169-a^2)/(286)\\\\-0.3* 286=290-a^2\\\\-85.8=290-a^2\\\\-85.8-290=-a^2\\\\375.8=a^2\\\\a=√(375.8)\\\\a=19.38](https://img.qammunity.org/2020/formulas/mathematics/high-school/rijt83716f9hinmwqbw4hn4xun0tbbccf8.png)
Hence, Third option is correct.