Precalculus, Chapter 6, 6.1, Section 6.1, Problem 31

The given in the triangle are A=120^o, a=25 and b=25.
To solve using Sine Law, let's use the formula:
(sinA)/a=sinB/b
Plug-in the values of the sides a and b. Also, plug-in the value of angle A.
sin(120^o)/25=(sinB)/25
Then, simplify the equation. To simplify, cancel the denominators.
sin(120^o)/25*25=(sinB)/25*25
sin (120^o)=sinB
120^o=B
Now two angles of the triangle are known, which are A=120^o and B=120^o .
Take note that the sum of three angles of the triangle is 180^o .
A+B+C= 180^o
However, the sum of these two angles A and B is:
A+B=120^o +120^o=240^o
which is larger than 180^o .
Therefore, it is not possible to form a triangle with the given measures A=120^o , a=25 and b=25 .