`A = 120^@, a = b = 25` Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

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...

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` .