Question 42529
I am showing the procedure to solve this type of problems. I am solving the first question. Do the rest yourself. That will nake you understand this things better.
For the first problem, A={{{130^o}}}, a=19, b=11.


The sine rule states:
{{{a/sin(A) = b/sin(B)}}}

 
{{{19/sin(130^o) = 11/sin(B)}}}
or {{{19*sin(B) = 11*sin(130^o)}}}
or {{{sin(B) = (11*sin(130^o))/19}}}
or {{{sin(B) = (11*0.766)/19}}}
or {{{sin(B) = 0.4435}}}
or {{{B = sin^-1(0.4435)}}}
or {{{B = 26.33^o}}}


Hence, {{{C = 180^o - A - B}}}
or {{{C = 180^o - 135^o - 26.33^o = 18.67^o}}}


Now, apply sine rule again
{{{a/sin(A) = c/sin(C)}}}
or {{{19/sin(135^o) = c/sin(18.67^o)}}}
or {{{c = (19*sin(18.67^o))/sin(135^o)}}}
or {{{c = (19*0.32)/0.766}}}
or {{{c = 7.94}}}


Now, we observe a > b + c [since 19 > 11 + 7.94].
But, for a triangle to exist, the sum of any two sides must be greater than the third.
So this triangle does not exist.
So there is no solution.



Solve the other problems in a similar way.
If the condition: "Sum of any two sides of a triangle is greater than the third" is not violated then the triangle exists and hence the solution is unique i.e. only one solution. Otherwise there is no solution.