There are 6 unknowns. We are given four equations, three explicitly and one implicitly.
\(\displaystyle A = 71.\)
\(\displaystyle b = 26.\)
\(\displaystyle a + c = 45 \implies c = 45 - a.\)
\(\displaystyle A + B + C = 180 \implies C = 180 - A - B = 109 - B.\)
Now consider the law of sines. Let sin(A) = sin(71) = p, which is a number not an unknown. Let q = sin(109) and r = cos(109). They are also numbers. Let s = sin(B), which is an unknown.
\(\displaystyle \dfrac{sin(A)}{a} = \dfrac{sin(B)}{b} \implies \dfrac{p}{a} = \dfrac{s}{26} \implies a = \dfrac{26p}{s}.\)
\(\displaystyle \dfrac{sin(A)}{a} = \dfrac{sin(C)}{c} = \dfrac{sin(109 - B)}{45 - a} \implies \dfrac{p}{a}= \dfrac{sin(109 - B)}{45 - a} \implies\)
\(\displaystyle a * sin(109 - B) = 45p - 45a \implies a\{sin(109 - B) + 45\} = 45p \implies a = \dfrac{45p}{sin(109 - B) + 45}.\)
Now we can use some identities.
\(\displaystyle sin(109 - B) = q * cos(B) - rs = q * \sqrt{1 - s^2} - rs.\)
\(\displaystyle \therefore \dfrac{s}{26} = a = \dfrac{45p}{q\sqrt{1 - s^2} - rs + 45} \implies 1170p = qs\sqrt{1 - s^2} - rs^2 + 45s \implies\)
\(\displaystyle 1170p + rs^2 - 45s = qs\sqrt{1 - s^2} \implies (1170p + rs^2 - 45s)^2 = q^2s^2 - q^2s^4.\)
If that quartic has a solution, then such a triangle exists. Of course there may be as many as four such triangles.