Cosine rule: y^2=x^2+z^2-2xzcosY: 6.9^2=x^2+9.7^2-19.4xcos42.6; cos42.6=0.736097 approx.
So we have a quadratic in x: x^2-14.28x+46.48=0.
So x=(14.28±√(203.93-185.92))/2=7.14±2.12 approx. We get two values for x: 9.26m and 5.02m; so there are two possible triangles fitting the given criteria.