I will calculate this problem in two ways
alternative method 1
we know that
The inscribed angle measures half of the arc it comprises.
∠TNE=(1/2)*arc TG------> (1/2)*70°----> 35°
the triangle TEN is a right triangle
∠TNE+∠GET=90°-----> by complementary angles
∠GET=90-35---> 55°
the answer is
∠GET=55°
alternative method 2
we know that
The measure of the external angle is the semi difference of the arcs that it covers.
∠GET=(1/2)*[arc TN-arc TG]---> (1/2)*[180-70]---> (1/2)*110---> 55°
the answer is
∠GET=55°