<span>I will calculate this problem in two ways </span> alternative method 1 we know that <span>The inscribed angle measures half of the arc it comprises. </span>∠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 <span>The measure of the external angle is the semi difference of the arcs that it covers. </span>∠GET=(1/2)*[arc TN-arc TG]---> (1/2)*[180-70]---> (1/2)*110---> 55°
<span>Line ET is tangent to circle A at T </span> ∴ ET ⊥ TN ∴ ∠ ETN = 90 <span>the measure of Arc TG = 70 </span> ∠ TNG = half <span>the measure of Arc TG = 0.5 *70 = 35 </span> <span>Δ ETN ⇒ The sum of all angles = 180 </span> <span>∴∠ GET = ∠ NET = 180 - (90 + 35 ) = 55 </span>
∴ The correct answer is option <span>A. 55°</span>
