Four points $B, A, E, L$ are on a straight line, as shown. $G$ is a point off the line so that $\angle B A G=120^{\circ}$ and $\angle G E L=80^{\circ}$. If the reflex angle at $G$ is $x^{\circ}$, then $x$ equals
(A) 340
(B) 200
(C) 300
(D) 240
(E) 310

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-076.jpg?height=382&width=482&top_left_y=796&top_left_x=1293)
Since the sum of the angles at any point on a line is $180^{\circ}$, then $\angle G A E=180^{\circ}-120^{\circ}=60^{\circ}$ and $\angle G E A=180^{\circ}-80^{\circ}=100^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-343.jpg?height=442&width=485&top_left_y=1324&top_left_x=885)

Since the sum of the angles in a triangle is $180^{\circ}$, $\angle A G E=180^{\circ}-\angle G A E-\angle G E A=180^{\circ}-60^{\circ}-100^{\circ}=20^{\circ}$.

Since $\angle A G E=20^{\circ}$, then the reflex angle at $G$ is $360^{\circ}-20^{\circ}=340^{\circ}$, so $x=340$.

ANSWER: (A)

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