In the diagram, IG is parallel to JE, $J\overset{\wedge }{\mathop{E}}\,F={{120}^{\circ }}$ and $F\overset{\wedge }{\mathop{H}}\,G={{130}^{\circ }}$. Find the angle marked t

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$\begin{align} & \text{Produce }\left| JE \right|\text{ to meet }\left| FH \right|\text{ to point }A \\ & F\overset{\wedge }{\mathop{E}}\,A={{180}^{\circ }}-J\overset{\wedge }{\mathop{E}}\,F={{180}^{\circ }}-{{120}^{\circ }}={{60}^{\circ }} \\ & Also \\ & F\overset{\wedge }{\mathop{E}}\,B=A\overset{\wedge }{\mathop{H}}\,B\text{ }\!\!\{\!\!\text{ corresponding angles }\!\!\}\!\!\text{ } \\ & F\overset{\wedge }{\mathop{A}}\,B={{130}^{\circ }} \\ & F\overset{\wedge }{\mathop{E}}\,A+E\overset{\wedge }{\mathop{F}}\,A={{130}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of the two opposite angles of a triangle }\!\!\}\!\!\text{ } \\ & {{60}^{\circ }}+t={{130}^{\circ }} \\ & t={{70}^{\circ }} \\\end{align}$
