Answer:
In the given figure, ZAOB and ZCOB are the angles of linear pair. So, \[\angle AOB+\angle BOC=180{}^\circ \] \[\left( 3x+{{10}^{^{0}}} \right)+\left( 2x-{{30}^{^{0}}} \right)={{180}^{^{0}}}\] \[3x+10{}^\circ +2x-30{}^\circ =180{}^\circ \] \[5x-20{}^\circ =180{}^\circ \] \[5x=180{}^\circ +20{}^\circ \] \[5x=200{}^\circ \] \[x=\frac{200{}^\circ }{5}\] Thus, \[x\text{ }=\text{ }40{}^\circ \] Now, \[\angle AOB=3x+10{}^\circ \] \[=\text{ }3\left( 40{}^\circ \right)+10{}^\circ \] \[=\text{ }120{}^\circ +10{}^\circ \]
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