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a)
$\widehat{{{A}_{1}}}+\widehat{{{A}_{3}}}=120{}^\circ $
$\to \widehat{{{A}_{1}}}+\widehat{{{A}_{1}}}=120{}^\circ $
$\to 2\widehat{{{A}_{1}}}=120{}^\circ $
$\to \widehat{{{A}_{1}}}=120{}^\circ :2$
$\to \widehat{{{A}_{1}}}=60{}^\circ =\widehat{{{A}_{3}}}$
$\to \widehat{{{A}_{2}}}=\widehat{{{A}_{4}}}=180{}^\circ -60{}^\circ =120{}^\circ $
b)
$\widehat{{{A}_{2}}}-\widehat{{{A}_{1}}}=30{}^\circ $
$\to 180{}^\circ -\widehat{{{A}_{1}}}-\widehat{{{A}_{1}}}=30{}^\circ $
$\to 2\widehat{{{A}_{1}}}=150{}^\circ $
$\to \widehat{{{A}_{1}}}=75{}^\circ =\widehat{{{A}_{3}}}$
$\to \widehat{{{A}_{2}}}=\widehat{{{A}_{4}}}=180{}^\circ -75{}^\circ =105{}^\circ $
c)
$3\widehat{{{A}_{1}}}=7\widehat{{{A}_{4}}}$
$\to 3\widehat{{{A}_{1}}}=7\left( 180{}^\circ -\widehat{{{A}_{1}}} \right)$
$\to 3\widehat{{{A}_{1}}}=1260{}^\circ -7\widehat{{{A}_{1}}}$
$\to 10\widehat{{{A}_{1}}}=1260{}^\circ $
$\to \widehat{{{A}_{1}}}=126{}^\circ =\widehat{{{A}_{3}}}$
$\to \widehat{{{A}_{2}}}=\widehat{{{A}_{4}}}=180{}^\circ -126{}^\circ =54{}^\circ $
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