Answer:
Given, \[\angle \]l = (2a + b)° and Z6 == (3a - b)° so, \[\angle \]l = \[\angle \]7 [alternate exterior angles] so, we have \[\angle \]7 = (2a +b) \[\therefore \] \[\angle \]6 + \[\angle \]7 = 180° [linear pair] \[{{(3a+b)}^{0}}+(2a+{{b}^{0}})\] \[3b-b+2a+b-180{}^\circ \] \[5a\text{ }=\text{ }180{}^\circ \] on dividing both sides by 5, we get \[a=\frac{{{180}^{\circ }}}{5}={{36}^{\circ }}\] \[\because \] \[\angle 1+\angle 2=180{}^\circ \] [linear pair] \[2a+b+\angle 2=180{}^\circ \] \[2\times 36{}^\circ +b+\angle 2=180{}^\circ \] \[b+\angle 2\text{ }=180{}^\circ -72{}^\circ \] \[b+\angle 2=108{}^\circ \] \[\therefore \] \[\angle 2=\left( {{108}^{\circ }}-b \right)\]
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