question_answer

In the given figure, PQ || RS. If \[\angle \]l = (2a + b)° and \[\angle 6=\left( 3a-b \right){}^\circ \], then find the measure of \[\angle 2\] in terms of b.

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] \[2a+b+\angle 2=180{}^\circ \] \[2\times 36{}^\circ +b+\angle 2=180{}^\circ \] \[b+\angle 2\text{ }=180{}^\circ -72{}^\circ \] \[b+\angle 2=108{}^\circ \] \[\therefore \]      \[\angle 2=\left( {{108}^{\circ }}-b \right)\]