Question

Question 35

In the given figure PQ|| RS. If $\angle 1=(2a+b{)}^{\circ }and\angle 6=(3a-b{)}^{\circ }$, then the measure of $\angle 2$ in terms of b is

(a) $(2+b{)}^{\circ }$
(b) $3-b{)}^{\circ }$
(c) $(108-b{)}^{\circ }$
(d) $(180-b{)}^{\circ }$

Answer 1

From the given figure,~$\angle 1=\angle 5\left[correspondingangles\right]\Rightarrow \angle 5=(2a+b{)}^{\circ }Also,\angle 5+\angle 6={180}^{\circ }\Rightarrow (2a+b{)}^{\circ }+(3a-b{)}^{\circ }={180}^{\circ }\Rightarrow (2a+3a)+(b-b)={180}^{\circ }\Rightarrow 5a={180}^{\circ }\Rightarrow a={36}^{\circ }Now,\angle 1+\angle 2={180}^{\circ }\Rightarrow \angle 2={180}^{\circ }-\angle 1\Rightarrow \angle 2={180}^{\circ }-(2a+b{)}^{\circ }\Rightarrow \angle 2={180}^{\circ }-2a-b\Rightarrow \angle 2={180}^{\circ }-(2\times {36}^{\circ })-b\Rightarrow \angle 2={180}^{\circ }-{72}^{\circ }-b\Rightarrow \angle 2=(108-b{)}^{\circ }$.