A) \[2{{x}^{{}^\circ }}+{{y}^{{}^\circ }}={{a}^{{}^\circ }}+{{b}^{{}^\circ }}\]
B) \[{{x}^{{}^\circ }}+\frac{1}{2}{{y}^{{}^\circ }}=\frac{{{a}^{{}^\circ }}+{{b}^{{}^\circ }}}{2}\]
C) \[{{x}^{{}^\circ }}+{{y}^{{}^\circ }}={{a}^{{}^\circ }}+{{b}^{{}^\circ }}\]
D) \[{{x}^{{}^\circ }}+{{a}^{{}^\circ }}={{y}^{{}^\circ }}+{{b}^{{}^\circ }}\]
Correct Answer: C
Solution :
(c): As \[{{a}^{{}^\circ }}+{{b}^{{}^\circ }}={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle A={{180}^{{}^\circ }}-b\] Also, \[\angle C+{{a}^{{}^\circ }}={{180}^{{}^\circ }}\](linear pair) \[\Rightarrow \]\[\angle C={{180}^{{}^\circ }}-{{a}^{{}^\circ }}\] But \[\angle A+\angle B+\angle C+\angle D~={{360}^{{}^\circ }}\] \[\Rightarrow \]\[\left( {{180}^{{}^\circ }}-{{b}^{{}^\circ }} \right)+x+\left( {{180}^{{}^\circ }}-{{a}^{{}^\circ }} \right)+{{y}^{{}^\circ }}={{360}^{{}^\circ }}\] \[\Rightarrow \]\[{{x}^{{}^\circ }}+{{y}^{{}^\circ }}={{a}^{{}^\circ }}+{{b}^{{}^\circ }}\]
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