question_answer

If \[p||q,\]then find the value of \[\angle 1,\,\angle 2\] and \[\angle 1+\,\angle 2\]respectively.

A)  \[{{41}^{o}},{{90}^{o}},{{131}^{o}}\]        

B) \[{{42}^{o}},{{89}^{o}},{{131}^{o}}\]   

C)       \[{{90}^{o}},{{41}^{o}},{{131}^{o}}\]   

D)       \[{{49}^{o}},{{41}^{o}},{{90}^{o}}\]      

Correct Answer: A

Solution :

Since \[p||q\] and l is transversal \[\angle PAC=\angle ECA\]       (Alternate angles) \[\Rightarrow \] \[\angle ECA={{139}^{o}}\]                         \[\Rightarrow \] \[\angle ECB+\angle BCA={{139}^{o}}\]           \[\Rightarrow \]  \[{{49}^{o}}+\angle BCA={{139}^{o}}\] \[\Rightarrow \] \[\angle BCA={{139}^{o}}-{{49}^{o}}\]                   \[\Rightarrow \] \[\angle BCA={{90}^{o}}\] Now, \[\angle BCA+\angle ACG={{180}^{o}}\](linear pair) \[\Rightarrow \]  \[{{90}^{o}}\text{+}\angle 2={{180}^{o}}\] \[\Rightarrow \] \[\angle 2={{180}^{o}}-{{90}^{o}}\Rightarrow \angle 2={{90}^{o}}\] Also, \[\angle DCE+\angle ECB+\angle BCA={{180}^{o}}\] (Angles on a striaght line) \[\Rightarrow \]            \[\angle DCE+{{49}^{o}}+{{90}^{o}}={{180}^{o}}\] \[\Rightarrow \]            \[\angle DCE={{180}^{o}}-{{139}^{o}}={{41}^{o}}\] Now, \[p||q\]and l is transversal \[\therefore \] \[\angle BAC=\angle ECD\](Corresponding angles) \[\angle 1={{41}^{o}}\] Hence, \[\angle 1+\angle 2={{41}^{o}}+{{90}^{o}}={{131}^{o}}\]