Given, \(p||q\) and cut by a transversal line.
\(\because\) \(125\degree+e=180\degree\) [Linear pair]
\(\therefore \;e=180\degree-125\degree=55\degree\) ……….(i)
Now \(e\) = \(=f=55\degree\) [Vertically opposite angles]
Also \(a=f=55\degree\) [Alternate interior angles]
\(a+b=180\degree\)[Linear pair]
\(\Rightarrow 55\degree+b=180\degree\) [From eq. (i)]
\(\Rightarrow b=180\degree-55\degree=125\degree\)
Now \(a=c=55\degree\) and \(b=d=125\degree\) [Vertically opposite angles]
Thus, \(a=55\degree,b=15\degree,c=55\degree,d=125\degree,e=55\degree\) and \(f=55\degree\)