In figure, lines XY and MN intersect at O. If \(\angle{POY}\) = \(90^\circ\) and a :b = 2:3. Find c.

It is given that, \(\angle{POY}\) = \(90^\circ\)

Now, by Linear pair axiom,
\(\angle{POY}\) + \(\angle{POX}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{POX}\) = \(180^\circ\) - \(\angle{POY}\)
\(\Rightarrow \) \(\angle{POX}\) = \(90^\circ\)
So, we get, a + b = \(90^\circ\) ....(i)
It is also given that,
a :b = 2:3

Let a = 2k and b = 3k,
From eq. (i),
2k + 3k = \(90^\circ\)
\(\Rightarrow \) 5k = \(90^\circ\)
\(\Rightarrow \) k = \(18^\circ\)

So, a = 2 × \(18^\circ\) = \(36^\circ\)
and b = 3 × \(18^\circ\) = \(54^\circ\)
Again by Linear pair axiom,
\(\angle{MOX}\) + \(\angle{XON}\) = \(180^\circ\)
\(\Rightarrow \) b + c = \(180^\circ\)
\(\Rightarrow \) c = \(180^\circ\) - \(54^\circ\)
\(\Rightarrow \) c = \(126^\circ\)