RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

(a) (a) quadrilateral ABCD, AP and BP are bisectors of ∠(A) and ∠(B), respectively.

$\therefore \angle \mathit{A}\mathit{P}\mathit{B}=180°\mathit{–}\left[\frac{1}{2}\angle \mathit{A}+\frac{1}{2}\angle \mathit{B}\right]$
We know that sum of all angles of a quadrilateral= 360°
⇒ ∠(A) + ∠(B) + ∠(C) + ∠D = 360°
∴$\frac{1}{2}\angle \mathit{A}+\frac{1}{2}\angle \mathit{B}+\frac{1}{2}\angle \mathit{C}+\frac{1}{2}\angle \mathit{D}=\frac{360°}{2}$
⇒$\Rightarrow \frac{1}{2}\angle \mathit{C}+\frac{1}{2}\angle \mathit{D}=180°\mathit{–}\left[\frac{1}{2}\angle \mathit{A}+\frac{1}{2}\mathit{B}\right]$
$\Rightarrow \frac{1}{2}\left(\angle \mathit{C}+\angle \mathit{D}\right)=\angle \mathit{A}\mathit{P}\mathit{B}$   [from eq. (i)]
$\Rightarrow \angle \left(\mathit{c}\right)+\angle \mathit{D}=2\angle \mathit{A}\mathit{P}\mathit{B}$