Madhya Pradesh Police Exam Model Paper 10

Given,
$\left(\angle \mathit{B}\mathit{–}\angle \mathit{A}\right)=40°$ ----(1)
$\left(\angle \mathit{C}\mathit{–}\angle \mathit{D}\right)=20°$ ----(2)
Adding (1) and (2), we get,
$\Rightarrow \angle \mathit{B}+\angle \mathit{C}=60°+\angle \mathit{A}+\angle \mathit{D}$
$\Rightarrow \angle \mathit{A}+\angle \mathit{D}=\angle \mathit{B}+\angle \mathit{C}\mathit{–}60°$ ----(3)
Now,
Sum of all angles of a trapezium = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
Substituting from (3),
$\Rightarrow 2\left(\angle \mathit{B}+\angle \mathit{C}\right)=360°+60°$
$\therefore \left(\angle \mathit{B}+\angle \mathit{C}\right)=420°/2=210°$