Madhya Pradesh Police Exam Model Paper 4

Given:
$\angle \mathit{C}\mathit{O}\mathit{B}=\left(3\mathit{x}-10\right)°$
$\angle \mathit{A}\mathit{O}\mathit{C}=\left(7\mathit{x}+40\right)°$
Calculation:
Sum of angles on a straight line is 180°
$\Rightarrow \angle \mathit{A}\mathit{O}\mathit{C}+\angle \mathit{C}\mathit{O}\mathit{B}=180°$
$\Rightarrow \left(7\mathit{x}+40\right)°+\left(3\mathit{x}-10\right)°=180°$
$\Rightarrow 10\mathit{x}+30°=180°$
$\Rightarrow 10\mathit{x}=180°-30°$
$\Rightarrow \mathit{x}=15°$
$\Rightarrow \angle \mathit{C}\mathit{O}\mathit{B}=\left(3\mathit{x}-10\right)°$
$\Rightarrow \left(3\times 15-10\right)°=35°$
∴ ∠COB = 35°
The correct option is 2 i.e. 35°.
The Sum of angles on a straight line is 180°.