MCQ
In Fig , if $A B, C D$ and $E F$ are three lines concurrent at $O$, then $y=$
- A$10^{\circ}$
- B$30^{\circ}$
- C$20^{\circ}$
- D$15^{\circ}$
✓
Answer
C. $20^{\circ}$
We find that $\angle A O E$ and $B O F$ are vertically opposite angles.$
\angle A O E=\angle B O F \Rightarrow \angle A O E=5 y .
$
Since $C O D$ is a straight line. Therefore,
$
\begin{array}{ll}
& \angle C O E+\angle E O A+\angle A O D=180^{\circ} \\
\Rightarrow \quad & 2 y+5 y+2 y=180^{\circ} \Rightarrow 9 y=180^{\circ} \Rightarrow y=20^{\circ}
\end{array}
$
We find that $\angle A O E$ and $B O F$ are vertically opposite angles.$
\angle A O E=\angle B O F \Rightarrow \angle A O E=5 y .
$
Since $C O D$ is a straight line. Therefore,
$
\begin{array}{ll}
& \angle C O E+\angle E O A+\angle A O D=180^{\circ} \\
\Rightarrow \quad & 2 y+5 y+2 y=180^{\circ} \Rightarrow 9 y=180^{\circ} \Rightarrow y=20^{\circ}
\end{array}
$
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