Question

Given circle with centre $O$ and $BC\parallel ED$, $m\left(arcBC\right)={94}^0$, $m\left(arcED\right)={86}^0$, $\angle ADE=8^0$. The m$\left(arcEB\right)$ is :

• A. ${120}^o$
• B. ${90}^o$
• C. ${16}^o$
• D. ${47}^o$

Answer 1

Answer: (B)

${90}^o$

Construct 2 perpendiculars from the center $O$ to the chords $BC$ and $DE$ which are $OM$ and $ON$ respectively.

Now, clearly $\angle BOC=2\angle BOM={94}^{\circ }$ and $\angle EOD=2\angle EON={86}^{\circ }$

Also, as $BC||DE$, $\angle BOM+\angle BOE+\angle EON={180}^{\circ }⟹m\left(\text{arc EB}\right)=\angle BOE={180}^{\circ }-{\frac{94}{2}}^{\circ }-{\frac{86}{2}}^{\circ }={90}^{\circ }$