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$\left(arcDC\right)$ is

• A. ${60}^o$
• B. ${120}^o$
• C. ${90}^o$
• D. ${170}^o$

Answer 1

The correct option is C ${90}^o$

In the given diagram, draw diameters $BD$ and $EC$ which bisects each other at centre $O$ and also we know that $BC||ED$

$therefore$, $\angle$$BOE=$$\angle$$COD=x$ (assume) $.....\left(i\right)$

According to Circle Chord property, measure of intercepted arc $=$ measure of the central angle

$\therefore$,$\angle$$COB=m\left(arcBC\right)={94}^{\circ }$ $.....\left(ii\right)$

$\angle$$EOD=m\left(arcED\right)={86}^{\circ }$ $.....\left(iii\right)$

Also, $\angle$$COD=m\left(arcDC\right)$ $.....\left(iv\right)$

Since, Sum of measure of central angles of a circle is equal to ${360}^{\circ }$

$\therefore$, $\angle$$BOE+$$\angle$$EOD+$$\angle$$COD+$$\angle$$COB={360}^{\circ }$

using equation $\left(i\right),\left(ii\right)and\left(iii\right)$ we get,

$x+{86}^{\circ }+x+{94}^{\circ }={360}^{\circ }$

$2x+{180}^{\circ }={360}^{\circ }$

$2x={360}^{\circ }-{180}^{\circ }$

$2x={180}^{\circ }$

$x=\frac{180}{2}$

$\therefore x={90}^{\circ }$

That is, $\angle$$COD=x={90}^{\circ }$

$\therefore$ $m\left(arcDC\right)={90}^{\circ }$ [from $eq.\left(iv\right)$