Question

In the given figure, O is the centre of the circle. Chord CD is parallel to the diameter AB. If $\angle$ABC = ${35}^0$, then find the value of $\angle$CED.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Given $\angle$ABC = ${35}^0$ and AB ll CD

$\therefore$ $\angle$BCD = $\angle$ABC = ${35}^0$ (Alternate intrior angle)

Join O to C and D

$\angle$AOC = 2$\angle$ABC and $\angle$BOD = 2 $\angle$BCD (The angle at the centre of a circle is twice the angle at the circumference if both angles stand on the same arc)

$\Rightarrow$ $\angle$AOC = 2$\times$ ${35}^0={70}^0$

Also, $\angle$AOC + $\angle$ COD + $\angle$BOD = ${180}^0$ (Angles on the straight line)

$\Rightarrow$${70}^0$ +$\angle COD+{70}^0={180}^0$

$\Rightarrow$$\angle$COD = ${40}^0$

$\Rightarrow$2$\angle$CED = ${40}^0$

$\Rightarrow$ $\angle$CED = ${20}^0$