Question

In the given figure, two chords $AB$ and $CD$ of a circle intersect each other at a point $E$ such that $\angle$ $BAC={45}^o$, $\angle$ $BED={120}^o$. Then $\angle$ $ABD=$

• A. $60$ ${}^o$
• B. $30$ ${}^o$
• C. $45$ ${}^o$
• D. $15$ ${}^o$

Answer 1

The correct option is D $15$ ${}^o$

Given- $O$ is the centre of a circle of which the chords $AB$ and $CD$ intersect at $E$.

$\angle BAC={45}^o$ and $\angle BED={120}^o$.

To find out- $\angle ABD=?$

Solution- $\angle BAC={45}^o=\angle BDC$

Since both the angles are on the same chord $BC$ ( the angles, subtended by a chord to different points of the circumference are equal).
So in $\Delta BDE$, we have

$\angle EBD$ or $\angle ABD={180}^o-(\angle BED+\angle BDE)={180}^o-({120}^o+{45}^o)={15}^o$.
Ans- Option D.