Question

In the given figure, O is the centre of the circle. If $\angle AOD={140}^o$ and $\angle CAB={50}^o$, then:

(i) $\angle EDB$ (ii) $\angle EBD$ are respectively:

• A. ${70}^o\&{50}^o$
• B. ${50}^o\&{110}^o$
• C. ${30}^o\&{70}^o$
• D. ${120}^o\&{130}^o$

Answer 1

The correct option is B ${50}^o\&{110}^o$

Given, $O$ is the centre of a circle in which a quadrilateral $ABCD$ has been inscribed.

Also, $AB$ & $CD$ are produced to meet at $E$.

Now, $\angle BAC={140}^o$ & $\angle AOD={50}^o$.

$\angle BOD={180}^o-\angle AOD$...(linear pair)

$⟹\angle BOD={180}^o-{140}^o={40}^o$.........(i).

Since, $OD$ and $OB$ are the radii of the same circle, $OD=OB$.

i.e $\Delta BOD$ is an isosceles one with $BD$ as base.

$\therefore \angle OBD=\angle ODB$

$⟹\angle OBD+\angle ODB=2\angle OBD$ $=2\angle ODB$.

Then, $\angle OBD+\angle ODB+\angle BOD={180}^o$ ...(angle sum property of triangles).

Using (i), we get,

$(2\angle OBD=2\angle ODB)={180}^o-{40}^o={140}^o$

$⟹(\angle OBD=\angle ODB)={70}^o.$

Again $\angle EBD={180}^o-\angle OBD$ ...(linear pair)

$={180}^o-{70}^o={110}^o$..........(iii).

Again $\angle BAC+\angle BDC={180}^o$ ...(sum of the opposite angles of a cyclic quadrilateral $={180}^o$).

$⟹\angle BDC={180}^o-\angle BAC={180}^o-{50}^o={130}^o$.

So $\angle ODC=\angle BDC-\angle ODB={130}^o-{70}^o={60}^o.$

Now we have $\angle ODC+\angle ODB+\angle EDB={180}^o$....(straight angle)

$\therefore \angle EDB={180}^o-(\angle ODC+\angle ODB)={180}^o-({60}^o+{70}^o)={50}^o$.....(iv).

So $\angle EDB$ & $\angle EBD$ are respectively ${50}^o$ and ${110}^o.$

Hence, option $B$ is correct.