Question

If O is the center of the circle, find the value of $x$ in each of the following figures.
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

Answer 1

We have to find $x$ in each figure.

(i) It is given that $\angle AOC={135}^{\circ }$
$\angle AOC+\angle COB={180}^{\circ }$ [Linear pair]

$\angle COB={180}^{\circ }-{135}^{\circ }={45}^{\circ }$

As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Now, $x=\frac{1}{2}\angle COB=22{\frac{1}{2}}^{\circ }$

Hence, $x=22{\frac{1}{2}}^{\circ }$

(ii) As we know that $\angle CAB=\angle CDB=x$ [Angles in the same segment]

line AB is diameter passing through center.

So, $\angle BCA={90}^{\circ }$ [Angle inscribed in a semicircle is a right angle]

$\angle CAB+\angle ABC+\angle BCA={180}^{\circ }$ [Angle sum property]

$\Rightarrow x+{40}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\Rightarrow x={50}^{\circ }$

(iii) It is given that

$\angle AOC={120}^{\circ }$

$\angle ABC=\frac{1}{2}$ Reflex $\angle AOC$

$\Rightarrow \angle ABC=\frac{1}{2}({360}^{\circ }-{120}^{\circ })$

So, $\angle ABC={120}^{\circ }$

And $\angle ABC+\angle CBD={180}^{\circ }$

${120}^{\circ }+\angle CBD={180}^{\circ }$

Then $\angle CBD={60}^{\circ }$

Hence, $x={60}^{\circ }$

(iv) $\angle DBC={65}^{\circ }$

$\angle ABC+\angle DBC={180}^{\circ }$ (Linear pair)

$\angle ABC={180}^{\circ }-{65}^{\circ }={115}^{\circ }$

And
$x=2\angle ABC$
Hence, $x=2\times {115}^{\circ }={230}^{\circ }$

(v) It is given that $\angle OAB={35}^{\circ }$

$△AOB$ is an isosceles triangle.

Therefore, $\angle ABO={35}^{\circ }$

And,

In $△AOB,\angle AOB+\angle OBA+\angle OAB={180}^{\circ }$

$\Rightarrow {70}^{\circ }+\angle AOB={180}^{\circ }$

$\Rightarrow \angle AOB={110}^{\circ }$
Reflex $\angle AOB=2$$\angle ACB$

$\angle ACB=\frac{1}{2}[{360}^{\circ }-{110}^{\circ }]$

$\angle ACB={125}^{\circ }$

Hence, $x={125}^{\circ }$

(vi) It is given that $\angle AOB={60}^{\circ }$

And $\angle COA+\angle AOB={180}^{\circ }$

$\Rightarrow \angle COA={180}^{\circ }-{60}^{\circ }$

$\Rightarrow \angle COA={120}^{\circ }$

$△OCA$ is an isosceles triangle.

So, $\angle CAO=\frac{1}{2}({180}^{\circ }-{120}^{\circ })={30}^{\circ }$

Hence, $x={30}^{\circ }$

(vii) $\angle BAC=\angle BDC={50}^{\circ }$ (Angle in the same segment)

In $△BDC$, we have

$\angle DBC+\angle BDC+\angle BCD={180}^{\circ }$

${70}^{\circ }+{50}^{\circ }+\angle BCD={180}^{\circ }$

$\angle BCD={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

Hence, $x={60}^{\circ }$

(viii)

As $OD=OB$ (Radius of the circle)
Therefore, $△DOB$ is an isosceles triangle.

$\angle ODB+\angle OBD+\angle DOB={180}^{\circ }$

${40}^{\circ }+{40}^{\circ }+\angle DOB={180}^{\circ }$

$\angle DOB={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

So, $\angle AOC=\angle DOB$ (Vertically opposite angles)

Hence, $x={100}^{\circ }$

(ix) It is given that $\angle ABD={50}^{\circ }$

$\angle DCA=\angle ABD={50}^{\circ }$ …… (1) (Angle in the same segment)
$\angle ADB=\angle ACB={32}^{\circ }$ ......(2) (Angle in the same segment)

Because $\angle DCA$ and $\angle ABD$ are on the same segment $AD$ of the circle.

Now from equations (1) and (2), we have

$\angle DCB={50}^{\circ }+{32}^{\circ }={82}^{\circ }$

Hence, $\angle DCB=x={82}^{\circ }$

(x) It is given that $\angle A={35}^{\circ }$

$\angle BAC=\angle BDC={35}^{\circ }$ (Angle in the same segment)

Now in $△BDC$, we have
$\angle BDC+\angle DCB+\angle CBD={180}^{\circ }$

$\Rightarrow {35}^{\circ }+{65}^{\circ }+\angle CBD={180}^{\circ }$

$\Rightarrow \angle CBD={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Hence, $x={80}^{\circ }$

(xi)

$\angle ABC=\angle ACD={40}^{\circ }$ (Angle in the same segment)
In $△PCD$, we have

$\angle CPD+\angle PCD+\angle PDC={180}^{\circ }$

${40}^{\circ }+{110}^{\circ }+\angle PDC={180}^{\circ }$

$\angle PDC={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

Hence, $x={30}^{\circ }$

(xii)

$\angle BAO=\angle CDO={52}^{\circ }$ (Angle in the same segment)

$△DOC$ is an isosceles triangle

So, $OD=OC$ (Radius of the same circle)

Then $\angle ODC=\angle OCD={52}^{\circ }$

Hence, $x={52}^{\circ }$