Question

Match the two columns;

$\begin{array}{cc}Column-I& Column-II\\ \text{(A) C is a point on the minor arc AB of the circle with centre O.}& \left(p\right){60}^{\circ }\\ \text{Given}\angle \text{ACB = x. Calculate}\angle \text{x if ACBO is a parallelogram.}& \\ \text{(B) Chord ED is parallel to the diameter AC of the circle.}& \left(q\right){120}^{\circ }\\ \text{Given}\angle CBE={55}^{\circ },\text{calculate}\angle DEC.& \end{array}$

$\begin{array}{cc}\text{(C) In the given figure,O is the centre of the circle.}& \left(r\right){35}^{\circ }\\ \text{If}\angle \text{ACB}={60}^{\circ }\text{find}\angle OAB.& \end{array}$



$\begin{array}{cc}\text{(D) In the given figure, O is the centre of a circle.}& \left(s\right){30}^{\circ }\\ \angle AOB={40}^{\circ }\text{and}\angle BDC={100}^{\circ }.\text{Find}\angle OBC& .\end{array}$

Choose the Correct option;

• A. A-s; B-p; C-q; D-r
• B. A-r; B-s; C-p; D-q
• C. A-q; B-p; C-r; D-s
• D. A-q; B-r; C-s; D-p

Answer 1

The correct option is D

A-q; B-r; C-s; D-p

(d) $\left(A\right)\to q;\left(B\right)\to r;\left(C\right)\to s;\left(D\right)\to p$
(A) Clearly, major arc BA subtends $anglex$ at a point on the remaining part of the circle.
$\therefore$ Reflex $\angle AOB=2x$


$\Rightarrow {360}^{\circ }-y=2x$
$\Rightarrow y={360}^{\circ }-2x$
Thus, $y={360}^{\circ }-2x$
Since ACBO is a parallelogram.
$\therefore x=y\Rightarrow x={360}^{\circ }-2x$
$\Rightarrow 3x={360}^{\circ }\Rightarrow x={120}^{\circ }$

(B) Consider the arc $CDE.\angle CBEand\angle CAE$
Are the angles in the same segment of arc CDE.
$\therefore \angle CAE=\angle CBE$
$\Rightarrow \angle CAE={55}^{\circ }[\angle CBE={55}^{\circ }]$
Since AC is the diameter of the circle and the angle in a semi - circle is a right angle
Therefore,
$\angle AEC={90}^{\circ }$. Now, in $\Delta ACE$, we have
$\angle ACE+\angle AEC+\angle CAE={180}^{\circ }$
$\Rightarrow \angle ACE+{90}^{\circ }+{55}^{\circ }={180}^{\circ }$
$\Rightarrow \angle ACE={35}^{\circ }$.


But $\angle DEC$ and $\angle ACE$ are alternate angles, because $AC||DE$.
$\therefore \angle DEC=\angle ACE={35}^{\circ }$.

(C) $\angle AOB={120}^{\circ }$
(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)
OA = OB (Radius of the circle)



$x+x+{120}^{\circ }={180}^{\circ }$
$\Rightarrow 2x={60}^{\circ }$
$\Rightarrow x={30}^{\circ }$

(D) $\angle AOB=2\times \angle ACB$
$\Rightarrow \angle ACB=\frac{1}{2}\times \angle AOB$
$=\frac{1}{2}\times {40}^{\circ }={20}^{\circ }$



In $\Delta BDC,\angle DBC+\angle BDC+\angle DCB={180}^{\circ }$
$\Rightarrow \angle DBC+{100}^{\circ }+{20}^{\circ }={180}^{\circ }$
$\therefore \angle DBC={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$Hence\angle OBC={60}^{\circ }$