Question

If O is the centre of the circle, find the value of x in each of the following figures:

Answer 1

(i) A circle with centre O

$\angle AOC={135}^{\circ }But\angle AOC+\angle COB={180}^{\circ }\left(Linearpair\right)\Rightarrow {135}^{\circ }+\angle COB={180}^{\circ }\Rightarrow \angle COB={180}^{\circ }-{135}^{\circ }={45}^{\circ }NowareBCsubtends\angle BOCatthecentreand\angle BPCattheremainingpartofthecircle\therefore \angle BOC=2\angle BPC\Rightarrow \angle BPC=\frac{1}{2}\angle BOC=\frac{1}{2}\times {45}^{\circ }=\frac{{45}^{\circ }}{2}\therefore \angle BPC=\frac{1}{2}\angle BOC=\frac{1}{2}\circ orx=22\frac{1}{2}\circ$

(ii) $\because CDandABarethediametersofthecirclewithcentreO\angle ABC={40}^{\circ }Butin\Delta OBC,OB=OC\left(Radiiofthecircle\right)\therefore \angle OCB=\angle OBC={40}^{\circ }Nowin\Delta BCD,\angle ODB+\angle OCB+\angle CBD={180}^{\circ }\Rightarrow x+{40}^{\circ }+{90}^{\circ }={180}^{\circ }\left(Anglesofatriangle\right)\Rightarrow x+{130}^{\circ }={180}^{\circ }\Rightarrow x={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

(iii) $IncirclewithcentreO,\angle AOC={120}^{\circ },ABisproducedtoD\because \angle AOC={120}^{\circ }and\angle AOC+convex\angle AOC={360}^{\circ }\Rightarrow {120}^{\circ }+convex\angle AOC={360}^{\circ }\therefore Convex\angle AOC={360}^{\circ }-{120}^{\circ }={240}^{\circ }\therefore areAPCSubtends\angle AOCatthecentreand\angle ABCattheremainingpartofthecircle\therefore \angle ABC=\frac{1}{2}\angle AOC=\frac{1}{2}\times {240}^{\circ }={120}^{\circ }But\angle ABCattheremainigpartofthecircle\therefore \angle ABC=\frac{1}{2}\angle AOC=\frac{1}{2}\times {240}^{\circ }={120}^{\circ }But\angle ABC+\angle CBD={180}^{\circ }\left(LinearPair\right)\Rightarrow {120}^{\circ }+x={180}^{\circ }\Rightarrow x={180}^{\circ }-{120}^{\circ }={60}^{\circ }\therefore x={60}^{\circ }$
(iv) A circle with centre O and $\angle CBD={65}^{\circ }But\angle ABC+\angle CBD={180}^{\circ }\left(Linearpair\right)\Rightarrow \angle ABC+{65}^{\circ }={180}^{\circ }\Rightarrow \angle ABC={180}^{\circ }-{65}^{\circ }={115}^{\circ }$
Now are AEC subtends $\angle x$ at the centre and $\angle ABC$ at the remaining part of the cirle
$\therefore \angle AOC=2\angle ABC\Rightarrow x=2\times {115}^{\circ }={230}^{\circ }\therefore x={230}^{\circ }$
(v) In circle with centre O AB is chord of the cirlce, $\angle OAB={35}^{\circ }In△OAB,$
$(OA=OB\therefore OBA=\angle OAB={35}^{\circ }$
But in $△OAB$
$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }\Rightarrow {35}^{\circ }+{35}^{\circ }+\angle AOB={180}^{\circ }\Rightarrow {70}^{\circ }+\angle AOB={180}^{\circ }\Rightarrow \angle AOB={180}^{\circ }-{70}^{\circ }={110}^{\circ }\therefore Convex\angle AOB={360}^{\circ }-{110}^{\circ }={250}^{\circ }$
But are AB subtends $\angle AOB$ at the centre and $\angle ACB$ at the remaining part of the circle
$\therefore \angle ACB=\frac{1}{52}\angle AOB\Rightarrow x=\frac{1}{2}\times {250}^{\circ }={125}^{\circ }\Rightarrow x={125}^{\circ }$
(vi) IN the circle with centr O, BOC is its diameter, $\angle AOB={60}^{\circ }$
Are AB subtends$\angle AOB$ at the centre of the circle and $\angle ACB$ at the remaining part of he circle
$\therefore \angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }Butin△OAC,OC=OA\therefore \angle OAC=\angle OCA=\angle ACB\Rightarrow x={30}^{\circ }$
(vii) IN the circle, $\angle BACand\angle BDC$ are in the same segment
$\angle DBC+\angle BCD+\angle BDC={180}^{\circ }\Rightarrow {70}^{\circ }+x+{50}^{\circ }={180}^{\circ }\Rightarrow x+{120}^{\circ }={180}^{\circ }\Rightarrow x={180}^{\circ }-{120}^{\circ }={60}^{\circ }\therefore x={60}^{\circ }$
(viii) IN circle with centre O,
$\angle OBD={40}^{\circ }$
AB and CD are diameters of the circle
$\angle DBAand\angle ACD$ are in the same segment
$\therefore \angle ACD=\angle DBA={40}^{\circ }in△OAC,OA=OC\therefore \angle OAC=\angle OCA={40}^{\circ }and\angle OAC+\angle OCA+\angle AOC={180}^{\circ }\Rightarrow {40}^{\circ }+{40}^{\circ }+x={180}^{\circ }\Rightarrow x+{80}^{\circ }={180}^{\circ }\Rightarrow x={180}^{\circ }-{80}^{\circ }={100}^{\circ }\therefore x={100}^{\circ }$
(ix) In the circle, ABCD is a cyclic quadrilatral $\angle ADB={32}^{\circ },\angle DAC={28}^{\circ },and\angle ABD={50}^{\circ },\angle ABDand\angle ACD$ are in the same segment of a circle
$\therefore \angle ABD=\angle ACD\Rightarrow \angle ACD={50}^{\circ }Similarly,\angle ADB=\angle ACB\Rightarrow \angle ACB={32}^{\circ }$
Now $\angle DCB=\angle ACD+\angle ACB={50}^{\circ }+{32}^{\circ }={82}^{\circ }x={82}^{\circ }$
(x) IN a circle,
$\angle BAC={35}^{\circ },\angle CBD={65}^{circ}\angle BACand\angle BDC$ are in the same segment
$\therefore \angle BAC=\angle BDC={35}^{\circ }In△BCD,\angle BDC+\angle BCD+\angle CBD={180}^{\circ }\Rightarrow {35}^{\circ }+x+{65}^{\circ }={180}^{\circ }\Rightarrow x+100={180}^{\circ }\Rightarrow x+{100}^{\circ }={180}^{\circ }\Rightarrow x-{180}^{\circ }-{100}^{\circ }={80}^{\circ }\therefore x={80}^{\circ }$
(xi) In the circle,
$\angle ABDand\angle ACD$ are in the same segment of a circle
$\therefore ABD=\angle ACD={40}^{\circ }Nowin△CPD,\angle CPD+\angle PCD+\angle PDC={180}^{\circ }\Rightarrow {110}^{\circ }+{40}^{\circ }+x={180}^{\circ }\Rightarrow x+{150}^{\circ }={180}^{\circ }\therefore x={180}^{\circ }-{150}^{\circ }={30}^{\circ }$
(xii) In the circle, two diameters AC and BD intersect each other at O
$\angle BAC={50}^{\circ }OA=OB\therefore \angle OBA=\angle OAB={52}^{\circ }\Rightarrow \angle ABD={52}^{\circ }$
But $\angle ABDand\angle ACD$are n the same segment of the circle
$\therefore \angle ABD=\angle ACD\Rightarrow {52}^{\circ }=x\therefore x={52}^{\circ }$