Figures given below will show the different part of a circle.
Now, we study about the central angle, inscribed angle, cyclic quadrilateral, tangents and related theorems.
In the circle PQR, radii OP and OQ form an angle at the centre O. So, acute ∠POQ and reflex ∠POQ are central angles. The part PQ of the circle is an arc.
∠POQ is subtended by \(\widehat{PQ}\) and reflex∠POQ is subtended by \(\widehat{PRQ}\). As the length of arc increases, the measure of the central angle subtended by the arc also increases. So the central angle can be measured with its corresponding opposite arc. Therefore∠POQ =\(\widehat{PQ}\) and reflex∠POQ = \(\widehat{PRQ}\)
The angle formed by joining the two chords at the circumference of a circle is called angle at the circumference or inscribed angle. In the adjoining figure, chord AB and chord CB meet at the point B on the circumference and ∠POQ is formed which is angle at circumference standing on the arc AC \((\widehat {AC})\). Inscribed angle also can be measured (expressed) in terms of its corresponding arc \( [\angle ABC = \frac {1}{2} \widehat {AC}]\)
In a circle ABC, chord AB divides the circle into two parts. The arc AB \((\widehat {AB}\) is formed by the chord AB. In this way, while reading an arc and its corresponding chord, we use the same letter but the length is not exactly same. In the adjoining figure \(\widehat{AB}\) is a minor arc and \(\widehat{ACB}\) is major arc. In short \(\widehat{ADB}\) is read as minor\(\widehat{AB}\) and \(\widehat{ACB}\) is read as major\( \widehat{AB}\).
If all the vertices of a quadrilateral are on the circumference of a circle, then the quadrilateral is called cyclic quadrilateral. In the adjoining figure, ABCD is a cyclic quadrilateral. In other words, the quadrilateral inscribed in a circle is called cyclic quadrilateral. The points A, B, C, D are concyclic ABCD is cyclic quadrilateral but ABCE and ADCE are not cyclic quadrilaterals.
A round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point (the centre)
.
Solution:
\(\angle QOR = 110°, \angle PSQ= ?\)
\begin{align*} \angle POQ+ \angle QOR &= 180° [\because \text{Sum of 2 adjacent angles}] \\ or, \angle POQ + 110° &= 180°\\ \therefore \angle POQ &= 70°\\ \\ \angle PSQ &= \frac{1}{2}\angle POQ [\because \text{The inscribed angles is half of the centre angle, when both are standing on the same arc.}]\\ \angle PSQ&=\frac{1}{2}\times 70°\\ \therefore \angle PSQ &= 35° \:\: _{Ans} \end{align*}
Solution:
\(\angle DAB = 92°\\ \angle FBE=10°\\ \angle ADC =?\)
\begin{align*} \angle DAB + \angle BCD &= 180° \:\:\: [\because\text {Sum of two opposite angle of cyclic quadrilateral} ] \\ 92°+\angle BCD &= 180°\\ or, \angle BCD &= 180°-92°\\ \therefore \angle BCD &= 88°\\ \angle CBE &= \angle BCD=88° \:\:\:[\because Alternative \: angles ]\\ \angle ADC &= \angle CBF\:\:\: [\because \text{ The exterior angle is equal to the opposite interior angle of cyclic quadrilateral} ] \\ \angle ADC &= 88°+10°\\ \therefore \angle ADC &= 98° \:\: \: _{ans} \end{align*}
Solution:
\(\angle BCD = 60°\),
\(\angle ACB = \angle BDC=x \)
\(\angle ADC=?\)
\(\angle BCD = \angle CAB = 60° \:\:\: [\because \text{Alternate segment angle}]\)
In \(\Delta ACD\)
\begin{align*} \angle CAD + \angle ACD + \angle ADC &= 180° \:\:[\because\text{Sum of 3 interior angles of triangle}] \\ 60° +x°+60°+x° &= 180° \\ or, 2x° + 120° &= 180°\\ or, x° &= \frac{180°-120°}{2}\\ \therefore x &= 30° _{Ans} \end{align*}
Solution:
AB=AD=BD
\begin{align*} \angle BAD &=60° \:\:\:[\because \text{Each angle of equilateral triangle}] \\ \angle BED &= \angle BAD = 60° \\ b&= 60° \\ \\ \angle BED + \angle BCD &= 180° \:\:\:[\because \text{Sum of opposite angle of cyclic quadrilateral}] \\ or, 60° + a &= 180°\\ or, a &= 180° - 60°\\ \therefore a &= 120° \end{align*}
Solution:
\(\angle QPR = 40° , \: \angle OQR=?\)
Now,
\begin{align*} \angle QOR &= 2\angle QPR \:\:\: [\because \text {The centre and inscribed angle are standing on the same arc}] \\ \angle QOR &= 2 \times 40° &= 80°\\ \angle OQR &= \angle ORQ \:\:\:[\because OQ=OR]\\ \angle OQR + \angle ORQ + \angle QOR &= 180° \:\:\:[\because \text{sum of 3 interior angles of triangles}] \\ \angle OQR + \angle OQR + 80° &= 180°\\ 2\angle OQR &= 180° - 80°\\ or, \angle OQR &= \frac{100}{2}\\ \therefore \angle OQR &= 50° \:\:_{Ans} \end{align*}
Solution:
\( \angle MON = 140° \\ \angle MPN = ? \)
Construction: Take a point A on the circumference of the circle. Join MA and NA.
\begin{align*} \angle MAN &= \frac{1}{2}\angle MON \:\:\: [\because \text{The inscribed angle in half of the centre angle when both angle standing on the same arc} ] \\ or, \angle MAN &= \frac{1}{2} \times 140\\ &= 70°\\ Now , \\ \angle MPN + \angle MAN &= 180° [\because \text{The sum of the two opposite angle of the cyclic quadrilateral}]\\ or, \angle MPN + 70° &= 180°\\ or, \angle MPN &= 180°-70°\\ \therefore \angle MPN &= 110°\:\: _{ans} \end{align*}
Solution:
\( \angle PQR = b°, \angle RPQ = \frac{3b°}{2}, \angle RSQ = y°, \angle PRQ = 40°\)
\(\angle PRQ = 90° \:\:\:\:\: [\because \text{Angle standing on diameter}]\\ In \Delta PQR, \)
\begin{align*} \angle PRQ + \angle PQR + \angle QPR &= 180° \:\:\: [\because \text{The sum of 3 interior angles of } \Delta ]\\ 90°+b°+\frac{3b}{2} &= 180°\\ or, \frac{2b+3b}{2} &= 180° - 90°\\ or, 5b &= 90 \times 2\\ or, b° &= \frac{180}{5}\\ \therefore b &= 36° \:\:\: _{Ans} \end{align*}
\begin{align*} \angle QPR &= \frac{3b}{2}\\ &= \frac{3 \times 36}{2} \\ &= 54°\\ \angle QSR &= \angle QPR = y° = 54° \:\:\: [\because \text{Standing on the same arc}] \end{align*}
Solution:
\( \angle ABC = 55° \\ \angle BDC = ? \\ \: \\ \angle ACB = 90° \: \:\:\: [\because \text{The angle made in semicircle}] \)
\begin{align*} \angle ABC + \angle ACB + \angle BAC &= 180° \:\: [\because \text{sum of 3 interior angle of }\Delta ] \\ 55° + 90° + \angle BAC &= 180° \\ \angle BAC &= 180° - 145°\\ \angle BAC &= 35° \end{align*}
\(BDC =\angle BAC= 35° [\because \text{Inscribed angle standing on same arc}]\\ \because \angle BDC = 35°\)
Solution:
Construction : Join PS and RS
\(\angle PQR = 90°\)
\begin{align*} \angle PQR + \angle PSR &= 180°\:\: [\because \text{Sum of opposite angles of cyclic quadrilateral.}] \\ 120° + \angle PSR &= 180° \\ \angle PSR &= 180° - 120°\\ \angle PSR &= 60°\\ \: \\ obtuse \: \: \angle POR &= 360° - 120° \:\: [\because \text{The angle made at a point is } 360°]\\ &= 240° \end{align*}
Solution:
\( \angle RQT = 98°\\ \angle PRQ = \angle RPQ = b° \:\: \: \: \: \: [\because PQ = RQ] \)
\begin{align*}\angle RQT &= \angle PRQ + \angle RPQ \:\:\: [\because \text{The exterior angle is equal to the sum of the two opposite interior angles }]\\ b+b&= 98°\\ or, 2b &= 98° \\ or, b &= \frac{98}{2}\\ \therefore b &= 49° \end{align*}
\begin{align*} \\ \: \\ \angle QPS + \angle SRQ &= 180° \:\: [\because \text{Sum of the opposite angles of the cyclic quadrilateral. }] \\ b + 44 ° + b+a &= 180° \\ or, 49°+44°+49° + a &= 180°\\ or, a &= 180°-142°\\ \therefore a &= 38° \end{align*}
Solution:
\( \angle DBC = 33°\\ \angle BAD = ? \\ \angle BDC = \angle DBC = 33° \: \: \: [\because BC = CD] \)
\begin{align*}\angle BDC + \angle DBC + \angle BCD &= 180° \:\:\: [\because \text{Sum of 3 angles of }\Delta] \\ 33° + 33° + \angle BCD &= 180°\\ \angle BCD &= 180° - 66° \\ \therefore \angle BCD &= 114° \\ \: \\ \angle BAD + 114° &= 180° \\ \angle BAD &= 180° - 114°\\ \therefore \angle BAD &= 66° \end{align*}
Solution:
\(\angle BAC = 2x° \\ \angle ACB = 3x°\\ \angle ABC = 90° \:\:\: [\because \text{The inscribed angle made in semi circle.}] \)
\begin{align*} \angle BAC + \angle BCA + \angle ABC &= 180° \: \: [\because \text{sum of 3 angles of } \Delta]\\ 2x + 3x + 90° &= 180°\\ or, 5x &= 180° - 90°\\ or, x&= \frac{90°}{5} \\ \therefore x &= 18° \\ \:\\ \angle BAC &= 2x°= 2 \times 18 = 36° \\ \angle BDC &= \angle BAC = 36° \: \: _{Ans} \end{align*}
Solution:
\( \angle DAB = 45° \\ \angle AEB = 60°\\ \angle BCD = 2\angle BAD \: \: \: [\because \text{Central angle is double the inscribed angle.} ] \\ \angle BCD = 2 \times 45° = 90°\)
\begin{align*} \angle BCD + \angle CBD + \angle BDC &= 180° \:\:\: [\because \text{Sum of angle of } \Delta ] \\ 90° + y° + y° &= 180°\\ or, 2y° &= 180° - 90°\\ or, y°&= \frac{90}{2}\\ \therefore y &= 45° \end{align*}
\begin{align*} \angle AEB &= \angle EBD + \angle EDB\\ or, 60° &= y +x\\ or, 60° - 45° &=x \\ \therefore x &=15° \end{align*}
Solution:
\(\angle AOC = 70°\)
\begin{align*}\angle AOC &= \angle OBC + \angle OCB \:\: [\because \text{The exterior angle is equal to the two opposite interior angle.}] \\ or, 70° &= x + x \:\:\: [\because OC = OB]\\ or, 2x &= 70°\\ or, x &= \frac{70}{2}\\ \therefore x &= 35° \: \\ \: \\ \angle CBD &= \angle BCO= 35° \:\: [\because Alternative\: angles] \\ \angle OBD &= \angle ODB = 35° + 35° = 70° \:\:\: [\because OB = OD] \end{align*}
\( y = \angle EBD + \angle EDB\\ y = 35° + 70° \\ y = 105° \)
Solution:
\(\angle POS = 60°\)
\begin{align*} \angle PQS &= \frac{1}{2} \angle POS \:\: [\because \text{The inscribed angle is half of the centre angle}] \\ &= \frac{1}{2} \times 60\\ &= 30° \\ \: \\ \angle OSQ &= \angle ORQ \:\: [\because \text{Both angle standing on same arc}] \\ x &= 30° \: \: [\because OS = OQ] \\ \angle ORQ &= \angle POS= 60° \:\: [\because OS \parallel QR]\\ \angle SQR &= 60° - 30° = 30° \\ \: \\ \angle STR &= \angle TQR + \angle RQT \:\: [\because \text{The exterior angle is equal to the sum of the two opposite interior angle.}]\\ y &= 60° + 30°\\ \therefore y &= 90° \end{align*}
Solution:
Construction: Join PA and RA
\begin{align*}\angle ROP + \angle OPQ &= 180° \:\:[\because \text{sum of co-interior angles}] \\ y + x &= 180° \:\: ............. (i) \\ \: \\ \angle RAP &= \frac{1}{2} \angle AOP \\ \angle RAP &= \frac{y}{2}\\ \: \\ \angle RQP &= y° \:\: [\text{Opposite angle of parallelogram}] \\ \angle RAP + \angle RQP &= 180° \: \: \: [\because \text{Sum of opposite angle of cyclic quadrilateral}]\\ \frac{y}{2} + y &= 180° \\ \frac{y + 2y}{2} &= 180°\\ or, 3y &= 180 \times 2 \\ or, y &= \frac{360}{3}\\ \therefore y &= 120° \end{align*}
Putting value of y in equation (i)
\(x + y = 180° \\ x = 180° - 120°\\ \therefore x = 60° \)
Solution:
\(\angle OAB = 30° \\ \angle OBA= \angle OAB = 30° \:\:\: \: \: [\because OA = OB] \)
\begin{align*} \angle AOB + \angle OBA + \angle OAB &= 180° \: \: \: \: [\because \text{Sum of 3 angles of }\Delta ] \\ \angle AOB + 30° + 30° &= 180°\\ \angle AOB &= 180° - 60° \\ \angle AOB &= 120° \end{align*} \begin{align*} \angle APB &= \frac{1}{2} \angle AOB \: \: [\because \text{The inscribed angle is half of the centre angle}] \\ \angle AOB &= \frac{120}{2} \\ \therefore \angle APB &= 60° \end{align*}
Solution:
\(\angle BAC = 35° \\ \angle OCB = ? \\ \: \\ \angle BOC = 2\angle BAC \\ \angle BOC = 2 \times 35° = 70° \\ \angle OBC = \angle OCB \)
\begin{align*} \angle OBC + \angle OCB + \angle BOC &= 180° \: \: \: [\because \text {Sum of 3 angles of } \Delta ]\\ or, \angle OCB + \angle OCB + 70° &= 180°\\ or, 2\angle OCB &= 180° - 70°\\ or, \angle OCB &= \frac{110}{2}\\ \therefore \angle OCB &= 55° \end{align*}
Solution:
\(\angle PQR = 90°\:[\because \text{Angle in the semi-circle}] \)
\begin{align*} \angle PQR + \angle QPR + \angle PRQ &= 180° \: \: [\because \text{The sum of 3 interior angle of }\Delta] \\ 90° + x + 2x &= 180°\\ or, 3x &= 180° - 90°\\ or, x &= \frac{90}{3}\\ \therefore x &=30° \end{align*}
Solution:
\begin{align*}\angle DAB &= \frac{1}{2}\angle BOD \: [\because \text{The inscribed angle is half of the centre angle.}] \\ x°&= \frac{1}{2} \times 130°\\ &= 65° \\ \: \\ x+y &=180° \: \: [\because \text{Opposite angle of cyclic quadrilateral}] \\ y&=180° - 65° \\ \therefore y &= 115° \: \: \: _{Ans} \end{align*}
Solution:
\begin{align*} \angle BCA &= \angle BDA = 25° \: \: [\because \text{Angles at the circumference on the same segment.}]\\ \angle ABD &= \angle ACD = 54° \:\: \: \: \: _{Ans} \end{align*}
Solution:
Construction: Join OB.
\begin{align*} \angle OBA &= \angle OAB = 30° \: \: \: [\because OB = OA ]\\ \angle OBC &= \angle OCB = 40° \: \: \: [\because OB = OC] \\ \: \\ \angle ABC &= \angle OBA + \angle OBC = 30° + 40° = 70°\\ \: \\ \angle AOC &= 2\angle ABC \\ &= 2 \times 70 = 140° \end{align*}
Solution:
\(\angle DBA = 40° \\ \angle ADB = 90° \: [\because \text{Angle made in the semi-circle.}] \)
\begin{align*}\angle DAB + \angle ADB + \angle DBA &= 180°\\ y + 90° + 40° &= 180°\\ or, y &= 180° - 130° \\ \therefore y&=50° \\ \: \\ x° + y° = 180° \: \: &[\because \text{Sum of opposite angles of cyclic quadrilateral}]\\ x + 50 = 180°\\ x =180° - 50° \\ \therefore x = 130° \: \: _{Ans} \end{align*}
Solution:
\(\angle VWZ = 58° \\ \angle VOZ = 2 \angle VWZ \: [\because \text{The central angle is double of inscribed angle}] \\ \: \: \: \: x° = 2 \times 58° \\ \therefore x = 116°\\ \: \\ y + 58° = 180° \: [\because \text{ sum of the opposite angles of the cyclic quadrilateral}]\\ y = 180° - 58°\\ \therefore x= 122°\)
If two arcs of a circle subtend equal angles at the center of the circle, the arcs are................. .
unequal
equal
half of each
twice
..................... arcs of a circle subtend equal angles at the centre of the circle.
unequal
half of each
one third
equal
Equal chords of a circle form................. arcs in the circle.
The angles at the centre of a circle is................. the angles of its circumference standing on the same arc.
twice
unequal
thrice
equal
Angles in the same segment of a circle are.................. .
equal
unequal
half of each
one third
The opposite angles of a cyclic quadrilateral are............................ .
complementary
isosceles
right angle
supplementary
You must login to reply
Aug 31, 2017
0 Replies
Successfully Posted ...
Please Wait...
Jul 28, 2017
0 Replies
Successfully Posted ...
Please Wait...
Dipen
Prove that the sum of angles subtended by each side of the cyclic quadrilateral at circumference is 6 right angles.
Mar 18, 2017
0 Replies
Successfully Posted ...
Please Wait...
apple
equal chords of circle are equidistant from the centre
Mar 15, 2017
0 Replies
Successfully Posted ...
Please Wait...