Subject: Compulsory Mathematics

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)

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

In the given O is the centre of the circle. If \(\angle QOR = 110°\), calculate the value of the \(\angle PSQ\)

**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*}

In the given figure, ABCD is a cyclic quadrilateral in which AB is produced to F. IF DC \(\parallel\) BE, \(\angle DAB\)= 90° and \(\angle FBE = 10°\), evaluate the \(\angle ADC \)

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*}

In the given figure, DCE is a tangent where C is a point of contact. If \(\angle ACB = \angle BDC=x \) and \(\angle BCD = 60°\), find the \(\angle ADC\)

**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*}

Find the value of 'a' and 'b' from the adjoining figure.

**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*}

In the given figure, O is the centre of a circle. If \( \angle QPR=40°\), find the measurement of \( \angle OQR.\)

**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*}

O is the circle of the given circle. If \(\angle MON = 140°\), what will be the value of \(\angle MPN.\)

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*}

In the given figure, M is the centre of the circle. If \(\angle PQR = b°, \angle RPQ = \frac{3b°}{2}\: and \: \angle RSQ = y°, \) find the value of y.

**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*}

In the given figure, O is the centre of the circle and AB is a diameter. If \(\angle ABC = 55°\), find the value of \(\angle BDC\)

**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*}

\(BDC =\angle BAC= 35° [\because \text{Inscribed angle standing on same arc}]\\ \because \angle BDC = 35°\)

In the given circle alongside, O is the centre. Find the obtuse and reflex angle at O.

**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*}

In the given figure, PQRS is a cyclic quadrilateral. If \(\angle RQT = 98°, \angle SPR = 44° \) and PQ = QR, find the value of a and b.

**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*}

In the given figure, O is the centre of circle and ABCD is a cyclic quadrilateral. BC produce to E. If BC = CD and \(\angle DBC = 33°\), find the value of \(\angle BAD\).

**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*}

In the adjoining figure, O is the centre of the circle. AC is a diameter. If \(\angle BAC = 2x \: and \: \angle ACB = 3x, \) find the value of \( \angle BDC.\)

**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*}

In the given figure, C is the centre of the circle. \(\angle DAB =45°, \angle AEB= 60°, \angle CBD=y \: and \: \angle ADB = x,\) find the value of x and y.

**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*}

In the given figure, O is the centre of the circle. If \(OC \parallel BD, \angle AOC = 70°, \angle OCB = x° \:\: and \:\: \angle CED= y° \) then find the value of x° and y°.

**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*}

\( y = \angle EBD + \angle EDB\\ y = 35° + 70° \\ y = 105° \)

In the given figure, O is the centre of circle. If \(SO \parallel RQ, \angle POS = 60°, \angle OSQ = x° and \angle STR = y° \) find the value of x and y.

**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*}

In the given figure O is the centre of a circle. If \(RQ \parallel OP, OR \parallel PQ, \angle ROP = y° \: and \: \angle OPQ = x° \) find the value of x and y.

**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)

\(x + y = 180° \\ x = 180° - 120°\\ \therefore x = 60° \)

In the given figure, O is the centre of a circle. If \(\angle OAB = 30°\), find the value of \(\angle APB.\)

**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*}

In the given figure O is the centre of the circle. If \( \angle BAC = 35°, \) find the \(\angle OCB. \)

**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*}

In the given circle. If PR is the diameter, find the value of x.

**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*}

Find the value of x and y.

**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*}

In the diagram, ABCD is a circle. If \(\angle ADB= 25° \: and \: \angle ACD = 54° \) Calculate \(\angle BCA \) and \(\angle ABD. \)

**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*}

In the given figure, O is the centre of the circle. \(\angle OAB = 30° \: and \: \angle OCB=40°.\) Find \(\angle AOC.\)

**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*}

In the adjoining figure, ABCD is a cyclic quadrilateral. AB is a diameter. If \(\angle DBA =40°\), find the value of x and y.

**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*}

In the given figure, find the value of x and y.

**Solution:**

\(\angle VWZ = 58° \\ \angle VOZ = 2 \angle VWZ \: [\because \text{The central angle is double of inscribed angle}] \\ \: \: \: \: x° = 2 \times 58° \\ \therefore x = 116°\\ \: \\ y + 58° = 180° \: [\because \text{ sum of the opposite angles of the cyclic quadrilateral}]\\ y = 180° - 58°\\ \therefore x= 122°\)

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