in the given figure the measure of oqp is

∠OAP = ∠OBP = 90° (a) 20° But AB + BC + AC = 12 + 8 + 10 = 30 (d) 1.9 (d) 2.5 cm = ∠1 + ∠2 = 90° …[OD is equally inclined to the tangents 10 = 14 – 2r Solution: A noise figure meter also has limitations. ∠1 = ∠2 CQ = CR ∴ AB = CD (Hence proved), Question 32. QD = QD [common side] PT and PQ are tangents from P The … ∴ ∠1 = ∠2 …[Equal angles opposite equal sides RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. CQ = CR = 6 – r ∠ABC, ∠BCD and ∠ADC respectively. (a) PA and PB are the tangents to the circle from P and ∠APB = 80° ∠OCA = 90° ..[Tangent is ⊥ to the radius through the point of contact (a) \(\frac { 3 }{ 2 }\) OQ = 25 cm AR = 23 – 5= 18 cm => 110° + ∠PTQ = 180° BD = BE AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. A is the centre of the circle. ∠DAC = ∠DBC = 45° To prove: AB passes through centre O or AOB is diameter of the circle. Solution: (a) 11 ⇒ ∠PRQ = 240 / 2 = 120 o. Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm In the figure, O is the centre of a circle of radius 5 cm. Let TP = x cm and TR = y cm B is the mid point of arc (ABC) ∴∠6 + 28 = 180° AB = \(10 \sqrt{2}\) cm, Question 5. OP and OT are joined Draw rays, lines, & line segments. ⇒ cos θ = \(\frac{r}{2 r}\) ⇒ cos θ = \(\frac{1}{2}\) = ED + DF + EF = ED + DK + EM + EF => 625 = r² + 576 \(4 \sqrt{3 x(x+14)}\) = 2(2x + 28) ∠OPQ = \(\frac{1}{2}\)(∠QPR) ..[Tangents drawn from an external point are equal ∴ r= \(\sqrt{625}\) = 25 cm, Question 21. Area of ∆ABC = Area of ∆AOB + ar. Solution: To prove: PR = RQ In the figure, if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then AF + BF + CD = AE + BD + CE (b) 18 cm OP, OQ and PQ are joined In the given figure, AP, AQ and BC are tangents to the circle. to a circle are equal PA = AE = x cm …[Tangent drawn from an external point In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. (b) 5 cm (b) In the figure, AP = PB (b) 2 Solution: To prove: OP ⊥ XY 14 – 9 = AD ⇒ AD = 5 cm. In ∆QOR, If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm). OA = 6 cm, OB = 3 cm and AP = 10 cm ∠BOC + ∠DOC = 180° 125° + ∠DOC = 180° ∠DOC = 18 ∠OTP – ∠OSP …[Each 90° 90° + 46° + 90° + ∠QOR = 360° (a) √7 cm Using a noise figure meter is the most straightforward way to measure noise figure. ∠TAP = ∠TPA = x (say) But AP = AQ = 12 cm (tangents from A to the circle) (2012D) ∠APO = ∠CPO = 30° Solution: ∠QOR + ∠QRO + ∠OQR = 180° …[∆ Rule ⇒ 30° + 2∠PQR = 180° Then length of chord BC which touches the inner circle at P is equal to Let AF = AE = x (d) 60° OQ ⊥ QP or ∠OQP = 90° ∠OQB = 90° ∠OPQ = ∠OQP (d) 6 cm (b) 13 cm (d) 90° Also, OA ⊥ PA z = 15 – 12 = 3 (a) a = 30°, b = 60° OR = 5 cm, Question 15. AB = AC (tangents drawn from A to the circle) [Tangent is ⊥ to the radius through the point of contact In the figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is [CBSE2014] (i) ∠6 + ∠8 = 180° AQ = 18 cm Find another word for measure. (b) TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110° ∴ ∠POQ = 120° …[∠POQ = 180o – (30° + 30°) Radius of circle is 5 cm, Question 38. Adding (i), (ii) and (iii), we get ⇒ 2x + 2y + 2z = 10 + 8 + 12 (b) 26 cm Solution: TP = TQ ∠QPB = 50° Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively. (d) 15 cm (∠5 + ∠7) + (∠6 + ∠8) = 360° (2011OD) (2013D) (c) 5 cm At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. ∠AOB + ∠COD + 180o = 360° … [From (i) Now in ∆POQ ⇒ 24x = 144 – 64 Also, join OB, which is perpendicular to AC. (d) 8 cm Calculate the nominal equivalent resistance, tolerance (5%), minimum and maximum circuit resistance for the circuit shown in Figure 3, and record in Table 6. (2013D) 55° + ∠TPQ = 90° AD = BD EM = EK = 9 cm Let AD = a, DB = b and EC = c, then Calculate the numerical value of x. (a) 1 cm ET = OT – OE …[Pythagoras’ theorem x + 3 = 10 Distance of A from the centre O = 5 cm, Question 4. (b) ∠OQP = 90° But AP = PB (given) ∴PQ || RS, Question 19. ∠AOB + 80o = 180° So, when the degree measure of the angle at the centre is 1°, then area of sector is 1 / 360 ° ×πr 2 . (b) 7.5 cm OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)² 3511 Views. But AB = AC … [Given In the figure, PQ and PR are two tangents to a circle with centre O. Join OQ and OA Solution: O is the centre of the circle ∠QPR = 60° Solution: Const. But OP = OQ : Take a point Q on XY other than P and join to OQuestion CR = CQ …. (c) In the figure, PA and PB are the tangents to the circle with centre O ∠1 + ∠2 + 100° = 180° Solution: (d) \(\frac { 1 }{ 4 }\) (5)² = (3)² + AR² ∠1 = ∠2 …[TP and TQ are equally inclined to OT From the given figure below, find the azimuths for the bearings A: 242 29 Measures: N02 34'26"E. Rebar Found/Cap PLS 1256 27.00 Ofset Found/Cap 5756 Measures: N97 0531 260.51 Dirt Drive 182.54 29 272 54 29" 87° 05 31 267°05'31 From the given figure below, find the azimuths for the bearings B: 182° 34' 26" 358° 33' 34" 2° 34' 2° 34' 26" … ∴ Reflex ∠POQ = 2∠PRO ⇒ x2 + 64 = 144 + x2 – 24x In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is Solution: (a) 11 cm ∠ABQ = ∠BQR = 70° [alternate angles] (b) 7 cm 2∠OAB = 50° Question 41. (i) ∠AOD + ∠BOC = 180° Similarly, If ∠PAB = 50°, then find ∠AOB. Solution: of ∆BOC + ar. ∴ OP ⊥ XY … [Shortest side is ⊥, Question 43. (c) In the figure, AP is the tangent to the circle with centre O such that Tangent at any point of a circle is perpendicular to the radius through the point of contact. OPBQ is a square PR = RQ … [Tangents drawn from an external point are equal Prove that the tangents drawn at the ends of a diameter of a circle are parallel. Thus 29+OQP+0c =180c +OQP=45c Thus (b) is correct option. RC = QC = 7 cm AS and AP are tangents from A (b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively AC = \(\frac{1}{2}\) (AB) … [⊥ from the centre bisects the chord (d) 50° Proof: ∠ACO = 90° … [Tangent is ⊥ to the radius through the point of contact Perimeter of ∆EDF PQ and RS are the tangents drawn to the circles ∴ OT bisects PQ CR = CO To prove: BD = CD ⇒ ∠TPQ = 90° – 55° = 35°, Question 4. ∴ AC = CB … [⊥ from the centre bisects the chord, Question 22. twice the radius of the inscribed circle). Solution: ∠RQS = 180° – 150o = 30°, Question 46. y + z = 8 cm …(ii) ∠1 + ∠BAT = 90° Now in right ∆OQP [Tangent is ⊥ to the radius through the point of contact] Solution: (b) QR Prove that the point of contact P bisects the base BC. (d) 15 cm QR = 3.8 + 3.8 = 7.6 cm, Question 43. PQ = PT (tangents from P to the circle) AQ = BR …. (c) a = 40°, b = 50° CQ and CR are tangents to the circle ∠OPR = 90° We have, OP = 5 cm, OT = 13 cm OQ² = OP² + PQ² (Pythagoras Theorem) Solution: (d) First, draw a circle of radius 5 cm having centre O. Part II: Join OP, OR, OQ, OA, OB and OC EB = ED …(ii) ∠OTP = 90° Also, OP is a bisector of line ∠APC (a) 20 cm Similarly PT and PR are tangents ⇒ 2(x + y + z) = 30 ∆OQP, cos 60° = \(\frac{PQ}{PO}\) Given: OD = 3 cm; OE = 3 cm; OF = 3 cm ar(∆ABC) = 54 cm2 If SQ = 6 cm and QR = 4 cm, then OR = (d) 10 cm Solution: In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. y + 5 = 8 (d) 25 cm In the larger circle, AB is a chord and OC ⊥ AB. Given: PT and PS are tangents from an external point P to the circle with centre O. If ∠QPR= 46, then ∠QOR equals (a) 67° (b) 134° (c) 44° (d) 46° Solution: (b) ∠OQP = 90° [Tangent is ⊥ to the radius through the point of contact] ∠ORP = 90° ∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.] (c) 44° Draw OM || CD. In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. = 3 + 7 = 10 cm, Question 15. Ellipsoid DA and DC are tangents to the first circle from D 2(∠1 + ∠2 + 25 + 26) = 360° => ∠OPQ + 60° = 90° In the figure, two equal circles, with centres 0 and O’, touch each other A at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. (2011OD) ∠OPQ = ∠OPT – ∠QPT CR = CQ = 3 cm Answer/ Explanation. (2015D) Up Next. (b) 2√7cm The tangents at P and Q intersect at a point T. Find the length of TP. In the figure, a ∆ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively 6 cm 9 cm of lengths 6 cm and 9 cm. OQ ⊥ AB and OQ bisects AB ∆BPO = ∆BQO … [SSS Congruency rule (b) a = 35°, b = 55° (c) In the figure, PT is the tangent to the circle with centre O. (2013OD) (b) 120° (a) AD = AB + BC + CA ∠AOB = 180° – 80° = 100°, Question 13. ∠PCA = ∠CBA …[Angle in the alternate segment Hence, the length of each tangent is 3√3 cm, Question 46. (d) 5 cm OT ⊥ PT ∠OAB = ∠OBA … [Angles opposite equal sides BD = BE = y …[Two tangents drawn from and an external point are equal AD = AE ……(i) (a) 60° (2011OD) ∠2 + ∠3 = 90° + 90° = 180° ∠QDA = ∠QDB [each 90°] (ii) ∠5 + ∠7 = 180° => (25)² = r² + (24)² (b) 100° (c) 9 cm In the figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. AB = 29 cm, AD = 23 cm, DS = 5 cm asked Dec 27, 2017 in Class IX Maths by navnit40 (-4,939 points) In the given figure, O is the centre of the circle. Measure each section separately and calculate the square footage of each. Solution: OF = OF …[Common Chord AC || BP, Given. (d) 90° => ∠QOP + 120° = 180° CF = 10 – x = 10 – 7 = 3 cm, Question 31. y = 8 – 5 In ∆ABC, Solution: (c) 24 cm Solution: : Join OC Now, ∠OQP + ∠OPQ + ∠POQ = 180° If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is [CBSE 2012] AB + CD = AD + BC AB + CD = (AS + DS) + (BQ + CQ) ∴ ∠AOB + ∠COD = 360° – 180o = 180° …(proved), 2nd method: AB = 12 cm, BC = 8 cm and AC = 10 cm In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. and AD = BC ..(i) Solution: Solution: In the figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. AB + CQ = AC + BQ (b) 50° Solution: But OP is the bisector of ∠AOB BC = BQ + QC In figure, if O P ∥ R S, ∠ O P Q = 1 1 0 o and ∠ Q R S = 1 3 0 o, then ∠ P Q R is equal to: (c) 3 cm (b) 6 cm (a) √91 cm (c) 30 cm² (a) 4 In the figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. (c) 70° If ∠QPT = 60°, find ∠PRQ. OA² = OP² + AP² ∠OQP = ∠OPQ = 30° … [In a ∆, equal sides have equal ∠s opp. Our mission is to provide a free, world-class education to anyone, anywhere. Radius = R Solution: ∠AOB + ∠OAB + ∠OBA = 180° … [∆ rule (2016D) AD = 7 cm, Question 21. (c) 2√3 cm If ∠PRQ = 120°, then prove that OR = PR + R. (2015OD) AP = AS ……(i) (Tangents drawn from an external point are equal in length Build a square around the circle and construct the octagon from that. In ∠TOS, r = 7 In right ∆OAQ, 8. Solution: To find: = \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) FH = FM Sometimes it is also known as the discrete density function. (b) 90° => ∠PTQ = 180° – 110° = 70°, Question 6. But ∠QOP + ∠POR = 180° (Linear pair) [Tangent at any point of circle is perpendicular to radius throughly the point of contact] ⇒ PT = \(\sqrt{144}\) V = 1.84 ´ 10 6 J m-3 K-1 => ∠OPB = 90° ∠PAB = ∠PBA … (i) …(Angles opposite to equal sides OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle Proof. (b) 7.6 (b) 134° ∴ ∆AO’D ~ ∴AOC …(AA similarity (a) 30° => BC² = 25 – 16 = 9 BC = 8 + 6 = 14 cm ⇒ 240 o = 2 ∠PRQ. CF = CE PA = PB = 12 cm …(i) (2014OD) ∠PAB = ∠PBA = ED + DH + FH + EF Solution: AB and AC are tangents ∴ AF = AE … (i) This happens with every point on the line XY except the point P. OQ is radius (d) 15 In the given figure, O is the centre of the circle. Given, ∠POQ = 130° In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. (c) 30° If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to => 60° + 90° + ∠OPQ = 180° = AP = PB = 4 cm To find : The measure of ∠POQ. ∠ORP = 90° (d) In the figure, ∆ABC is the circumscribed a circle ⇒ ∠OPQ = 90° – 60° = 30° AQ and AR the tangents to the circle AQ = AR = 4 cm PT = 3.8 cm 2AP = Perimeter of ∆ (c) 14 cm Solution: AB2 = 102 + 102 (b) 45° In ∆OPQ, OP = OQ => QP = 12 cm In ∆QDA and ∆QDB ∠3 = 90° …[Given (5)² = AQ² + (3)² AB is joined Etc. (a) In the figure, a circle touches the sides of a quadrilateral ABCD Hence, ∠QAB = 70° [from Eq. O’R = 13 cm [sum of cointerior is 180°] PQ = 10 cm = x + y = 10 …(i) ∠1 = ∠2 … [From (i) & (ii) (d) In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm But ∠POQ + ∠PTQ = 180° Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. In ABOC, 120° + ∠OTS + ∠OTS = 180° … [From (i) = (5)2 + (12)2 Solution: = 12 cm Construct a regular octagon given the perpendicular distance from one side of the octagon to the opposite (i.e. Then OT = (y + 6) cm 130° + 2∠OAB = 180° … [From (i) Find ∠DOC, ∠DCO and ∠OAB. ∴ AD = 7 cm, BE = 5 cm, CF = 3 cm, Question 28. ⇒ 2∠1 = 80° …[∵ Tangent is I to the radius through the point of contact The probability mass function is often the primary means of defining a discrete probability distribution, and such functions exist for either scalar or … Then the area of the quadrilateral PQOR is [NCERT Exemplar] x = 12 – 5 => 35° = a ∴ ∠3 = ∠4 = 90° AB = AR + BR = 4 + 6 = 10 cm (d) 62\(\frac { 1 }{ 2 }\)° The radius of the circle is [NCERT Exemplar] (2014OD) Let AD = AF = x Join OR and OQ In the figure, if AP = 10 cm, then BP = and ∠TBP = ∠TPB = y (say) (c), Question 51. 90° = ∠ORQ + 75° ⇒ cos θ = cos 60° ⇒ θ = 60° The radius of the circle inscribed in the triangle (in cm) is OT = OS = r … [Given Measure: an action planned or taken to achieve a desired result. (c) 3 cm ∠O’AD = ∠OAC …(Common x + x + x + y + y = 180° (b) In the given figure, 136 = (3)² + BP2 ∴ AB + CD = BC + AD (Hence proved), Question 25. (c) In the figure, AB and AC are the tangents to the circle from A ∴ CD = BD, Question 26. If OA = 15 cm, then AP + AQ = => ∠PTQ + 90° + 90° + 130° = 360° ∠OPB = 90° (d) 3√3 cm (d) 4AD = AB + BC + CA In the figure DE and DF are tangents from an external point D to a circle with centre A. ∴ ∠DOE = 90° …(proved), Question 50. ∠ABQ = 70° [∠BQR = 70°] QB = BP = AQ ∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180° x = \(\frac{80}{24}=\frac{10}{3}\) cm = AC = radius = 4 cm, Question 10. ∠POQ + ∠QPO = 90° ∠PRO = \(\frac{1}{2}\) ∠PRQ = \(\frac{1}{2}\) × 120° = 60° OQ ⊥ PQ (d) 32.5 cm² AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm 4 + CR =11 cm If ∠QPR = 46°, then calculate ∠QOR. Write the measure of ∠OAB. ∴ ∠1 + ∠4 + ∠3 + ∠2 =180°…(iii) ..[From (1) & (ii) BA = 4 cm ∆AET, (d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent In the figure, if AP = PB, then of ∆AOC ∠5 + ∠7 = 180°. = (12 – x) cm Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Radius of the circle = 11 cm, Question 39. => ∠COD = 180° – ∠AOB = 180° – 125° = 55°, Question 52. Similarly QA = QB ⇒ OR = PR + QR (∵ PR = RQ) …(Hence proved), Question 35. AB + CD = (BQ + CQ) + (AS + DS) QR is the common tangent ⇒ ∠PRQ + ∠RPQ + ∠POR = 180°…[∆ Rule AC || BC (d) AB = BC 50° + 50° + ∠APB = 180° ∠TQP = 60° AB = AE + EB = AE + AE = 2AE = 2x : OT = OS …[radii of same circle In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. ∠1 = ∠2 => 2 ∠PBA = 150° => AR = 4 cm In given figure, CP and CQ are tangents to a circle with centre O. OP² = OB² + BP² a + b + b + c + c + a = 30 In figure, a quadrilateral ABCD is drawn to circum- DA scribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, RA and S respectively. ∴ QM = 3 cm, RN = 5 cm, PL = 7 cm, Question 40. => QP² = 169 – 25 = 144 = 12² Solution: The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Find the Value of 'X' Giving Reasons - Mathematics If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is (NCERT Exemplar) 1st method: ⇒ 2∠3 = 180° ⇒ ∠3 = 90° r2 = (7)2 + (24)2 (c) 2 OP is the radius and PR is the tangent OPR = 90° 2∠1 = 110° ⇒ ∠1 = 55° (c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. PR = 12 cm = x + z = 12 …(iii) Now, in isosceles triangle POQwe have +++POQO++ PQ OQP =180c Equal sides subtend equal angles in isosceles triangle. PQ = √119, Question 2. But these are corresponding angles, ∠TQP = ∠TPQ = 60° (2016OD) Putting the value of z in (ii) & (iii), From (ii) But PR = PQ (tangents from P) AR = AQ (tangents to the circle from A) In the figure, there are two concentric, circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. => r = 2 AP and AR are the tangents from A to the circle Solution: ∠ABO + ∠ACO + ∠BAC + ∠BOC = 360° ⇒ ∠QSR = \(\frac{150^{\circ}}{2}\) = 750 … [Used ∠SRQ = 75° as solved above Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°.
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