{"id":170569,"date":"2024-05-29T10:17:55","date_gmt":"2024-05-29T04:47:55","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170569"},"modified":"2024-05-29T14:32:39","modified_gmt":"2024-05-29T09:02:39","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-ex-3-1","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-ex-3-1\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.1"},"content":{"rendered":"

Well-structured ML Aggarwal Class 11 Solutions<\/a> Chapter 3 Trigonometry Ex 3.1 facilitate a deeper understanding of mathematical principles.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.1<\/h2>\n

Very short answer\/objective questions (1 to 5) :<\/span><\/p>\n

Question 1.
\nDraw diagrams for the following angles:
\n(i) – 135\u00b0
\n(ii) 740\u00b0.
\nIn which quadrant do they lie ?
\n(iii) Find another positive angle whose initial and final sides are same as that of – 135\u00b0, and indicate on the same diagram.
\nSolution:
\n(i) Let OX be the initial side and OP be the terminal side.<\/p>\n

\"ML<\/p>\n

Clearly – 135\u00b0 lies in III rd quadrant.<\/p>\n

(ii) Let OX be the initial side and OP be the terminal side.<\/p>\n

\"ML<\/p>\n

Clearly 740\u00b0 lies in Ist quadrant.<\/p>\n

(iii) Let OX be the initial side and OP be the terminal side.
\nClearly – 135\u00b0 lies in 3rd quadrant.<\/p>\n

\"ML<\/p>\n

Clearly another positive angle whose initial and final sides are same as that of – 135\u00b0 be
\n360\u00b0 – 135\u00b0 = 225\u00b0.<\/p>\n

\"ML<\/p>\n

Question 2.
\nIf \u03b8 lies in second quadrant, in which quadrant the following will lie?
\n(i) \\(\\frac{\\theta}{2}\\)
\n(ii) 2 \u03b8
\n(iii) – \u03b8
\nSolution:
\nGiven \u03b8 lies in second quadrant.
\n\u2234 90\u00b0 \u2264 \u03b8 \u2264 180\u00b0<\/p>\n

(i) \u2234 45\u00b0 \u2264 \\(\\frac{\\theta}{2}\\) \u2264 90\u00b0
\n\u2234 \\(\\frac{\\theta}{2}\\) lies in first quadrant.<\/p>\n

(ii) \u2234 180\u00b0 \u2264 2\u03b8 \u2264 360\u00b0
\n\u2234 2 \u03b8 lies in 3rd and 4th quadrant.<\/p>\n

(iii) – 90\u00b0 \u2265 – \u03b8 \u2265 – 180\u00b0
\n\u21d2 180\u00b0\u2264 – \u03b8 \u2264 – 90\u00b0
\nClearly – \u03b8 lies in 3rd quadrant.<\/p>\n

Question 3.
\nExpress the following angles in radian measure :
\n(i) 240\u00b0
\n(ii) – 315\u00b0
\n(ii\u00ee) 570\u00b0
\nSolution:
\nWe know that,
\n\u03c0 radians = 180\u00b0
\n\u21d2 1\u00b0 = \\(\\frac{\\pi}{180}\\) rad<\/p>\n

(i) 240\u00b0 = 240 \u00d7 \\(\\frac{\\pi}{180}\\) rad
\n= \\(\\frac{4 \\pi}{3}\\) rad<\/p>\n

(ii) – 315\u00b0 = – 315\u00b0 \u00d7 \\(\\frac{\\pi}{180}\\) rad
\n= – \\(\\frac{7 \\pi}{4}\\) rad<\/p>\n

(iii) 570\u00b0 = 570 \u00d7 \\(\\frac{\\pi}{180}\\) rad
\n= \\(\\frac{19 \\pi}{6}\\) rad<\/p>\n

\"ML<\/p>\n

Question 4.
\nExpress the following angles in degree measure :
\n(i) \\(\\frac{5 \\pi}{3}\\)
\n(ii) \\(\\frac{13 \\pi}{4}\\)
\n(iii) – \\(\\frac{24 \\pi}{5}\\)
\nSolution:
\nWe know that,
\n\u03c0 radians = 180\u00b0
\n\u21d2 1 rad = \\(\\frac{180^{\\circ}}{\\pi}\\)<\/p>\n

(i) \\(\\frac{5 \\pi}{3}=\\frac{5 \\pi}{3} \\times \\frac{180^{\\circ}}{\\pi}\\)
\n= 300\u00b0<\/p>\n

(ii) \\(\\frac{13 \\pi}{4}=\\frac{13 \\pi}{4} \\times \\frac{180^{\\circ}}{\\pi}\\)
\n= 585\u00b0<\/p>\n

(iii) \\(-\\frac{24 \\pi}{5}=-\\frac{24 \\pi}{5} \\times \\frac{180^{\\circ}}{\\pi}\\)
\n= – 864\u00b0<\/p>\n

Question 5.
\nExpress the following angles in radian measure :
\n(i) 35\u00b0
\n(ii) 520\u00b0
\n(iii) 40\u00b0 20\u2019
\n(iv) – 37\u00b0 30\u2019.
\nSolution:
\nWe know that,
\n\u03c0 radians = 180\u00b0
\n\u21d2 1\u00b0 = \\(\\frac{\\pi}{180}\\) rad<\/p>\n

35\u00b0 = 35 \u00d7 \\(\\frac{\\pi}{180}\\)
\n= \\(\\frac{7 \\pi}{36}\\) rad<\/p>\n

(ii) 520\u00b0 = 520 \u00d7 \\(\\frac{\\pi}{180}\\)
\n= \\(\\frac{26 \\pi}{9}\\) radians<\/p>\n

(iii) 40\u00b0 20\u2019 = 40\u00b0 + 20\u2019
\n= 40\u00b0 + \\(\\left(\\frac{20}{60}\\right)^{\\circ}\\)
\n= 40\u00b0 + \\(\\left(\\frac{1}{3}\\right)^{\\circ}\\)
\n= \\(\\left(\\frac{121}{3}\\right)^{\\circ}=\\left(\\frac{121}{3} \\times \\frac{\\pi}{180}\\right)\\) rad
\n= \\(\\frac{121 \\pi}{540}\\) rad<\/p>\n

(iv) – 37\u00b0 30′ = – (37\u00b0 + 30\u2019)
\n= – \\(\\left[37^{\\circ}+\\left(\\frac{30}{60}\\right)^{\\circ}\\right]\\)
\n[\u2235 1\u00b0 = 60\u2019
\n1\u2019 = \\(\\left(\\frac{1}{60}\\right)^0\\)]
\n= \\(\\left[37^{\\circ}+\\left(\\frac{1}{2}\\right)^{\\circ}\\right]\\)
\n= – \\(\\left(\\frac{75}{2}\\right)^{\\circ}\\)
\n= \\(\\left[\\frac{75}{2} \\times \\frac{\\pi}{180}\\right]\\)
\n= – \\(\\frac{5 \\pi}{24}\\) rad<\/p>\n

\"ML<\/p>\n

Short answer questions (5 to 10) :<\/span><\/p>\n

Question 6.
\nFind the degree measures corresponding to the following radian measures :
\n(i) 6
\n(ii) \\(\\frac{3}{4}\\)
\n(iii) – 3
\nSolution:
\nWe know that,
\n\u03c0 radians = 180\u00b0<\/p>\n

(i) 6 rad = 6 \u00d7 \\(\\frac{180^{\\circ}}{\\pi}\\)<\/p>\n

\"ML<\/p>\n

(ii) \\(\\frac{3}{4}\\) rad
\n= \\(\\left(\\frac{3}{4} \\times \\frac{180^{\\circ}}{\\pi}\\right)\\)<\/p>\n

\"ML<\/p>\n

(iii) – 3 rad<\/p>\n

\"ML<\/p>\n

Question 7.
\nA wheel makes 360 revolutions in a minute. Through how many radians does it turn in one second ?
\nSolution:
\nSince the angle traced out by wheel in one revolution = 2\u03c0 rad
\n\u2234 angle traced out by wheel in 360 revolutions in 1 minute = (2\u03c0 \u00d7 360) rad
\nThus, angle traced out by wheel in 360 revolutions in 1 second = \\(\\left(\\frac{2 \\pi \\times 360}{60}\\right)\\) rad
\n= 12\u03c0 radians<\/p>\n

\"ML<\/p>\n

Question 8.
\nFind the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of length :
\n(i) 10 cm
\n(ii) 15 cm.
\nSolution:
\n(i) Given radius of pendulum = r = 75 cm
\narc length that tip of pendulum describes = s = 10 cm
\nWe know that \u03b8 = \\(\\frac{s}{r}\\)
\nwhere \u03b8 radians be the angle through which the pendulum swings.
\n\u03b8 = \\(\\frac{10}{75}\\) rad
\n= \\(\\frac{2}{15}\\) rad<\/p>\n

(ii) Here s = 15 cm ; r = 75 cm
\n\u2234 \u03b8 = \\(\\frac{s}{r}\\)
\n= \\(\\frac{15}{75}=\\frac{1}{5}\\) rad<\/p>\n

Question 9.
\nFind the radius of the circle in which a central angle of 45\u00b0 makes an arc of length 187 cm. (use \u03c0 = \\(\\frac{22}{7}\\)).
\nSolution:
\nLet r cm be the radius of circle
\nHere, arc length = s
\n= 187 cm
\nand \u03b8 = 45\u00b0
\n= 45 \u00d7 \\(\\frac{\\pi}{180}\\)
\n= \\(\\frac{\\pi}{4}\\) rad
\n[\u2235 \u03c0 radians = 180\u00b0]
\nWe know that,
\n\u03b8 = \\(\\frac{s}{r}\\)
\n\u21d2 r = \\(\\frac{s}{\\theta}\\)
\n= \\(\\left(\\frac{187 \\times 4}{\\pi}\\right)\\) cm
\n\u21d2 r = \\(\\left(\\frac{187 \\times 4 \\times 7}{22}\\right)\\)
\n= 238 cm<\/p>\n

Question 10.
\nFind the length of an arc of a circle of diameter 20 cm which subtends an angle of 45 at the centre.
\nSolution:
\nGiven diameter of circle = 20 cm
\n\u2234 radius of circle = r = 10 cm
\nLet s be the required arc length of a circle with radius 10 cm
\nand \u03b8 = 45\u00b0 = \\(\\frac{\\pi}{4}\\) rad
\nWe know that,
\n\u03b8 = \\(\\frac{s}{r}\\)
\n\u21d2 s = 10 \u00d7 \\(\\frac{\\pi}{4}\\)
\n= \\(\\left(\\frac{10 \\times 22}{7 \\times 4}\\right)\\) cm
\n= \\(\\frac{5 \\pi}{2}\\) cm<\/p>\n

\"ML<\/p>\n

Long answer questions (11 to 20) :<\/span><\/p>\n

Question 11.
\nAn engine is travelling along a circular railway track of radius 1500 metres with a specd of 60 km\/h. Find the angle in degrees turned by the engine in 10 seconds.
\nSolution:
\nGiven radius of circular railway track = 1500 m
\ngiven speed of engine = 60 km \/ h
\nThus, the distance covered by engine in 1 hour = 60 km
\n\u2234 distance covered by engine in 1 second = \\(\\frac{60 \\times 1000}{60 \\times 60}\\)
\nThus disiance covered by engine in 10 seconds = s
\n= \\(\\frac{60 \\times 1000}{60 \\times 60}\\)
\n= \\(\\frac{500}{3}\\) m
\nLet \u03b8 in radians be the angle turned by engine in 10 seconds
\nThen \u03b8 = \\(\\frac{s}{r}\\)
\n\u03b8 = \\(\\frac{500}{3 \\times 1500}\\)
\n= \\(\\frac{5}{45}=\\frac{1}{9}\\) rad
\n[\u2235 \u03c0 rad = 180\u00b0]<\/p>\n

Question 12.
\nIf the arcs of the same length in two circles subtend angles of 65\u00b0 and 110\u00b0 at their respective centres, find the ratio of their radii.
\nSolution:
\nLet r1<\/sub> and r2<\/sub> be the radii of two given circles
\nand let their arcs of same length say s and subtends an angles 65\u00b0 and 110\u00b0 at respective centres.
\nNow 65\u00b0 = 65 \u00d7 \\(\\frac{\\pi}{180}=\\frac{13 \\pi}{36}\\)
\n[\u2235 \u03c0 rad = 180\u00b0]
\nand 110\u00b0 = 110 \u00d7 \\(\\frac{\\pi}{180}=\\frac{11 \\pi}{18}\\)
\nWe know that
\n\u03b8 = \\(\\frac{s}{r}\\)
\n\u21d2 r = \\(\\frac{s}{\\theta}\\)
\ni.e. r1<\/sub> = \\(\\frac{s}{13 \\pi}\\) \u00d7 36 ………………..(1)
\nand r2<\/sub> = \\(\\frac{s}{11 \\pi}\\) \u00d7 18 …………………(2)
\nOn dividing (1) and (2) ; we have
\n\\(\\frac{r_1}{r_2}=\\frac{36}{13} \\times \\frac{11}{18}\\)
\n= \\(\\frac{22}{13}\\)
\nThus,
\nr1<\/sub> : r2<\/sub> = 22 : 13<\/p>\n

\"ML<\/p>\n

Question 13.
\nLarge hand of a clock is 21 cm long. flow much distance does its extremity move in 20 minutes ?
\nSolution:
\nThe angle traced out by large hand of clock in 60 minutes = 2 \u03c0 rad
\n\u2234 angle traced out by large hand of clock in 20 minutes = \u03b8
\n= \\(\\frac{2 \\pi}{60}\\) \u00d7 20
\n= \\(\\frac{2 \\pi}{3}\\) radians
\ngiven radius of clock = length of large hand = r = 21 cm
\nLet s be the required length of the arc moved by the tip of minute hand
\nThen s = r\u03b8
\n= (21 \u00d7 \\(\\frac{2 \\pi}{3}\\))
\n= 14 \u03c0
\n= (14 \u00d7 \\(\\frac{22}{7}\\)) cm
\n= 44 cm<\/p>\n

Question 14.
\nFind the angles in degrees through which a pendulum swings if Its length is 50 cm and the tip describes an arc of length :
\n(i) 10 cm
\n(ii) 16 cm
\n(iii) 26 cm (use \u03c0 = \\(\\frac{22}{7}\\))
\nSolution:
\n(i) The pendulum describes a circle of radius 50 cm and its tip describes an arc of length 10 cm.
\nLet \u03b8 radians be the angle through which the pendulum swings.
\nHere r = 50 cm ;
\ns = 10 cm
\nThen \u03b8 = \\(\\frac{s}{r}\\)
\n= \\(\\frac{10}{50}=\\frac{1}{5}\\) rad
\n= \\(\\frac{1}{5} \\times \\frac{180^{\\circ}}{\\pi}\\)
\n= \\(\\left(\\frac{1}{5} \\times \\frac{180}{22} \\times 7\\right)^{\\circ}\\)
\n= \\(\\left(\\frac{126}{11}\\right)^{\\circ}=\\left(11 \\frac{5}{11}\\right)^{\\circ}\\)
\n= 11\u00b0 + \\(\\left(\\frac{5}{11} \\times 60\\right)^{\\prime}\\)
\n= 11\u00b0 + \\(\\left(27 \\frac{3}{11}\\right)^{\\prime}\\)
\n= 11\u00b0 + 27\u2019 + \\(\\left(\\frac{3}{11} \\times 60\\right)^{\\prime \\prime}\\)
\n= 11\u00b0 + 27\u2019 + 16″
\n= 11\u00b0 27\u2019 16″<\/p>\n

(ii) The pendulum describes a circle of radius 50 cm and its tip describes an arc of length 16 cm.
\nLet \u03b8 (in rad) be the angle through which the pendulum swings.
\nHere r = 50 cm ;
\ns = 16 cm<\/p>\n

\"ML<\/p>\n

(iii) The pendulum describes a circle of radius 50 cm and its tip describes an arc of length 26 cm.
\nLet \u03b8 (in radians) be the angle through which the pendulum swings.
\nHere r = 50 cm ;
\ns = 26 cm
\nThen \u03b8 = \\(\\frac{s}{r}\\)
\n= \\(\\frac{26}{50}=\\frac{13}{25}\\) rad
\n= \\(\\left(\\frac{13}{25} \\times \\frac{180}{22} \\times 7\\right)^{\\circ}\\)
\n\u03b8 = 29\u00b0 + \\(\\left(\\frac{1638}{55}\\right)^{\\circ}=\\left(29 \\frac{43}{55}\\right)^{\\circ}\\)
\n= 29\u00b0 + \\(\\left(\\frac{43}{55} \\times 60\\right)^{\\prime}\\)
\n= 29\u00b0 + \\(\\left(46 \\frac{10}{11}\\right)^{\\prime}\\)
\n= 29\u00b0 + 46′ + \\(\\left(\\frac{10}{11} \\times 60\\right)^{\\prime \\prime}\\)
\n= 29\u00b0 46′ 55″<\/p>\n

\"ML<\/p>\n

Question 15.
\nFind the length of an arc of a circle of radius 75 cm that spans a central angle of measure 126\u00b0. Take \u03c0 = 3.1416.
\nSolution:
\nLet s be the length of arc of circle of radius 75 cm that subtends an angle of 126\u00b0 at the centre.
\nHere, r = 75 cm ;
\n\u03b8 = 126\u00b0
\n= (126 \u00d7 \\(\\frac{\\pi}{180}\\))
\n= \\(\\frac{126 \\times 3.1416}{180}\\)
\n[\u2235 \u03c0 radians = 180\u00b0]
\n\u2234 \u03b8 = 2.19912 rad.
\nThen \u03b8 = \\(\\frac{s}{r}\\)
\n\u21d2 s = r\u03b8
\n= (75 \u00d7 2.19912) cm
\n= 164.934 cm<\/p>\n

Question 16.
\nFind the angle \u00a1n radians between the hands of a clock at 3.30 A.M.
\nSolution:
\nHere, Given time be 3.30 A.M.
\nHour hand and minute hand were together at 12 night.<\/p>\n

\"ML<\/p>\n

Time of rotation for hour hand = 3 hour 30 minute
\n= (3 + \\(\\frac{30}{60}\\)) hr
\n= \\(\\frac{7}{2}\\) hr
\nangle trace out by hour hand in 12 hours = 360\u00b0
\n\u2234 angle trace out by hour hand in \\(\\frac{7}{2}\\) hr
\n\u03b8 = \\(\\left(\\frac{360}{12} \\times \\frac{7}{2}\\right)^0\\) = 105\u00b0
\nat 3.00 A.M., i.e. after 3 hours, the minute hand is at 12.
\nthus the angle trace out by minute hand at 3.00 A.M. be 0.
\nNow angle trace out by minute hand in 60 minutes = 360\u00b0
\n\u2234 angle trace out by minute hand in 30 minutes \u03a6 = \\(\\left(\\frac{360}{60} \\times 30\\right)^{\\circ}\\) = 180\u00b0
\nThus, required angle between the hands of clock at 3.30 A.M. = \u03a6 – \u03b8
\n= 180\u00b0 – 105\u00b0 = 75\u00b0
\n= 75 \u00d7 \\(\\frac{\\pi}{180}\\) rad
\n= \\(\\frac{5 \\pi}{12}\\) rad<\/p>\n

\"ML<\/p>\n

Question 17.
\nThe circular measures of two angles of a triangle are \\(\\frac{1}{2}\\) and \\(\\frac{1}{3}\\). Find the third angle in degree measure. Take \u03c0 = \\(\\frac{22}{7}\\).
\nSolution:
\nSince the sum of all angles of triangle = 180\u00b0 or \u03c0 rad
\nor Now \\(\\frac{1}{2}\\) rad = \\(\\left(\\frac{1}{2} \\times \\frac{180}{\\pi}\\right)^{\\circ}\\)
\n[\u2235 \u03c0 rad = 180\u00b0]
\n\u2234 \\(\\frac{1}{2}\\) rad = \\(\\left(\\frac{90 \\times 7}{22}\\right)^{\\circ}\\)
\n= \\(\\left(\\frac{315}{11}\\right)^{\\circ}\\)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 18.
\nThe difference between two acute angles of a right angled triangle is \\(\\frac{\\pi}{5}\\) in radian measure. Find these angles in degrees.
\nSolution:
\nLet the other two required angles in radians be \u03b8 and \u03a6.
\nSince the sum of angles of triangle be \u03c0 rad.
\n\u2234 \u03b8 + \u03a6 + 90\u00b0 = 180\u00b0
\n\u03b8 + \u03a6 = 90\u00b0
\n= \\(\\frac{\\pi}{2}\\) rad
\nalso given,
\n0 – \u03a6 = \\(\\frac{\\pi}{5}\\) rad
\nOn adding (1) and (2) ; we have<\/p>\n

\"ML<\/p>\n

Question 19.
\nThe angles of a triangle are in A.P. and the greatest angle \u00a1s double the least. Find all the angles in circular measure.
\nSolution:
\nLet the required angles of triangle in A.P. be (a – d)\u00b0, a\u00b0 and (a + d\u00b0).
\nThen, a – d + a + a + d = 180
\n\u21d2 a = 60
\nNow least angle = (a – d)\u00b0 = (60 – d)\u00b0
\nand greatest anale = (a + d)\u00b0 = (60 + d)\u00b0
\naccording to question ; we have
\n(a + d)\u00b0 = 2 (a – d)\u00b0
\n\u21d2 a + d = 2a – 2d
\n\u21d2 a = 3d
\n\u21d2 3d = 60
\n\u21d2 d = 20
\nThus the required angle are (60 – 20)\u00b0, 60\u00b0, (60 + 20)\u00b0 i.e. 40\u00b0, 60\u00b0, 80\u00b0
\nsince \u03c0 rad = 180\u00b0
\n\u21d2 1\u00b0 = \\(\\frac{\\pi}{180}\\) rad
\ni.e. angles in radians are ;
\n(40 \u00d7 \\(\\frac{\\pi}{180}\\)) ; 60 \u00d7 \\(\\frac{\\pi}{180}\\), 8o \u00d7 \\(\\frac{\\pi}{180}\\)
\ni.e. \\(\\frac{2 \\pi}{9}, \\frac{\\pi}{3}, \\frac{4 \\pi}{9}\\)<\/p>\n

\"ML<\/p>\n

Question 20.
\nEstimate the diameter of the sun supposing that it subtends an angle of 32′ at the eye of an observer. Given that the distance of the sun is 91 \u00d7 106<\/sup> km.
\nTake \u03c0 = \\(\\frac{22}{7}\\).
\nSolution:
\nLet r km be the radius of Sun that subtends an angle of 32′ at the eye of an observer.
\nGiven distance of sum from observer r = 91 \u00d7 106<\/sup> km
\nHere \u03b8 = 32\u2019
\n= \\(\\left(\\frac{32}{60}\\right)^{\\circ}=\\left(\\frac{8}{15}\\right)^{\\circ}\\)
\n[\u2235 1\u00b0 = 60 rad
\n\u21d2 1′ = \\(\\left(\\frac{1}{60}\\right)^{\\circ}\\)]
\n\u21d2 \u03b8 = \\(\\left(\\frac{8}{15} \\times \\frac{\\pi}{180}\\right)\\) rad
\n= \\(\\frac{2 \\pi}{675}\\) rad
\n[\u2235 \u03c0 rad = 180\u00b0]
\nWe know that,
\n\u03b8 = \\(\\frac{s}{r}\\)
\n\u21d2 s = r\u03b8
\n= (91 \u00d7 106<\/sup> \u00d7 \\(\\frac{2 \\pi}{675}\\)) km
\n\u21d2 s = 847407.4 km<\/p>\n

\"ML<\/p>\n

Hence the required diameter of the sun = arc length AB = 847407.4 km.<\/p>\n","protected":false},"excerpt":{"rendered":"

Well-structured ML Aggarwal Class 11 Solutions Chapter 3 Trigonometry Ex 3.1 facilitate a deeper understanding of mathematical principles. ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.1 Very short answer\/objective questions (1 to 5) : Question 1. Draw diagrams for the following angles: (i) – 135\u00b0 (ii) 740\u00b0. In which quadrant …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170569"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170569"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170569\/revisions"}],"predecessor-version":[{"id":170584,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170569\/revisions\/170584"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170569"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170569"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170569"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}