{"id":6978,"date":"2023-05-25T16:49:44","date_gmt":"2023-05-25T11:19:44","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=6978"},"modified":"2023-05-26T17:12:15","modified_gmt":"2023-05-26T11:42:15","slug":"ap-ssc-10th-class-maths-solutions-chapter-11-intext-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-ssc-10th-class-maths-solutions-chapter-11-intext-questions\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry InText Questions"},"content":{"rendered":"

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry InText Questions<\/h2>\n

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 11 Trigonometry InText Questions and Answers.<\/p>\n

10th Class Maths 11th Lesson Trigonometry InText Questions and Answers<\/h3>\n

Do This<\/span><\/p>\n

Identify \u201cHypotenuse\u201d, \u201cOpposite side\u201d and \u201cAdjacent side\u201d for the given angles in the given triangles.\u00a0 (Page No. 271)<\/p>\n

Question 1.
\nFor angle R
\nAnswer:
\n\"AP
\nIn the \u25b3PQR
\nOpposite side = PQ
\nAdjacent side = QR
\nHypotenuse = PR<\/p>\n

\"AP<\/p>\n

Question 2.
\ni) For angle X
\nii) For angle Y\u00a0 \u00a0 \u00a0 \u00a0 (Page No. 271)
\nAnswer:
\n\"AP
\nIn the \u25b3XYZ,
\ni) For angle X
\nOpposite side = YZ
\nAdjacent side = XZ
\nHypotenuse = XY
\nii) For angle Y
\nOpposite side = XZ
\nAdjacent side = YZ
\nHypotenuse = XY<\/p>\n

Question 3.
\nFind (i) sin C (ii) cos C and (iii) tan C in the given triangle.\u00a0 \u00a0 (Page No. 274)
\n\"AP
\nAnswer:
\nBy Pythagoras theorem
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n132<\/sup> = AB2<\/sup> + 52<\/sup>
\nAB2<\/sup> = 169 – 25
\nAB2<\/sup> = \u221a144
\n\u21d2 AB = \u221a144 = 12
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 4.
\nIn a triangle XYZ, \u2220Y is right angle, XZ = 17 cm and YZ = 15 cm, then find (i) sin X (ii) cos Z (iii) tan X.\u00a0 \u00a0 (Page No. 274)
\nAnswer:
\nGiven \u25b3XYZ, \u2220Y is right angle.
\n\"AP
\nBy Pythagoras theorem
\nXZ2<\/sup> = YZ2<\/sup> + XY2<\/sup>
\n172<\/sup> = 152<\/sup> + XY2<\/sup>
\nXY2<\/sup> = 172<\/sup> – 152<\/sup> = 289 – 225
\nXY2<\/sup> = 64
\nXY = \u221a64 = 8
\n\"AP<\/p>\n

Question 5.
\nIn a triangle PQR with right angle at Q, the value of \u2220P is x, PQ = 7 cm and QR = 24 cm, then find sin x and cos x.\u00a0 \u00a0 (Page No. 274)
\nAnswer:
\nGiven right angled triangle is PQR with right angle at Q. The value of \u2220P is x.
\n\"AP
\nBy Pythagoras theorem
\nPR2<\/sup> = PQ2<\/sup> + QR2<\/sup>
\nPR2<\/sup> = 72<\/sup> + 242<\/sup>
\nPR2<\/sup> = 49 + 576
\nPR2<\/sup> = 625
\nPR2<\/sup> = \u221a625 = 25
\nsin x = \\(\\frac{QR}{PR}\\) = \\(\\frac{24}{25}\\)
\ncos x = \\(\\frac{PQ}{PR}\\) = \\(\\frac{7}{25}\\)<\/p>\n

\"AP<\/p>\n

Try This<\/span><\/p>\n

Question 1.
\nWrite lengths of \u201cHypotenuse\u201d, \u201cOpposite side\u201d and \u201cAdjacent side\u201d for the given angles in the given triangles.
\n1. For angle C
\n2. For angle A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Page No. 271)
\nAnswer:
\n\"AP
\nBy Pythagoras theorem
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n(5)2<\/sup> = AB2<\/sup> + 42<\/sup>
\n25 = AB2<\/sup> + 16
\nAB2<\/sup> = 25 – 16
\nAB2<\/sup> = 9
\nAB = \u221a9 = 3
\nFor angle C:
\nOpposite side = AB = 3 cm
\nAdjacent side = BC = 4 cm
\nHypotenuse = AC = 5 cm
\nFor angle A:
\nOpposite side = BC = 4 cm
\nAdjacent side = AB = 3 cm
\nHypotenuse = AC = 5 cm<\/p>\n

Question 2.
\nIn a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC – CA = 7 cm, then find sin A and tan B.\u00a0 \u00a0 (Page No. 274)
\nAnswer:
\nIn a right angle triangle ABC, right angle is at C.
\n\"AP
\nBC = \\(\\frac{30}{2}\\) = 15
\nBC = 15
\nSubstituting BC = 15 in equation (1)
\nBC + CA = 23
\nCA = 23 – BC = 23 – 15
\nCA = 8
\nBy Pythagoras theorem
\nAB2<\/sup> = AC2<\/sup> + BC2<\/sup>
\n= 82<\/sup> + 152<\/sup>
\n= 64 + 225 = 289
\n= \u221a289 = 17
\nsin A = \\(\\frac{BC}{AB}\\) = \\(\\frac{15}{17}\\)
\ncos B = \\(\\frac{AC}{BC}\\) = \\(\\frac{8}{15}\\)<\/p>\n

Question 3.
\nWhat will be the ratios of sides for sec A and cot A?\u00a0 \u00a0 (Page No. 275)
\nAnswer:
\nsec A = \\(\\frac{\\text { Hypotenuse }}{\\text { Adjacent side of the angle } \\mathrm{A}}\\)
\ncot A = \\(\\frac{\\text { Adjacent side of the angle } A}{\\text { Opposite side of the angle } A}\\)<\/p>\n

\"AP<\/p>\n

Think & Discuss<\/span><\/p>\n

Question 1.
\nDiscuss with your friends
\ni) sin x = \\(\\frac{4}{3}\\) does exist for some value of angle x?
\nii) The value of sin A and cos A is always less than 1. Why?
\niii) tan A is product of tan and A.\u00a0 \u00a0 \u00a0 \u00a0 (Page No. 274)
\nAnswer:
\ni) The value of sin 0 always lies between 0 and 1. Here sin x = \\(\\frac{4}{3}\\) which is greater than 1. So, it does not exist.
\nii)
\n\"AP
\nWe observe in above sin A, cos A, hypotenuse is in denominator which is greater than other two sides
\n\u2234 sin A = \\(\\frac{\\text { Opposite side }}{\\text { Hypotenuse }}\\) = \\(\\frac{Nr}{Dr}\\)
\nhere denominator is more than numerator. Hence its value will be less than 1.
\ncos A = \\(\\frac{\\text { Adjacent side }}{\\text { Hypotenuse }}\\)
\nhere also adjacent side is always less than hypotenuse. Hence its value is also less than or equal to 1.<\/p>\n

iii) The symbol tan A is used as an abbreviation for \u201cthe tan of the angle A\u201d.
\ntan A is not the product of \u201ctan\u201d and A. \u201ctan\u201d separated from A\u2019 has no meaning.<\/p>\n

Question 2.
\nIs \\(\\frac{\\sin \\mathrm{A}}{\\cos \\mathrm{A}}\\) equal to tan A?\u00a0 \u00a0 \u00a0 (Page No. 275)
\nAnswer:
\nYes, \\(\\frac{\\sin \\mathrm{A}}{\\cos \\mathrm{A}}\\) = tan A
\nProof:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 3.
\nIs \\(\\frac{\\cos \\mathrm{A}}{\\sin \\mathrm{A}}\\) equal to cot A?\u00a0 \u00a0 \u00a0(Page No. 275)
\nAnswer:
\nYes, \\(\\frac{\\cos \\mathrm{A}}{\\sin \\mathrm{A}}\\) = tan A
\nProof:
\n\"AP
\ncot A = \\(\\frac{\\text { Adjacent side }}{\\text { Opposite side }}\\)<\/p>\n

Do this<\/span><\/p>\n

Question 1.
\nFind cosec 60\u00b0, sec 60\u00b0 and cot 60\u00b0.\u00a0 \u00a0 \u00a0 \u00a0(Page No. 279)
\nAnswer:
\n\"AP
\nConsider an equilateral triangle ABC. Since each angle is 60\u00b0 in an equilateral triangle, we have \u2220A = \u2220B = \u2220C = 60\u00b0 and the sides of equilateral triangle is AB = BC = CA = 2a units.
\nDraw the perpendicular line AD from vertex A to BC as shown in the given figure.
\nPerpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
\nTherefore, \u2220BAD = \u2220CAD = 30\u00b0. Since point D divides the side BC into equal halves.
\nBD = \\(\\frac{1}{2}\\) BC = \\(\\frac{2a}{2}\\) = a units.
\nConsider right angle triangle ABD in the above given figure.
\nWe have AB = 2a and BD = a
\nThen AD2<\/sup> = AB2<\/sup> – BD2<\/sup>
\n(By Pythagoras theorem)
\n= (2a)2<\/sup> – (a)2<\/sup> = 3a2<\/sup>
\nTherefore, AD = a\u221a3
\nFrom definitions of trigonometric ratios,
\n\"AP<\/p>\n

\"AP<\/p>\n

Try this<\/span><\/p>\n

Question 1.
\nFind sin 30\u00b0, cos 30\u00b0, tan 30\u00b0, cosec 30\u00b0, sec 30\u00b0 and cot 30\u00b0 by using the ratio concepts.\u00a0 \u00a0 \u00a0(Page No. 279)
\nAnswer:
\n\"AP
\nConsider an equilateral triangle ABC. Since each angle is 60\u00b0 in an equilateral triangle, we have \u2220A = \u2220B = \u2220C = 60\u00b0 and the sides of equilateral triangle is AB = BC = CA = 2a units.
\nDraw the perpendicular line AD from vertex A to BC as shown in the given figure.
\nPerpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
\nTherefore, \u2220BAD = \u2220CAD = 30\u00b0. Since point D divides the side BC into equal halves.
\nBD = \\(\\frac{1}{2}\\) BC = \\(\\frac{2a}{2}\\) = a units.
\nConsider right angle triangle ABD in the above given figure.
\nWe have AB = 2a and BD = a
\nThen AD2<\/sup> = AB2<\/sup> – BD2<\/sup>
\n(By Pythagoras theorem)
\n= (2a)2<\/sup> – (a)2<\/sup> = 3a2<\/sup>
\nTherefore, AD = \\(\\sqrt{3 a^{2}}\\) = \u221a3a
\nBD = a, AD = \u221a3a and hypotenuse = AB = 2a and \u2220DAB = 30\u00b0.
\nsin 30\u00b0 = \\(\\frac{BD}{AB}\\) = \\(\\frac{a}{2a}\\) = \\(\\frac{1}{2}\\)
\n\"AP<\/p>\n

Question 2.
\nFind the values for tan 90\u00b0, cosec 90\u00b0, sec 90\u00b0 and cot 90\u00b0.\u00a0 \u00a0 \u00a0(Page No. 281)
\nAnswer:
\nFrom the adjacent figure, the trigonometric ratios of \u2220A, gets larger and larger in \u25b3ABC till it becomes 90\u00b0.
\n\"AP
\nAs \u2220A get larger and larger, \u2220C gets smaller and smaller. Therefore, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when \u2220A is very close to 90\u00b0, \u2220C becomes very close to 0\u00b0 and the side AC almost coincides with side BC.
\n\"AP
\n\u2234 AB = 0 and AC = BC = r
\nFrom trigonometric ratios
\nsin A = \\(\\frac{BC}{AC}\\)
\nsin A = \\(\\frac{AB}{AC}\\)
\nIf A = 90\u00b0, then AB = 0 and AC = BC = r,
\n\"AP<\/p>\n

\"AP<\/p>\n

Think & Discuss<\/span><\/p>\n

Question 1.
\nDiscuss with your friend about the following conditions:
\nWhat can you say about cosec 0\u00b0 = \\(\\frac{1}{\\sin 0^{\\circ}}\\)? Is it defined? Why?\u00a0 \u00a0 (Page No. 280)
\nAnswer:
\nsin 0\u00b0 = 0
\ncosec 0\u00b0 = \\(\\frac{1}{\\sin 0^{\\circ}}\\) = \\(\\frac{1}{0}\\) = not defined.
\nIt is not defined.
\nReason:
\nDivision by \u20180\u2019 is not allowed, hence \\(\\frac{1}{0}\\) is indeterminate.<\/p>\n

Question 2.
\nWhat can you say about cot 0\u00b0 = \\(\\frac{1}{\\tan 0^{\\circ}}\\). Is it defined? Why?\u00a0 \u00a0 (Page No. 281)
\nIs it defined? Why?
\nAnswer:
\ntan 0\u00b0 = 0
\ncot 0\u00b0 = \\(\\frac{1}{\\tan 0^{\\circ}}\\) = \\(\\frac{1}{0}\\) = undefined.
\nReason:
\nDivision by \u20180\u2019 is not allowed, hence \\(\\frac{1}{0}\\) is indeterminate.<\/p>\n

Question 3.
\nsec 0\u00b0 = 1. Why?\u00a0 \u00a0 \u00a0 (Page No. 281)
\nAnswer:
\nsec 0\u00b0 = \\(\\frac{1}{\\cos 0^{\\circ}}\\) [\u2235 cos 0\u00b0 = 1]
\n= \\(\\frac{1}{1}\\) = 1<\/p>\n

Question 4.
\nWhat can you say about the values of sin A and cos A, as the value of angle A increases from 0\u00b0 to 90\u00b0? (Observe the above table)
\ni) If A \u2265 B, then sin A \u2265 sin B. Is it true?
\nii) If A \u2265 B, then cos A \u2265 cos B. Is it true? Discuss.\u00a0 \u00a0 \u00a0 (Page No. 282)
\nAnswer:
\ni) Given statement
\n\u201cIf A \u2265 B, then sin A \u2265 sin B\u201d
\nYes, this statement is true.
\nBecause, it is clear from the table below that the sin A increases as A increases.
\n\"AP<\/p>\n

ii) Given statement
\n\u201cIf A \u2265 B, then cos A \u2265 cos B\u201d
\nNo, this statement is not true.
\nBecause, it is clear from the table below that cos A decreases as A increases.
\n\"AP<\/p>\n

\"AP<\/p>\n

Think & Discuss<\/span><\/p>\n

Question 1.
\nFor which value of acute angle
\n(i) \\(\\frac{\\cos \\theta}{1-\\sin \\theta}\\) + \\(\\frac{\\cos \\theta}{1+\\sin \\theta}\\) = 4 is true?
\nFor which value of 0\u00b0 \u2264 \u03b8 \u2264 90\u00b0, above equation is not defined?\u00a0 \u00a0 (Page No. 285)
\nAnswer:
\n\"AP
\n\"AP
\n\u21d2 cos \u03b8 = cos 60\u00b0 [from trigonometric ratios table]
\n\u21d2 \u03b8 = 60\u00b0
\nGiven statement is true for the acute angle i.e., \u03b8 = 60\u00b0.<\/p>\n

Question 2.
\nCheck and discuss the above relations in the case of angles between 0\u00b0 and 90\u00b0, whether they hold for these angles or not? So,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Page No. 286)
\ni) sin (90\u00b0 – A) = cos A
\nii) cos (90\u00b0 – A) = sin A
\niii) tan (90\u00b0 – A) = cot A and
\niv) cot (90\u00b0 – A) = tan A
\nv) sec (90\u00b0 – A) = cosc A
\nvi) cosec (90\u00b0 – A) = sec A
\nAnswer:
\nLet A = 30\u00b0
\ni) sin (90\u00b0 – A) = cos A
\n\u21d2 sin (90\u00b0 – 30\u00b0) = cos 30\u00b0
\n\u21d2 sin 60\u00b0 = cos 30\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

ii) cos (90\u00b0 – A) = sin A
\n\u21d2 cos (90\u00b0 – 30\u00b0) = sin 30\u00b0
\n\u21d2 cos 60\u00b0 = sin 30\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n

iii) tan (90\u00b0 – A) = cot A
\n\u21d2 tan (90\u00b0 – 30\u00b0) = cot 30\u00b0
\n\u21d2 tan 60\u00b0 = cot 30\u00b0 = \u221a3<\/p>\n

iv) cot (90\u00b0 – A) = tan A
\n\u21d2 cot (90\u00b0 – 30\u00b0) = tan 30\u00b0
\n\u21d2 cot 60\u00b0 = tan 30\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\)<\/p>\n

v) sec (90\u00b0 – A) = cosec A
\n\u21d2 sec (90\u00b0 – 30\u00b0) = cosec 30\u00b0
\n\u21d2 sec 60\u00b0 = cosec 30\u00b0 = 2<\/p>\n

vi) cosec (90\u00b0 – A) = sec A
\n\u21d2 cosec (90\u00b0 – 30\u00b0) = sec 30\u00b0
\n\u21d2 cosec 60\u00b0 = sec 30\u00b0 = \\(\\frac{2}{\\sqrt{3}}\\)<\/p>\n

So, the above relations hold for all the angles between 0\u00b0 and 90\u00b0.<\/p>\n

\"AP<\/p>\n

Do this<\/span><\/p>\n

(Page No. 290)<\/p>\n

Question 1.
\ni) If sin A = \\(\\frac{15}{17}\\) then find cos A.
\nAnswer:
\nGiven sin A = \\(\\frac{15}{17}\\)
\ncos A = \\(\\sqrt{1-\\sin ^{2} A}\\) [From Identity -I]
\n\"AP<\/p>\n

ii) If tan x = \\(\\frac{5}{12}\\), then find sec x. (AS j)
\nAnswer:
\nGiven tan x = \\(\\frac{5}{12}\\)
\nWe know that sec2<\/sup> x – tan2<\/sup> x = 1
\nsec2<\/sup> x = 1 + tan2<\/sup> x
\n\"AP<\/p>\n

iii) If cosec \u03b8 = \\(\\frac{25}{7}\\), then find cot \u03b8.
\nAnswer:
\nGiven cosec \u03b8 = \\(\\frac{25}{7}\\)
\nWe know that cosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1
\ncot2<\/sup> \u03b8 = cosec2<\/sup> \u03b8 – 1
\n\"AP<\/p>\n

Try This<\/span><\/p>\n

(Page No. 290)<\/p>\n

Question 1.
\nEvaluate the following and justify your answer.<\/p>\n

i) \\(\\frac{\\sin ^{2} 15^{\\circ}+\\sin ^{2} 75^{\\circ}}{\\cos ^{2} 36^{\\circ}+\\cos ^{2} 54^{\\circ}}\\)
\nAnswer:
\nGiven \\(\\frac{\\sin ^{2} 15^{\\circ}+\\sin ^{2} 75^{\\circ}}{\\cos ^{2} 36^{\\circ}+\\cos ^{2} 54^{\\circ}}\\)
\n\"AP
\n[\u2235 sin (90\u00b0 – \u03b8) = cos \u03b8; cos (90\u00b0 – \u03b8) = sin \u03b8]
\n= \\(\\frac{1}{1}\\) = 1 [By sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1]<\/p>\n

ii) sin 5\u00b0 cos 85\u00b0 + cos 5\u00b0 sin 85\u00b0
\nAnswer:
\nGiven sin 5\u00b0 cos 85\u00b0 + cos 5\u00b0 sin 85\u00b0
\n= sin 5\u00b0 . cos (90\u00b0 – 5\u00b0) + cos 5\u00b0 . sin (90\u00b0 – 5\u00b0)
\n= sin 5\u00b0 . sin 5\u00b0 + cos 5\u00b0 . cos 5\u00b0
\n[\u2235 sin (90\u00b0 – \u03b8) = cos \u03b8; cos (90\u00b0 – \u03b8) = sin \u03b8]
\n= sin2<\/sup> 5\u00b0 + cos2<\/sup> 5\u00b0
\n= 1 [\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1]<\/p>\n

iii) sec 16\u00b0 cosec 74\u00b0 – cot 74\u00b0 tan 16\u00b0.
\nAnswer:
\nGiven sec 16\u00b0 cosec 74\u00b0 – cot 74\u00b0 tan 16\u00b0
\n= sec 16\u00b0 . cosec (90\u00b0 – 16\u00b0) – cot (90\u00b0 – 16\u00b0) . tan 16\u00b0
\n= sec 16\u00b0. sec 16\u00b0 – tan 16\u00b0 . tan 16\u00b0 [\u2235 cosec (90\u00b0 – \u03b8) = sec \u03b8; cot (90\u00b0 – \u03b8) \u2014 tan \u03b8]
\n= sec2<\/sup> 16\u00b0 – tan2<\/sup> 16\u00b0
\n= 1 [\u2235 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1]<\/p>\n

\"AP<\/p>\n

Think & Discuss<\/span><\/p>\n

(Page No. 290)<\/p>\n

Question 1.
\nAre these identities true for 0\u00b0 \u2264 A \u2264 90\u00b0? If not for which values of A they are true?
\ni) sec2<\/sup> A – tan2<\/sup> A = 1
\nAnswer:
\nGiven identity is sec2<\/sup> A – tan2<\/sup> A = 1
\nLet A = 0\u00b0
\nL.H.S. = sec2<\/sup> 0\u00b0 – tan2<\/sup> 0\u00b0
\n= 1 – 0 = 1 = R.H.S.
\nLet A = 90\u00b0
\ntan A and sec A are not defined.
\nSo it is true.
\n\u2234 For all given values of \u2018A\u2019 such that 0\u00b0 \u2264 A \u2264 90\u00b0 this trigonometric identity is true.<\/p>\n

ii) cosec2<\/sup> A – cot2<\/sup> A = 1
\nAnswer:
\nGiven identity is cosec2<\/sup> A – cot2<\/sup> A = 1
\nLet A = 0\u00b0
\ncosec A and cot A are not defined for A = 0\u00b0.
\nTherefore identity is true for A = 0\u00b0.
\nLet A = 90\u00b0
\ncosec A = cosec 90\u00b0 = 1
\ncot A = cot 90\u00b0 = 0
\nL.H.S. = l2<\/sup> – 02<\/sup> = 1 – 0 = 1 = R.H.S.
\n\u2234 This identity is true for all values of A, such that 0\u00b0 \u2264 A \u2264 90\u00b0.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry InText Questions AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry InText Questions and Answers. 10th Class Maths 11th Lesson Trigonometry InText Questions and Answers Do This Identify \u201cHypotenuse\u201d, \u201cOpposite side\u201d and \u201cAdjacent side\u201d for the given angles in the … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6978"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=6978"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6978\/revisions"}],"predecessor-version":[{"id":40683,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6978\/revisions\/40683"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6978"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6978"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6978"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}