{"id":1148,"date":"2024-04-17T10:46:25","date_gmt":"2024-04-17T05:16:25","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=1148"},"modified":"2024-04-17T14:17:20","modified_gmt":"2024-04-17T08:47:20","slug":"samacheer-kalvi-10th-maths-guide-chapter-6-unit-exercise-6","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-10th-maths-guide-chapter-6-unit-exercise-6\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Unit Exercise 6"},"content":{"rendered":"

Students can download Maths Chapter 6 Trigonometry Unit Exercise 6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Unit Exercise 6<\/h2>\n

Question 1.
\nProve that
\n(i) cot2<\/sup> A \\(\\left(\\frac{\\sec A-1}{1+\\sin A}\\right)\\) + sec2<\/sup> A \\(\\left(\\frac{\\sin A-1}{1+\\sec A}\\right)\\) = 0
\n(ii) \\(\\frac{\\tan ^{2} \\theta-1}{\\tan ^{2} \\theta+1}\\) = 1 – 2 cos2<\/sup> \u03b8
\nAnswer:
\n\"Samacheer
\n\"Samacheer
\nHence it is proved
\n\"Samacheer
\n= 1 – cos2<\/sup> \u03b8
\nL.H.S = R.H.S
\nHence it is proved.<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nProve that
\n\\(\\left(\\frac{1+\\sin \\theta-\\cos \\theta}{1+\\sin \\theta+\\cos \\theta}\\right)^{2}=\\left(\\frac{1-\\cos \\theta}{1+\\cos \\theta}\\right)\\)
\nAnswer:
\n\"Samacheer
\nLHS = RHS
\nHence it is proved<\/p>\n

Question 3.
\nIf x sin2<\/sup> \u03b8 + y cos2<\/sup> \u03b8 = sin \u03b8 cos \u03b8 and x sin \u03b8 = y cos \u03b8, then prove that x2<\/sup> + y2<\/sup> = 1.
\nAnswer:
\nGiven x sin2<\/sup> \u03b8 + y cos2<\/sup> \u03b8 = sin \u03b8 cos \u03b8
\nx sin \u03b8 = y cos \u03b8 ……..(1)
\nx sin3<\/sup> \u03b8 + y cos3<\/sup> \u03b8 = sin \u03b8 cos \u03b8
\nx sin \u03b8 (sin2<\/sup> \u03b8) + y cos \u03b8 (cos2<\/sup> \u03b8) = sin \u03b8 cos \u03b8
\nx sin \u03b8 (sin2<\/sup> \u03b8) + x sin \u03b8 (cos2<\/sup> \u03b8) = sin \u03b8 cos \u03b8
\nx sin \u03b8 (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) = sin \u03b8 cos \u03b8
\nx sin \u03b8 = sin \u03b8 cos \u03b8
\nx = cos \u03b8
\nsubstitute x = cos \u03b8 in (1)
\ncos \u03b8 sin \u03b8 = y cos \u03b8 y = sin \u03b8
\nL. H. S = x2<\/sup> + y2<\/sup> = cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\nL.H. S = R.H.S
\nHence it is proved.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nIf a cos \u03b8 – b sin \u03b8 = c, then prove that (a sin \u03b8 + b cos \u03b8) = \u00b1 \\(\\sqrt{a^{2}+b^{2}-c^{2}}\\).
\nAnswer:
\nGiven a cos \u03b8 – b sin \u03b8 = c
\nSquaring on both sides
\n(a cos \u03b8 – b sin \u03b8)2<\/sup> = c2<\/sup>
\na2<\/sup> cos2<\/sup> \u03b8 + b2<\/sup> sin2<\/sup> \u03b8 – 2 ab cos \u03b8 sin \u03b8 = c2<\/sup>
\na2<\/sup> (1 – sin2<\/sup> \u03b8) + b2<\/sup> (1 – cos2<\/sup> \u03b8) – 2 ab cos \u03b8 sin \u03b8 = c2<\/sup>
\na2<\/sup> – a2<\/sup> sin2<\/sup> \u03b8 + b2<\/sup> – b2<\/sup> cos2<\/sup> \u03b8 – 2 ab cos \u03b8 sin \u03b8 = c2<\/sup>
\n– a2<\/sup> sin2<\/sup> \u03b8 – B2<\/sup> – cos2<\/sup> \u03b8 – 2 ab cos \u03b8 sin \u03b8 = – a2<\/sup> – b2<\/sup> + c2<\/sup>
\na2<\/sup> sin2<\/sup> \u03b8 + b2<\/sup> cos2<\/sup> \u03b8 + 2 ab cos \u03b8 sin \u03b8 = a2<\/sup> + b2<\/sup> – c2<\/sup>
\n(a sin \u03b8 + b cos \u03b8)2<\/sup> – a2<\/sup> + b2<\/sup> – c2<\/sup>
\na sin \u03b8 + b cos \u03b8 = \u00b1 \\(\\sqrt{a^{2}+b^{2}-c^{2}}\\)
\nHence it is proved.<\/p>\n

Question 5.
\nA bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45\u00b0 . The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30\u00b0 . Determine the speed at which the bird flies. (\\(\\sqrt { 3 }\\) = 1.732)
\nAnswer:
\n\"Samacheer
\nA is the initial position of the bird B is the final position of the bird Let the speed of the bird be “s”
\nDistance = speed \u00d7 time
\n\u2234 AB = 2x
\nLet CD be x
\n\u2234 CE = x + 2s
\nIn the \u2206 CDA, tan 45\u00b0 = \\(\\frac { AD }{ CD } \\)
\n1 = \\(\\frac { 80 }{ x } \\)
\nx = 80 ……..(1)
\nIn the \u2206 BCE
\ntan 30\u00b0 = \\(\\frac { BE }{ CE } \\)
\n\\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac { 80 }{ x+2s } \\)
\nx + 2s = 80 \\(\\sqrt { 3 }\\)
\nx = 80 \\(\\sqrt { 3 }\\) – 2 s ………(2)
\nFrom (1) and (2) we get
\n80 \\(\\sqrt { 3 }\\) – 2 s = 80
\n80 \\(\\sqrt { 3 }\\) – 80 = 2 s \u21d2 80 (\\(\\sqrt { 3 }\\) – 1) = 2 s
\ns = \\(\\frac{80(\\sqrt{3}-1)}{2}\\) = 40 (1.732 – 1) = 40 \u00d7 0.732 = 29.28
\nSpeed of the flying bird = 29.28 m\/sec<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nAn aeroplane is flying parallel to the Earth\u2019s surface at a speed of 175 m\/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth\u2019s surface is 37\u00b0 at a given point. After what period of time does the angle of elevation increase to 53\u00b0? (tan 53\u00b0 = 1.3270, tan 37\u00b0 = 0.7536)
\nAnswer:
\nLet C is the initial and D is the final position of the aeroplane.
\nLet the time taken by the aeroplane be “t”
\n\u2234 CD = 175 t (Distance = speed \u00d7 time)
\n\"Samacheer
\nLet AB be x
\n\u2234 AE = x + 175 t
\nIn the right \u2206 ABC
\n\"Samacheer
\n\"Samacheer
\n\u2234 Time taken is 1. 97 seconds<\/p>\n

Question 7.
\nA bird is flying from A towards B at an angle of 35\u00b0, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48\u00b0 and distance 32 km away.
\n(i) How far is B to the North of A?
\n(ii) How far is B to the West of A?
\n(iii) How far is C to the North of B?
\n(iv) How far is C to the East of B?
\n(sin 55\u00b0 = 0.8192, cos 55\u00b0 = 0.5736, sin 42\u00b0 = 0.6691, cos 42\u00b0 = 0.7431)
\nAnswer:
\n\"Samacheer
\n(i) To find the distance of B to the north of A
\nIn \u2206 ABB,
\n\"Samacheer
\nDistance of B to the North of A = 24. 58 km<\/p>\n

(ii) Distance from B to the west of A is AB’
\nIn \u2206 ABB’
\ncos 55\u00b0 = \\(\\frac{\\mathrm{AB}^{\\prime}}{\\mathrm{AB}}\\)
\n0.5736 = \\(\\frac{A B^{\\prime}}{30}\\)
\n\u2234 AB’ = 0.5736 \u00d7 30 = 17. 21 km
\nDistance of B to the West of A is 17. 21 km<\/p>\n

(iii) Distance from C to the North of B is CD
\nIn the right \u2206 BCD, sin 42\u00b0 = \\(\\frac { CD }{ BC } \\)
\n0.6691 = \\(\\frac { BD }{ 32 } \\)
\n\u2234 CD = 0.6691 \u00d7 32 = 21.41 km
\nDistance of C to the North B is 21. 41 km<\/p>\n

(iv) The distance of C to the East of B is BD
\nIn the right \u2206 BDC, cos 42\u00b0 = \\(\\frac { BD }{ BC } \\)
\n0.7431 = \\(\\frac { BD }{ 32 } \\)
\n\u2234 BD = 0.7431 \u00d7 32
\n= 23.78 km
\nDistance of C to the East of B is 23.78 km.<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nTwo ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60\u00b0 and 45\u00b0 respectively. If the distance between the ships is 200 \\(\\left(\\frac{\\sqrt{3}+1}{\\sqrt{3}}\\right)\\) meters, find the height of the lighthouse
\nAnswer:
\nLet A and B the position of the first ship and the second ship
\nDistance = 200 \\(\\left(\\frac{\\sqrt{3}+1}{\\sqrt{3}}\\right)\\) m
\nLet the height of the light house CD be “h”
\n\"Samacheer
\nIn the right \u2206 ACD, tan 60\u00b0 = \\(\\frac { CD }{ AD } \\)
\n\\(\\sqrt { 3 }\\) = \\(\\frac { h }{ AD } \\)
\n\u2234 AD = \\(\\frac{h}{\\sqrt{3}}\\) ……….(1)
\nIn the right \u2206 BCD
\ntan 45\u00b0 = \\(\\frac { DC }{ BD } \\)
\n1 = \\(\\frac { h }{ BD } \\)
\n\u2234 BD = h
\nDistance between the two ships = AD + BD
\n200 (\\(\\frac{\\sqrt{3}+1}{\\sqrt{3}}\\)) = \\(\\frac{h}{\\sqrt{3}}\\) + h \u21d2 200 (\\(\\sqrt { 3 }\\) + 1) = h + \\(\\sqrt { 3 }\\) h
\n200 (\\(\\sqrt { 3 }\\) + 1) = h(1 + \\(\\sqrt { 3 }\\)) \u21d2 h = \\(\\frac{200(\\sqrt{3}+1)}{(1+\\sqrt{3})}\\)
\nh = 200
\nHeight of the light house = 200 m<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nA building and a statue are in opposite side of a street from each other 35 m apart. From a point on the roof of building the angle of elevation of the top of statue is 24\u00b0 and the angle of depression of base of the statue is 34\u00b0 . Find the height of the statue. (tan 24\u00b0 = 0.4452, tan 34\u00b0 = 0.6745)
\nAnswer:
\nLet the height of the statue be “h” m
\nLet AD be x
\n\u2234 EC = h – x
\nIn the right \u2206 ABD,
\n\"Samacheer
\ntan 34\u00b0 = \\(\\frac { AD }{ AB } \\)
\n0.6745 = \\(\\frac { x }{ 35 } \\)
\n\u2234 x = 0.6745 \u00d7 35 \u21d2 x = 23.61 m
\nIn the right \u2206 DEC \u21d2 tan 24\u00b0 = \\(\\frac { EC }{ DE } \\)
\n0.4452 = \\(\\frac { h-x }{ 35 } \\) \u21d2 h – x = 0.4452 \u00d7 35
\nh – 23.61 = 15. 58 \u21d2 h = 15.58 + 23.61 = 39.19 m
\nHeight of the statue = 39.19 m<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 6 Trigonometry Unit Exercise 6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Unit Exercise …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[2],"tags":[],"class_list":["post-1148","post","type-post","status-publish","format-standard","hentry","category-class-10"],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/1148"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=1148"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/1148\/revisions"}],"predecessor-version":[{"id":39948,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/1148\/revisions\/39948"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=1148"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=1148"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=1148"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}