{"id":9691,"date":"2024-02-23T11:44:46","date_gmt":"2024-02-23T06:14:46","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9691"},"modified":"2024-02-24T17:55:47","modified_gmt":"2024-02-24T12:25:47","slug":"ts-10th-class-maths-solutions-chapter-11-trigonometry-ex-11-1","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-11-trigonometry-ex-11-1\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.1"},"content":{"rendered":"

Students can practice\u00a010th Class Maths Study Material Telangana<\/a>\u00a0Chapter 11 Trigonometry Ex 11.1 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.1<\/h2>\n

Question 1.
\nIn a right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A. (AS1<\/sub>)
\nSolution:
\nGiven that \u2206ABC, AB = 8 cm; BC = 15 cm; CA = 17 cm
\n\"TS<\/p>\n

Question 2.
\nThe sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and \u2220Q = 90\u00b0 respectively. Then find tan Q – tan R.
\n(AS1<\/sub>)
\nSolution:
\nGiven that in \u2206PQR, PQ = 7 cm
\nQR = 25 cm and RP = 24 cm
\nPR = \\(\\sqrt{\\mathrm{QR}^2-\\mathrm{PQ}^2}\\) = \\(\\sqrt{25^2-7^2}\\)
\n= \\(\\sqrt{625-49}\\) = \\(\\sqrt{576}\\) – 24
\nIn the text given problem is wrong. We take
\n\u2220P = 90\u00b0 instead of \u2220Q = 90\u00b0
\ntan P = \\(\\frac{\\text { Side opposite to } \\angle \\mathrm{P}}{\\text { Side adjacent to } \\angle \\mathrm{P}}\\)
\n= \\(\\frac{\\mathrm{PR}}{\\mathrm{PQ}}\\) = \\(\\frac{24}{7}\\)
\ntan R = \\(\\frac{\\text { Side opposite to } \\angle \\mathrm{R}}{\\text { Side adjacent to } \\angle \\mathrm{R}}\\)
\n= \\(\\frac{\\mathrm{PQ}}{\\mathrm{PR}}\\) = \\(\\frac{7}{24}\\)
\n\"TS
\n\u2234 tan P – tan R = \\(\\frac{24}{7}\\) – \\(\\frac{7}{24}\\)
\n= \\(\\frac{576-49}{168}\\) = \\(\\frac{27}{168\/}\\)<\/p>\n

\"TS<\/p>\n

Question 3.
\nIn a right tingle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and \u2220BAC = \u03b8. Then, find cos \u03b8 and tan \u03b8.
\nSolution:
\nIn \u2206ABC, \u2220B = 90\u00b0
\na = BC = 24 units
\nb = AC = 25 units
\nIn \u2206ABC,
\n\u2234 AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\u2220B = 90\u00b0 (Pythagoras theorem)
\n\"TS
\n252<\/sup> = AB2<\/sup> + 242<\/sup>
\n\u21d2 AB2<\/sup> = 252<\/sup> – 242<\/sup>
\n= 625 – 576 = 49
\n\u21d2 AB = \\(\\sqrt{49}\\) = 7
\nc = AB = 7 units
\ncos \u03b8 = \\(\\frac{\\text { Side adjacent to } \\theta}{\\text { Hypotenuse }}\\)
\n= \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}\\) = \\(\\frac{7}{25}\\)
\nsin \u03b8 = \\(\\frac{\\text { Side opposite to } \\theta}{\\text { Hypotenuse }}\\)
\n= \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}\\) = \\(\\frac{24}{25}\\)
\n\u2234 tan \u03b8
\n= \\(\\frac{\\sin \\theta}{\\cos \\theta}=\\frac{\\mathrm{BC}}{\\mathrm{AC}} \\times \\frac{\\mathrm{AC}}{\\mathrm{AB}}=\\frac{24}{25} \\times \\frac{25}{7}=\\frac{24}{7}\\)<\/p>\n

Question 4.
\nIf cos A = \\(\\frac{12}{13}\\), then find sin A and tan A. (AS1<\/sub>) (A.P. Mar. ’16)
\nSolution:
\nGiven that, cos A = \\(\\frac{12}{13}\\)
\nand cos A = \\(\\frac{\\text { Adjacent side }}{\\text { Hypotenuse }}\\)
\nFor angle A, adjacent side = AB = 12 k Hypotenuse = AC = 13 k (Where k is a positive number)
\n\"TS
\nNow, we have in \u2206ABC
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (Pythagoras theorem)
\n\u21d2 (13k)2<\/sup> = (12k)2<\/sup> + BC2<\/sup>
\n\u21d2 BC2<\/sup> = (13k)2<\/sup> – (12k)2<\/sup>
\n= 169 k2<\/sup> – 144 k2<\/sup>
\n= 25 k2<\/sup>
\n\u2234 BC = \\(\\sqrt{25 \\mathrm{k}^2}\\) = 5 k
\nBC = 5 k = Opposite side
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf 3 tan A = 4, then find sin A and cos A. (AS1<\/sub>)
\nSolution:
\nGiven that, 3 tan A = 4
\n\"TS
\nOpposite side to \u2220A = BC = 4k
\nAdjacent side to \u2220A = AB = 3k
\nNow we have in \u2206ABC, \u2220B = 90\u00b0
\n\u2234 AC2<\/sup> = AB2<\/sup> + BC2<\/sup> (By Pythagoras theorem)
\n= (3k)2<\/sup> + (4k)2<\/sup>
\n= 9k2<\/sup> + 16k2<\/sup>
\n= 25k2<\/sup>
\n\u2234 AC = \\(\\sqrt{25 \\mathrm{k}^2}\\) = 5k
\nAC = 5k = Hypotenuse
\nsin A = \\(\\frac{\\text { Opposite side }}{\\text { Hypotenuse }}\\)
\n= \\(\\frac{4 \\mathrm{k}}{5 \\mathrm{k}}\\) = \\(\\frac{4}{5}\\)
\ncos A = \\(\\frac{\\text { Adjacent side }}{\\text { Hypotenuse }}\\)
\n= \\(\\frac{3 \\mathrm{k}}{5 \\mathrm{k}}\\) = \\(\\frac{4}{5}\\)<\/p>\n

\"TS<\/p>\n

Question 6.
\nIf \u2220A and \u2220X are acute angles such that cos A = cos X, then show that \u2220A = \u2220X. (AS2<\/sub>)
\nSolution:
\nIn the given triangle,
\n\"TS
\n\u21d2 \\(\\frac{\\mathrm{AC}}{\\mathrm{AX}}\\) = \\(\\frac{\\mathrm{XC}}{\\mathrm{AX}}\\)
\n\u21d2 AC = XC
\n\u21d2 \u2220A = \u2220X
\n(\u2234 Angles opposite to equal sides are also equal)<\/p>\n

Question 7.
\nGiven cot \u03b8 = \\(\\frac{7}{8}\\), then evaluate
\ni) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\nii) \\(\\frac{(1+\\sin \\theta)}{\\cos \\theta}\\) (AS1<\/sub>)
\nSolution:
\nGiven,
\n\"TS
\nLet AB = 7k and BC = 8k
\nIn a right angled triangle,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (By Pythagoras theorem)
\n= (7k)2<\/sup> + (8k)2<\/sup>
\n= 49 k2<\/sup> + 64 k2<\/sup>
\nAC2<\/sup> = 113 k2<\/sup>
\nAC = \\(\\sqrt{113}\\) k
\nNow,
\n\"TS
\n\"TS<\/p>\n

Question 8.
\nIn a right angle triangle ABC, right angle Is at B, If tan A = \\(\\sqrt{3}\\), then find the value of
\ni) sin A cos C + cos A sin C
\nii) cos A cos C – sin A sin C (AS1<\/sub>)
\nSolution:
\nGiven, tan A = \\(\\frac{\\sqrt{3}}{1}\\)
\n\"TS
\nLet opposite side = \\(\\sqrt{3}\\)k and adjacent side = 1 k
\nIn right angled \u2206ABC
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (By Pythagoras theorem)
\n\u21d2 AC2<\/sup> = (1k)2<\/sup> + (\\(\\sqrt{3}\\)k)2<\/sup>
\n\u21d2 AC2<\/sup> = 1k2<\/sup> + 3k2<\/sup>
\n\u21d2 AC2<\/sup> = 4k2<\/sup> \u21d2 AC = \\(\\sqrt{4 \\mathrm{k}^2}\\)
\nAC = 2k
\nNow,
\nsin A = \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{\\sqrt{3} \\mathrm{k}}{2 \\mathrm{k}}=\\frac{\\sqrt{3}}{2}\\)
\ncos A = \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{1 \\mathrm{k}}{2 \\mathrm{k}}=\\frac{1}{2}\\)
\n\"TS
\nii) cos A . cos C – sin A . sin C
\n= \\(\\frac{1}{2}\\) . \\(\\frac{\\sqrt{3}}{2}\\) – \\(\\frac{\\sqrt{3}}{2}\\) . \\(\\frac{1}{2}\\)
\n= \\(\\frac{\\sqrt{3}}{4}\\) – \\(\\frac{\\sqrt{3}}{4}\\) = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice\u00a010th Class Maths Study Material Telangana\u00a0Chapter 11 Trigonometry Ex 11.1 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.1 Question 1. In a right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9691"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9691"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9691\/revisions"}],"predecessor-version":[{"id":9831,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9691\/revisions\/9831"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9691"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9691"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9691"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}