{"id":1533,"date":"2020-09-15T05:32:26","date_gmt":"2020-09-15T00:02:26","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1533"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-11-ex-11-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-11-ex-11-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1<\/h2>\n

Question 1.
\nIn \u2206ABC, right-angled at B, AB = 24 cm., BC = 7 cm. Determine:
\ni) sin A, cos A
\nii) sin C, cos C
\nSolution:
\n\"KSEEB
\nIn \u22a5\u2206ABC, \u2220B = 90\u00b0
\nAs per Pythagoras theorem
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n= (24)2<\/sup> + (7) 2<\/sup>
\n= 576 + 49
\nAC2<\/sup> = 625
\n\u2234 AC = 25 cm
\n\"KSEEB<\/p>\n

Question 2.
\nIn the given figure, find tan P – cot R.
\nSolution:
\n\"KSEEB
\nIn \u22a5\u2206PQR, \u2220Q = 90\u00b0
\n\u2234 PQ2<\/sup> + QR2<\/sup> = PR2<\/sup>
\n(12)2<\/sup> + QR2<\/sup> = (13)2<\/sup>
\n144 + QR2<\/sup> = 169
\nQR2<\/sup> = 169 – 144
\nQR2<\/sup> = 25
\n\u2234 QR = 5 cm.
\n\"KSEEB<\/p>\n

Question 3.
\nIf sin A = \\(\\frac{3}{4}\\) calculate cos A and tan A.
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 4.
\nGiven 15 cot A = 8, find sin A and sec A.
\nSolution:
\n\"KSEEB<\/p>\n

Question 5.
\nGiven sec \u03b8 = \\(\\frac{13}{12}\\), calculate all other trigonometric ratios
\nSolution;
\n\"KSEEB<\/p>\n

Question 6.
\nIf \u2220A and \u2220B are acute angles such that cos A = cos B, then show that \u2220A = \u2220B.
\nSolution:
\nIn \u22a5\u2206ABC, \u2220A and \u2220B are acute angles. CD \u22a5 AB is drawn.
\n\"KSEEB
\nCD is common.
\nS.S.S. Postulate :
\n\u2234 \u2206ADC ~ \u2206ADB
\n\u2234 \u2220A = \u2220B
\n\u2235 \u201cAngles of similar triangles are equiangular.\u201d<\/p>\n

Question 7.
\nIf cot \u03b8 = \\(\\frac{7}{8}\\), evaluate :
\n(i) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\n(ii) cot2<\/sup>\u03b8
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 8.
\nIf 3 cot A = 4, check whether \\(\\frac{1-\\tan ^{2} A}{1+\\tan ^{2} A}\\) = cos2<\/sup> A – sin2<\/sup> A or not
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 9.
\nIn \u2206ABC, right-angled at B, if tan A =\\(\\frac{1}{\\sqrt{3}}\\) find the value of:
\ni) sin A cos C + cos A sin C
\nii) cos A cos C – sin A sin C
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 10.
\nIn \u2206PQR, right-angled at Q, PR + QR = 25 cm. and PQ = 5 cm. Determine the values of sin P, cos P and tan P
\nSolution:
\nPQ = 5 cm
\nPR + QR = 25 cm
\n\u2234 PR = 25 – QR
\n\"KSEEB
\nPR2<\/sup> = PQ2<\/sup> + QR2<\/sup>
\nQR2<\/sup> = PR2<\/sup> – PQ2<\/sup>
\n= (25 – QR)2<\/sup> – (5)2<\/sup>
\nQR2<\/sup> = 625 – 50QR + QR2<\/sup> – 25
\n50QR = 600
\n\u2234 QR = 12 cm.
\n\u2234 PR = 25 – QR = 25 – 12 = 13 cm.
\n\u2234 QR = 12 cm
\n\u2234 PR = 25 – QR = 25 – 12 = 13 cm
\n\"KSEEB<\/p>\n

Question 11.
\nState whether the following are true or false. Justify your answer.
\n(i) The value of tan A is always less than 1.
\nAnswer:
\nFalse. 60\u00b0 = \\(\\sqrt{3}\\) > 1<\/p>\n

(ii) If sec A = \\(\\frac{12}{5}\\) for some value of angle A.
\nAnswer:
\nTrue. because sec A > 1.<\/p>\n

(iii) cos A is the abbreviation used for the cosecant of angle A.
\nAnswer:
\nFalse. Because cos A is simplified as cos.<\/p>\n

(iv) cot A is the product of cot and A.
\nAnswer:
\nFalse. cot is \u2220A or meaningless. Here,
\ncot A = \\(\\frac{\\text { Adjacent side }}{\\text { Opposite side }}\\)<\/p>\n

(v) sin \u03b8 = \\(\\frac{4}{3}\\) for some angle \u03b8
\nAnswer:
\nFalse. because sin \u03b8 \u226f 1<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1. Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1 Question 1. In …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1533"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1533"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1533\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1533"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1533"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1533"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}