{"id":8408,"date":"2024-11-06T09:49:59","date_gmt":"2024-11-06T04:19:59","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=8408"},"modified":"2024-11-06T09:49:59","modified_gmt":"2024-11-06T04:19:59","slug":"ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-1\/","title":{"rendered":"NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions – Ex 3.1"},"content":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0<\/span><\/strong><\/span><\/h2>\n

The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 3 Trigonometric Functions <\/strong>Exercise 3.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter 3 (Trigonometric Functions)\u00a0<\/span><\/strong><\/span><\/h2>\n\n\n\n
Exercise – 3.1<\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1. Find the radian measures corresponding to the following degree measures:
\n<\/strong>(i) 25\u00b0
\n(ii) \u201347\u00b030\u2032
\n(iii) 240\u00b0
\n(iv) 520\u00b0 <\/strong><\/span><\/p>\n

Solution –\u00a0 <\/strong><\/span><\/p>\n

(i) 25\u00b0
\n<\/strong>As we know that 180\u00b0 = \u03c0 radian<\/span>
\nSo,<\/span>
\n1\u00b0 = \u03c0\/180\u00b0 radian<\/span>
\nThen, 25\u00b0 = (\u03c0\/180\u00b0) \u00d7 25\u00b0<\/span>
\n= 5\u03c0\/36 radians<\/span>
\nHence, 25\u00b0 equals to 5\u03c0\/36 radians.<\/span><\/p>\n

(ii) \u201347\u00b030\u2032
\n<\/strong>As we know that 180\u00b0 = \u03c0 radian<\/span>
\nSo,<\/span>
\n1\u00b0 = \u03c0\/180\u00b0<\/span>
\nAnd 60\u2032 = 1\u00b0<\/span>
\n30\u2032 = (1\/2)\u00b0<\/span>
\nSo, -47\u00b030\u2032 = -47 (1\/2)\u00b0<\/span>
\n-47(1\/2)\u00b0 = (\u03c0\/180) \u00d7 (-95\/2)<\/span>
\n= (-19\u03c0\/72) radian.<\/span>
\nHence, -47\u00b030\u2032 is equals to -19\u03c0\/72 radian.<\/span><\/p>\n

(iii) 240\u00b0
\n<\/strong>As we know that 180\u00b0 = \u03c0 radian<\/span>
\n1\u00b0 = \u03c0\/180\u00b0 radian<\/span>
\nSo<\/span>
\n240\u00b0 = (\u03c0\/180\u00b0) \u00d7 240\u00b0<\/span>
\n= 4\u03c0\/3 radians<\/span>
\nHence, 240\u00b0 equals to 4\u03c0\/3 radians.<\/span><\/p>\n

(iv) 520\u00b0
\n<\/strong>As we know that 180\u00b0 = \u03c0 radian<\/span>
\n1\u00b0 = \u03c0\/180\u00b0 radian<\/span>
\nSo,<\/span>
\n520\u00b0 = (\u03c0\/180\u00b0) \u00d7 520\u00b0<\/span>
\n= 26\u03c0\/9 radians<\/span>
\nHence, 520\u00b0 equals to 26\u03c0\/9 radians.<\/span><\/p>\n

2. Find the degree measures corresponding to the following radian measures (Use \u03c0 = 22\/7)
\n<\/strong>(i) 11\/16
\n<\/strong>(ii) -4
\n<\/strong>(iii) 5\u03c0\/3
\n<\/strong>(iv) 7\u03c0\/6 <\/strong><\/span><\/p>\n

Solution –\u00a0 <\/strong><\/span><\/p>\n

(i) 11\/16
\n<\/strong>= (11\/16) (180\u00b0\/\u03c0) \u00a0{as 180\u00b0 = \u03c0 radian, then 1 radian = 180\u00b0\/\u03c0} <\/span>
\n= (11\/16) \u00d7 (180\u00b0 \u00d7 7\/22) <\/span>
\n= (11 \u00d7 180\u00b0 \u00d7 7\/16 \u00d7 22) <\/span>
\n= 315\/8\u00b0 <\/span>
\n= 39 (3\/8)\u00b0 <\/span>
\n= 39(3\/8)\u00b0 <\/span>
\n= 39\u00b0 + (3\/8)\u00b0 <\/span><\/p>\n

Again 1\u00b0 = 60\u2032 <\/span>
\nSo <\/span>
\n(3\/8)\u00b0 = 60\u2032 \u00d7 (3\/8)<\/span>
\n= 22 (1\/2)\u2019 <\/span>
\n= 22 (1\/2)\u2019\u00a0 <\/span>
\n= 22\u2032 + 1\/2\u2032 <\/span><\/p>\n

Again 1\u2032 = 60\u2033 <\/span>
\n= (1\/2)\u2019 = 30\u2033 <\/span><\/p>\n

Hence, 11\/16 radian results to 39\u00b0 22\u2032 30\u2033.<\/span><\/p>\n

(ii) -4
\n<\/strong>= -4 \u00d7 (180\u00b0\/\u03c0) {as 180\u00b0 = \u03c0 radian, then 1 radian = 180\u00b0\/\u03c0}.\u00a0 <\/span>
\n= -4 \u00d7180\u00b0 \u00d7 7\/22 <\/span>
\n= -229\u00b0 (1\/11)\u00a0 <\/span>
\n= -229 (1\/11)\u00b0= -229\u00b0 + (1\/11)\u00b0 <\/span><\/p>\n

Again(1\/11)\u00b0 = (1\/11) \u00d7 60\u2032. {as 1\u00b0 = 60\u2032}\u00a0 <\/span>
\n= 5(5\/11)\u2019 <\/span><\/p>\n

\u00a0Also, 5 (5\/11)\u2019 = 5\u2032 + (5\/11)\u2019 <\/span>
\n(5\/11)\u2019 = (5\/11) \u00d7 60\u2033 {as 1\u2032 = 60\u2033} <\/span>
\n= 27\u2033\u00a0 <\/span>
\nSo, -229(1\/11) = -229\u00b0 5\u201927\u201d\u00a0 <\/span>
\nHence, -4 radian results to -229\u00b0 5\u2032 27\u2033.<\/span><\/p>\n

(iii) 5\u03c0\/3
\n<\/strong>= (5 \u03c0\/3) \u00d7 (180\/\u03c0) {as 180\u00b0 = \u03c0 radian, then 1 radian =180\u00b0\/\u03c0}.\u00a0 <\/span>
\n= (5 \u00d7 180\/3)\u00b0 <\/span>
\n= 300\u00b0 <\/span>
\nHence, 5\u03c0\/3 results to 300\u00b0.<\/span><\/p>\n

(iv) 7\u03c0\/6
\n<\/strong>= (7\u03c0\/6) \u00d7 (180\u00b0\/\u03c0) \u00a0{as 180\u00b0 = \u03c0 radian, then 1 radian =180\u00b0\/\u03c0}. <\/span>
\n= (7 \u00d7 180\/6)\u00b0 <\/span>
\n= 210\u00b0 <\/span>
\nHence, 7\u03c0\/6 radian results to 210\u00b0. <\/span><\/p>\n

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
\n<\/strong>Solution –\u00a0<\/strong>It is given that
\nTotal revolutions made by the wheel in one \u00a0minute is 360.\u00a0 <\/span>
\n1 second = 360\/6 = 60 <\/span>
\nWe know that <\/span>
\nWhen a \u00a0wheel revolves once it covers \u00a02\u03c0 radian of distance.\u00a0 <\/span>
\nIn one minute, it will turn an angle of 360 \u00d7 2\u03c0 radian = 720 \u03c0 radian <\/span>
\nIn one second, it will turn an angle of 720 \u03c0 radian\/60 = 12 \u03c0 radian {as 1 minute = 60 seconds}\u00a0 <\/span>
\nHence, in one second, the wheel turns an angle of 12\u03c0 radian.<\/span><\/p>\n

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use \u03c0 = 22\/7).
\n<\/strong><\/span>Solution –\u00a0 <\/strong>The radius of circle (r) = 100 cm.\u00a0 <\/span>
\nLength of the arc (l) = 22 cm.\u00a0 <\/span>
\nLet us consider the angle subtended by the arc is \u03b8.\u00a0 \u00a0<\/span>
\nAlso, we know that \u03b8 = l\/r <\/span>
\nThe angle subtended (\u03b8) = 22\/100 radian <\/span>
\nFor finding the degree measure we have to multiply 180\u00b0\/\u03c0 with radian measure <\/span>
\nSo, <\/span>
\n\u03b8 = (22\/100) \u00d7 (180\/\u03c0)\u00a0 <\/span>
\n\u03b8 = (22\/100) \u00d7 (180 \u00d7 7\/22)\u00a0 <\/span>
\n\u03b8 = (22 \u00d7 180 \u00d7 7\/22 \u00d7 100)\u00a0 <\/span>
\n\u03b8 = 126\/10 degree <\/span>
\n\u03b8 = 12 (3\/5) degree
\n<\/span>We know that 1\u00b0 = 60\u2032 <\/span>
\n(3\/5)\u00b0 = 60\u2032 \u00d7 (3\/5) <\/span>= 36\u2032 <\/span>
\nSo
\n12 (3\/5)\u00b0 = 12\u00b0 36\u2032\u00a0 <\/span>
\nHence, the degree measure of the angle subtended at the Centre of a circle is 12\u00b0 36\u2032 <\/span><\/p>\n

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
\n<\/strong>Solution –
\n<\/strong>Diameter of circle (d) = 40 cm <\/span>
\nRadius (r) = d\/2 = 40\/2 = 20 cm <\/span>
\nLet us consider AB as the chord of circle having length 20 cm, and Centre at O.\u00a0<\/span>
\n\"NCER <\/span>
\nIt forms a triangle OAB,\u00a0 <\/span>
\nHaving Radius = OA = OB = 20 cm <\/span>
\nAlso, chord AB = 20 cm <\/span>
\nHence, In \u0394OAB OA = OB = AB. (equilateral triangle.) <\/span>
\nSo angle subtend = (\u03c0\/3) radian <\/span>
\nWe know that \u03b8 = l\/r (where \u03b8 = angle subtended by the arc, l = length of arc, r = radius)\u00a0 <\/span>
\nPutting values of r and \u03b8 we get <\/span>
\n\u03c0\/3 = l\/20 <\/span>
\nSo.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span>
\nl = 20 \u03c0\/3 <\/span>
\nHence, length of the arc is 20\u03c0\/3 cm.\u00a0<\/span><\/p>\n

6. If in two circles, arcs of the same length subtend angles 60\u00b0 and 75\u00b0 at the centre, find the ratio of their radii.
\n<\/strong><\/span>Solution –\u00a0<\/strong>Given that\u00a0 <\/span>
\nAngle subtend by 1st arc (\u03b81<\/sub>) = 60 <\/span>
\nAngle subtend by 2nd arc (\u03b82<\/sub>) = 75 <\/span>
\nWe know that \u03b8 = l\/r <\/span>
\nFor 1st arc \u03b81<\/sub>\u00a0= l1<\/sub>\/r1
\n<\/sub>For 2nd arc \u03b82<\/sub>\u00a0= l2<\/sub>\/r2
\n<\/sub>\u03b81<\/sub>\/\u03b82<\/sub>\u00a0= (l1<\/sub>\/r1<\/sub>)\/(l2<\/sub>\/r2<\/sub>)\u00a0 <\/span>
\n\u03b81<\/sub>\/\u03b82<\/sub>\u00a0= (l\/r1<\/sub>)\/(l\/r2<\/sub>) \u00a0{here l1<\/sub>\u00a0= l2<\/sub> = l}\u00a0 <\/span>
\n\u03b81<\/sub>\/\u03b82<\/sub>\u00a0= r2<\/sub>\/r1
\n<\/sub>60\/75 = r2<\/sub>\/r1
\n<\/sub>r2<\/sub>\/r1<\/sub> = 4\/5 <\/span>
\nr1<\/sub>\/r2<\/sub> = 5\/4 <\/span>
\nHence, ratio of their radius is 5:4.\u00a0 <\/span><\/p>\n

7. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length
\n<\/strong>(i) 10 cm
\n(ii) 15 cm
\n(iii) 21 cm <\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>In a circle of radius r unit, if an arc of length l unit subtends an angle \u03b8 radian at the centre, then \u03b8 = 1\/r
\nWe know that r = 75 cm <\/span><\/p>\n

(i)\u00a0<\/strong>Given that <\/span>
\nLength of an arc (l) = 10 cm <\/span>
\nRadius which represents length of pendulum(r) = 75 <\/span>
\nAs We know that \u03b8 = l\/r <\/span>
\nSo \u03b8 = 10\/75 = 2\/15 rad <\/span>
\nHence, \u03b8 = 2\/15 rad<\/span><\/p>\n

(ii)<\/strong> Given that <\/span>
\nLength of an arc (l) = 15 cm <\/span>
\nRadius which represents length of pendulum (r) = 75 <\/span>
\nAs We know that \u03b8 = l\/r <\/span>
\nSo \u03b8 = 15\/75 = 1\/5 rad <\/span>
\nHence, \u03b8 = 1\/5 rad <\/span><\/p>\n

(iii)<\/strong> Given that <\/span>
\nLength of an arc (l) = 21 cm <\/span>
\nRadius which represents length of pendulum(r) = 75 <\/span>
\nAs We know that \u03b8 = l\/r <\/span>
\nSo \u03b8 = 21\/75 = 7\/25 rad <\/span>
\nHence, \u03b8 = 7\/25 radian<\/span><\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0 The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 3 Trigonometric Functions Exercise 3.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 3 (Trigonometric Functions)\u00a0 Exercise – 3.1 1. Find the radian measures corresponding to the following degree measures: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1674],"tags":[1675,1679,1692,1693,1678,1680,1681],"class_list":["post-8408","post","type-post","status-publish","format-standard","hentry","category-class-11-maths","tag-class-11-ncert-mathematics-solutions","tag-class-11-ncert-solutions","tag-ncert-class-11-mathematics-chapter-3-trigonometric-functions-solutions","tag-ncert-class-11-mathematics-exercise-3-1-solutions","tag-ncert-class-11-mathematics-solutions","tag-ncert-solutions-class-11-mathematics","tag-ncert-solutions-class-11-maths"],"yoast_head":"\nNCERT Class 11 Maths Solutions - Exercise 3.1 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 11 Maths\u00a0The NCERT Solutions in English Language for Class 11 Mathematics Chapter - 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