{"id":5943,"date":"2024-06-25T12:31:35","date_gmt":"2024-06-25T07:01:35","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=5943"},"modified":"2024-06-25T17:39:04","modified_gmt":"2024-06-25T12:09:04","slug":"samacheer-kalvi-11th-business-maths-guide-chapter-4-ex-4-5","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-11th-business-maths-guide-chapter-4-ex-4-5\/","title":{"rendered":"Samacheer Kalvi 11th Business Maths Guide Chapter 4 Trigonometry Ex 4.5"},"content":{"rendered":"

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.5 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide<\/a>\u00a0Pdf 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 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.5<\/h2>\n

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.5 Text Book Back Questions and Answers<\/h3>\n

Question 1.
\nThe degree measure of \\(\\frac{\\pi}{8}\\) is:
\n(a) 20\u00b060′
\n(b) 22\u00b030′
\n(c) 22\u00b060′
\n(d) 20\u00b030′
\nAnswer:
\n(b) 22\u00b030′
\nHint:
\nWe know that, one radian = \\(\\frac{180^{\\circ}}{\\pi}\\)
\n\u2234 \\(\\frac{\\pi}{8}=\\frac{180^{\\circ}}{\\pi} \\times \\frac{\\pi}{8}\\) degrees
\n= \\(\\frac{45^{\\circ}}{2}\\)
\n= 22.5\u00b0
\n= 22\u00b030′<\/p>\n

Question 2.
\nThe radian measure of 37\u00b030′ is:
\n(a) \\(\\frac{5 \\pi}{24}\\)
\n(b) \\(\\frac{3 \\pi}{24}\\)
\n(c) \\(\\frac{7 \\pi}{24}\\)
\n(d) \\(\\frac{9 \\pi}{24}\\)
\nAnswer:
\n(a) \\(\\frac{5 \\pi}{24}\\)
\nHint:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nIf tan \u03b8 = \\(\\frac{1}{\\sqrt{5}}\\) and \u03b8 lies in the first quadrant then cos \u03b8 is:
\n(a) \\(\\frac{1}{\\sqrt{6}}\\)
\n(b) \\(\\frac{-1}{\\sqrt{6}}\\)
\n(c) \\(\\frac{\\sqrt{5}}{\\sqrt{6}}\\)
\n(d) \\(\\frac{-\\sqrt{5}}{\\sqrt{6}}\\)
\nAnswer:
\n(c) \\(\\frac{\\sqrt{5}}{\\sqrt{6}}\\)
\nHint:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 4.
\nThe value of sin 15\u00b0 is:
\n(a) \\(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\)
\n(b) \\(\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}\\)
\n(c) \\(\\frac{\\sqrt{3}}{\\sqrt{2}}\\)
\n(d) \\(\\frac{\\sqrt{3}}{2 \\sqrt{2}}\\)
\nAnswer:
\n(b) \\(\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}\\)
\nHint:
\nsin 15\u00b0 = sin(45\u00b0 – 30\u00b0)
\n= sin 45\u00b0 cos 30\u00b0 – cos 45\u00b0 sin 30\u00b0
\n= \\(\\frac{1}{\\sqrt{2}} \\times \\frac{\\sqrt{3}}{2}-\\frac{1}{\\sqrt{2}} \\times \\frac{1}{2}\\)
\n= \\(\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nThe value of sin(-420\u00b0)
\n(a) \\(\\frac{\\sqrt{3}}{2}\\)
\n(b) \\(-\\frac{\\sqrt{3}}{2}\\)
\n(c) \\(\\frac{1}{2}\\)
\n(d) \\(\\frac{-1}{2}\\)
\nAnswer:
\n(b) \\(-\\frac{\\sqrt{3}}{2}\\)
\nHint:
\nsin(-420\u00b0) = -sin(420\u00b0) [\u2235 sin(-\u03b8) = -sin \u03b8]
\n= -sin(360\u00b0 + 60\u00b0)
\n= -sin 60\u00b0
\n= \\(-\\frac{\\sqrt{3}}{2}\\)<\/p>\n

Question 6.
\nThe value of cos(-480\u00b0) is:
\n(a) \u221a3
\n(b) \\(-\\frac{\\sqrt{3}}{2}\\)
\n(c) \\(\\frac{1}{2}\\)
\n(d) \\(\\frac{-1}{2}\\)
\nAnswer:
\n(d) \\(\\frac{-1}{2}\\)
\nHint:
\ncos(-480\u00b0) = cos 480\u00b0 [\u2235 cos(-\u03b8) = cos \u03b8]
\n= cos(360\u00b0 + 120\u00b0)
\n= cos 120\u00b0
\n= cos(180\u00b0 – 60\u00b0)
\n= -cos 60\u00b0
\n= \\(\\frac{-1}{2}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nThe value of sin 28\u00b0 cos 17\u00b0 + cos 28\u00b0 sin 17\u00b0
\n(a) \\(\\frac{1}{\\sqrt{2}}\\)
\n(b) 1
\n(c) \\(\\frac{-1}{\\sqrt{2}}\\)
\n(d) 0
\nAnswer:
\n(a) \\(\\frac{1}{\\sqrt{2}}\\)
\nHint:
\nsin 28\u00b0 cos 17\u00b0 + cos 28\u00b0 sin 17\u00b0 = sin(28\u00b0 + 17\u00b0)
\nThis is of the form sin(A + B), A = 28\u00b0, B = 17\u00b0
\n= sin 45\u00b0
\n= \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

Question 8.
\nThe value of sin 15\u00b0 cos 15\u00b0 is:
\n(a) 1
\n(b) \\(\\frac{1}{2}\\)
\n(c) \\(\\frac{\\sqrt{3}}{2}\\)
\n(d) \\(\\frac{1}{4}\\)
\nAnswer:
\n(d) \\(\\frac{1}{4}\\)
\nHint:
\nsin 15\u00b0 cos 15\u00b0 = \\(\\frac{1}{2}\\) (2 sin 15\u00b0 cos 15\u00b0)
\n= \\(\\frac{1}{2}\\) (sin 30\u00b0)
\n= \\(\\frac{1}{2}\\left(\\frac{1}{2}\\right)\\)
\n= \\(\\frac{1}{4}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nThe value of sec A sin(270\u00b0 + A) is:
\n(a) -1
\n(b) cos2<\/sup> A
\n(c) sec2<\/sup> A
\n(d) 1
\nAnswer:
\n(a) -1
\nHint:
\nsec A (sin(270\u00b0 + A)) = \\(\\frac{1}{\\cos A}\\) (-cos A) = -1<\/p>\n

Question 10.
\nIf sin A + cos A = 1 then sin 2A is equal to:
\n(a) 1
\n(b) 2
\n(c) 0
\n(d) \\(\\frac{1}{2}\\)
\nAnswer:
\n(c) 0
\nHint:
\nGiven sin A + cos A = 1
\nSquaring both sides we get
\nsin2<\/sup> A + cos2<\/sup> A + 2 sin A cos A = 1
\n1 + sin 2A = 1
\nsin 2A = 0<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nThe value of cos2<\/sup> 45\u00b0 – sin2<\/sup> 45\u00b0 is:
\n(a) \\(\\frac{\\sqrt{3}}{2}\\)
\n(b) \\(\\frac{1}{2}\\)
\n(c) 0
\n(d) \\(\\frac{1}{\\sqrt{2}}\\)
\nAnswer:
\n(c) 0
\nHint:
\ncos2<\/sup> 45\u00b0 – sin2<\/sup> 45\u00b0
\n= cos 2 \u00d7 45\u00b0 (\u2235 cos2<\/sup> A – sin2<\/sup> A = cos 2A)
\n= cos 90\u00b0
\n= 0<\/p>\n

Question 12.
\nThe value of 1 – 2 sin2<\/sup> 45\u00b0 is:
\n(a) 1
\n(b) \\(\\frac{1}{2}\\)
\n(c) \\(\\frac{1}{4}\\)
\n(d) 0
\nAnswer:
\n(d) 0
\nHint:
\n1 – 2 sin2<\/sup> 45\u00b0
\n= cos(2 \u00d7 45\u00b0) [\u2235 cos 2A = 1 – 2 sin2<\/sup> A]
\n= cos 90\u00b0
\n= 0<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nThe value of 4 cos3<\/sup> 40\u00b0 – 3 cos 40\u00b0 is
\n(a) \\(\\frac{\\sqrt{3}}{2}\\)
\n(b) \\(-\\frac{1}{2}\\)
\n(c) \\(\\frac{1}{2}\\)
\n(d) \\(\\frac{1}{\\sqrt{2}}\\)
\nAnswer:
\n(b) \\(-\\frac{1}{2}\\)
\nHint:
\n4 cos3<\/sup> 40\u00b0 – 3 cos 40\u00b0
\n= cos (3 \u00d7 40\u00b0) [\u2235 cos 3A = 4 cos3<\/sup> A – 3 cos A]
\n= cos 120\u00b0
\n= cos (180\u00b0 – 60\u00b0)
\n= -cos 60\u00b0
\n= \\(-\\frac{1}{2}\\)<\/p>\n

Question 14.
\nThe value of \\(\\frac{2 \\tan 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}\\) is:
\n(a) \\(\\frac{1}{2}\\)
\n(b) \\(\\frac{1}{\\sqrt{3}}\\)
\n(c) \\(\\frac{\\sqrt{3}}{2}\\)
\n(d) \u221a3
\nAnswer:
\n(d) \u221a3
\nHint:
\nWe know that sin 2A = \\(\\frac{2 \\tan A}{1+\\tan ^{2} A}\\)
\n\\(\\frac{2 \\tan 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}\\) = sin(2 \u00d7 30\u00b0) = sin 60\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)
\n= tan 2A
\n= tan 60\u00b0
\n= \u221a3<\/p>\n

\"Samacheer<\/p>\n

Question 15.
\nIf sin A = \\(\\frac{1}{2}\\) then 4 cos3<\/sup> A – 3 cos A is:
\n(a) 1
\n(b) 0
\n(c) \\(\\frac{\\sqrt{3}}{2}\\)
\n(d) \\(\\frac{1}{\\sqrt{2}}\\)
\nAnswer:
\n(b) 0
\nHint:
\nGiven sin A = \\(\\frac{1}{2}\\)
\nsin A = sin 30\u00b0
\n\u2234 A = 30\u00b0
\n[\u2235 4 cos3<\/sup> A – 3 cos A = cos 3A]
\n= cos(3 \u00d7 30\u00b0)
\n= cos 90\u00b0
\n= 0<\/p>\n

Question 16.
\nThe value of \\(\\frac{3 \\tan 10^{\\circ}-\\tan ^{3} 10^{\\circ}}{1-3 \\tan ^{2} 10^{\\circ}}\\) is:
\n(a) \\(\\frac{1}{\\sqrt{3}}\\)
\n(b) \\(\\frac{1}{2}\\)
\n(c) \\(\\frac{\\sqrt{3}}{2}\\)
\n(d) \\(\\frac{1}{\\sqrt{2}}\\)
\nAnswer:
\n(a) \\(\\frac{1}{\\sqrt{3}}\\)
\nHint:
\n\\(\\frac{3 \\tan 10^{\\circ}-\\tan ^{3} 10^{\\circ}}{1-3 \\tan ^{2} 10^{\\circ}}\\) = tan(3 \u00d7 10\u00b0) [\u2235 tan 3A = \\(\\frac{3 \\tan A-\\tan ^{3} A}{1-3 \\tan ^{2} A}\\)]
\n= tan 30\u00b0
\n= \\(\\frac{1}{\\sqrt{3}}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 17.
\nThe value of cosec-1<\/sup> (\\(\\frac{2}{\\sqrt{3}}\\)) is:
\n(a) \\(\\frac{\\pi}{4}\\)
\n(b) \\(\\frac{\\pi}{2}\\)
\n(c) \\(\\frac{\\pi}{3}\\)
\n(d) \\(\\frac{\\pi}{6}\\)
\nAnswer:
\n(c) \\(\\frac{\\pi}{3}\\)
\nHint:
\nLet cosec-1<\/sup> (\\(\\frac{2}{\\sqrt{3}}\\))
\n\\(\\frac{2}{\\sqrt{3}}\\) = cosec A
\ncosec A = \\(\\frac{2}{\\sqrt{3}}\\)
\nsin A = \\(\\frac{\\sqrt{3}}{2}\\) = sin 60\u00b0
\n\u2234 A = 60\u00b0 = \\(\\frac{\\pi}{3}\\)<\/p>\n

Question 18.
\nsec-1<\/sup> (\\(\\frac{2}{3}\\)) + cosec-1<\/sup> (\\(\\frac{2}{3}\\)) =
\n(a) \\(\\frac{-\\pi}{2}\\)
\n(b) \\(\\frac{\\pi}{2}\\)
\n(c) \u03c0
\n(d) -\u03c0
\nAnswer:
\n(b) \\(\\frac{\\pi}{2}\\)
\nHint:
\nWe know that sec-1<\/sup> x + cosec-1<\/sup> x = \\(\\frac{\\pi}{2}\\)
\n\u2234 sec-1<\/sup> (\\(\\frac{2}{3}\\)) + cosec-1<\/sup> (\\(\\frac{2}{3}\\)) = \\(\\frac{\\pi}{2}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 19.
\nIf \u03b1 and \u03b2 be between 0 and \\(\\frac{\\pi}{2}\\) and if cos(\u03b1 + \u03b2) = \\(\\frac{12}{13}\\) and sin(\u03b1 – \u03b2) = \\(\\frac{3}{5}\\) then sin 2\u03b1 is:
\n(a) \\(\\frac{16}{15}\\)
\n(b) 0
\n(c) \\(\\frac{56}{65}\\)
\n(d) \\(\\frac{64}{65}\\)
\nAnswer:
\n(c) \\(\\frac{56}{65}\\)
\nHint:
\n\"Samacheer
\nGiven that cos(\u03b1 + \u03b2) = \\(\\frac{12}{13}\\)
\n\u2234 sin(\u03b1 + \u03b2) = \\(\\frac{5}{13}\\)
\n\"Samacheer
\nAlso given that sin(\u03b1 – \u03b2) = \\(\\frac{3}{5}\\)
\n\u2234 cos(\u03b1 – \u03b2) = \\(\\frac{4}{5}\\)
\nsin 2\u03b1 = sin[(\u03b1 + \u03b2) + (\u03b1 – \u03b2)]
\n= sin(\u03b1 + \u03b2) cos(\u03b1 – \u03b2) + cos(\u03b1 + \u03b2) sin(\u03b1 – \u03b2)
\n= \\(\\frac{5}{13} \\times \\frac{4}{5}+\\frac{12}{13} \\times \\frac{3}{5}\\)
\n= \\(\\frac{20}{65}+\\frac{36}{65}\\)
\n= \\(\\frac{56}{65}\\)<\/p>\n

Question 20.
\nIf tan A = \\(\\frac{1}{2}\\) and tan B = \\(\\frac{1}{3}\\) then tan(2A + B) is equal to:
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4
\nAnswer:
\n(c) 3
\nHint:
\nGiven tan A = \\(\\frac{1}{2}\\), tan B = \\(\\frac{1}{3}\\)
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 21.
\ntan(\\(\\frac{\\pi}{4}\\) – x) is:
\n(a) \\(\\left(\\frac{1+\\tan x}{1-\\tan x}\\right)\\)
\n(b) \\(\\left(\\frac{1-\\tan x}{1+\\tan x}\\right)\\)
\n(c) 1 – tan x
\n(d) 1 + tan x
\nAnswer:
\n(b) \\(\\left(\\frac{1-\\tan x}{1+\\tan x}\\right)\\)
\nHint:
\n\"Samacheer
\n[\u2235 tan \\(\\frac{\\pi}{4}\\) = 1]<\/p>\n

Question 22.
\n\\(\\sin \\left(\\cos ^{-1} \\frac{3}{5}\\right)\\) is:
\n(a) \\(\\frac{3}{5}\\)
\n(b) \\(\\frac{5}{3}\\)
\n(c) \\(\\frac{4}{5}\\)
\n(d) \\(\\frac{5}{4}\\)
\nAnswer:
\n(c) \\(\\frac{4}{5}\\)
\nHint:
\n\"Samacheer
\nLet cos-1<\/sup> (\\(\\frac{3}{5}\\)) = A
\n\\(\\frac{3}{5}\\) = cos A
\nsin A = \\(\\frac{4}{5}\\)
\nNow sin(cos-1 (\\(\\frac{3}{5}\\))) = sin A = \\(\\frac{4}{5}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 23.
\nThe value of \\(\\frac{1}{cosec(-45^{\\circ})}\\) is:
\n(a) \\(\\frac{-1}{\\sqrt{2}}\\)
\n(b) \\(\\frac{1}{\\sqrt{2}}\\)
\n(c) \u221a2
\n(d) -\u221a2
\nAnswer:
\n(a) \\(\\frac{-1}{\\sqrt{2}}\\)
\nHint:
\n\\(\\frac{1}{cosec(-45^{\\circ})}\\) = sin(-45\u00b0)
\n= -sin 45\u00b0
\n= \\(\\frac{-1}{\\sqrt{2}}\\)<\/p>\n

Question 24.
\nIf p sec 50\u00b0 = tan 50\u00b0 then p is:
\n(a) cos 50\u00b0
\n(b) sin 50\u00b0
\n(c) tan 50\u00b0
\n(d) sec 50\u00b0
\nAnswer:
\n(b) sin 50\u00b0
\nHint:
\np sec 50\u00b0 = tan 50\u00b0
\np(\\(\\frac{1}{\\cos 50^{\\circ}}\\)) = \\(\\frac{\\sin 50^{\\circ}}{\\cos 50^{\\circ}}\\)
\n\u2234 p = sin 50\u00b0<\/p>\n

\"Samacheer<\/p>\n

Question 25.
\n(\\(\\frac{cos x}{cosec x})-\\sqrt{1-\\sin ^{2} x} \\sqrt{1-\\cos ^{2} x}\\) is:
\n(a) cos2<\/sup> x – sin2<\/sup> x
\n(b) sin2<\/sup> x – cos2<\/sup> x
\n(c) 1
\n(d) 0
\nAnswer:
\n(d) 0
\nHint:
\n(\\(\\frac{cos x}{cosec x})-\\sqrt{1-\\sin ^{2} x} \\sqrt{1-\\cos ^{2} x}\\)
\n= cos x \u00d7 sin x – \\(\\sqrt{\\cos ^{2} x} \\sqrt{\\sin ^{2} x}\\)
\n= cos x \u00d7 sin x – cos x \u00d7 sin x
\n= 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.5 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide\u00a0Pdf 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 11th Business Maths Solutions Chapter 4 Trigonometry …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[6],"tags":[],"class_list":["post-5943","post","type-post","status-publish","format-standard","hentry","category-class-11"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/5943"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=5943"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/5943\/revisions"}],"predecessor-version":[{"id":40453,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/5943\/revisions\/40453"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=5943"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=5943"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=5943"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}