{"id":22480,"date":"2020-02-05T15:45:53","date_gmt":"2020-02-05T10:15:53","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=22480"},"modified":"2021-07-02T15:25:48","modified_gmt":"2021-07-02T09:55:48","slug":"1st-puc-basic-maths-question-bank-chapter-13","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-question-bank-chapter-13\/","title":{"rendered":"1st PUC Basic Maths Question Bank Chapter 13 Angles and Trigonometric Ratios"},"content":{"rendered":"

Students can Download Basic Maths Chapter 13 Angles and Trigonometric Ratios Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Basic Maths Question Bank Chapter 13 Angles and Trigonometric Ratios<\/h2>\n

Question 1.
\nFind the length of the arc and the area of the sector formed by an arc of a circle of radius 8 cm subtending an angle 25\u00b0 as the centre.
\nAnswer:
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 2.
\nFind the angle between minute hand and the hourhand of a clock when the time is 5.40.
\nAnswer:
\nAngle described by the hourhand is 12 hours = 360\u00b0
\n\u2234 per hour = \\(\\frac{360}{12}\\) = 30\u00b0
\n\"1st
\nNow the angle described by minute hand is 60\u00b0 minutes = 360\u00b0
\ni.e., Angle in 40\u00b0 minutes = 40 \u00d7 6\u00b0 = 240\u00b0,
\n\"1st<\/p>\n

Question 3.
\nA circular wire of radius 6 cm is cut and bent so as to lie along the circumference of a loop whose radius in 66 cm. Find in degrees the angle which subtended at the centre of the loop.
\nAnswer:
\nThe length of the arc subtended is the circumference of the circle of radius 6cm.
\n\u2234 S = 2\u03c0r = 2\u03c0 \u00d7 6 = 127 cm
\nradius of the loop = 66cm
\n\"1st<\/p>\n

Question 4.
\nThe angles of a triangle are in A.P and the greatest is double the least. Express the angles in degrees and radians.
\nAnswer:
\nLet the angle, be (a – d)\u00b0 a\u00b0 and (a + d)\u00b0, (a > d > 0\u00b0)
\n\u2234 (a – d) + a + (a + d) = 180\u00b0 \u21d2 3a = 180\u00b0 \u21d2 a = 60\u00b0
\ni. e., (a – d) is the least and (a + d) is the greatest,
\ngiven (a + d) = 2 (a – d) = a + d = 2a – 2d
\n\u21d2 a – 3d = 0 \u21d2 60\u00b0 – 3d = 0 \u21d2 d = 20\u00b0
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 5.
\nExpress 25\u00b0. 30′ 10″ in radians.
\nAnswer:
\n\"1st<\/p>\n

Trigonometric Ratios of Acute Angle<\/span><\/p>\n

Question 1.
\nProve that tan \u03b1\\(\\sqrt{1-\\sin ^{2} \\alpha}\\) = sin \u03b1
\nAnswer:
\nLHS : tan \u03b1 . \\(\\sqrt{1-\\sin ^{2} \\alpha}\\) = tan \u03b1 . \\(\\sqrt{\\cos ^{2} \\alpha}\\)
\n= \\(\\frac{\\sin \\alpha}{\\cos \\alpha}\\) \u00d7 cos \u03b1 = sin \u03b1 = RHS<\/p>\n

Question 2.
\nProve that \\(\\frac{1-\\cos \\theta}{\\sin \\theta}=\\frac{\\sin \\theta}{1+\\cos \\theta}\\)
\nAnswer:
\n\"1st<\/p>\n

Question 3.
\nProve that \\(\\sqrt{\\frac{1-\\sin A}{1+\\sin A}}\\) = sec A – tan A.
\nAnswer:
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 4.
\nShow that \\(\\frac{\\tan A}{1-\\cot A}+\\frac{\\cot A}{1-\\tan A}\\) = 1 + sec A cosec A.
\nAnswer:
\n\"1st<\/p>\n

Question 5.
\nIf x = a cos \u03b8 + sin \u03b8. y = a sin \u03b8 – b cos \u03b8. Show that x2<\/sup> + y2<\/sup> = a2<\/sup> + b2<\/sup>
\nAnswer:
\nConsider x2<\/sup> = (a cos \u03b8 + b sin \u03b8)2<\/sup>
\n\u21d2 x2<\/sup> = a2<\/sup>cos2<\/sup>\u03b8 + 2ab cos \u03b8 sin \u03b8 + b2<\/sup> sin2<\/sup>\u03b8 …….(1)
\nAgain consider
\ny2<\/sup> = (a sin \u03b8 – b cos \u03b8)2<\/sup> => y2<\/sup> = a2<\/sup> sin2<\/sup>\u03b8 – ab cos \u03b8 sin \u03b8 + b2<\/sup> cos2<\/sup> \u03b8 …….. (2)
\n(1) + (2) \u21d2 x2<\/sup> + y2<\/sup> = a2<\/sup> + b2<\/sup><\/p>\n

\"1st<\/p>\n

Question 6.
\nIf sin \u03b8 = \\(\\frac{3}{5}\\) and \u03b8 is acute, find the value of
\nAnswer:
\nBy data sin \u03b8 = \\(\\frac{3}{5}\\)
\ncos \u03b8 = \\(\\frac{3}{5}\\) . tan \u03b8 = \\(\\frac{3}{4}\\), cot = \\(\\frac{4}{3}\\)
\n\"1st<\/p>\n

Trigonometric Ratios of some Standard Angles<\/span><\/p>\n

Question 1.
\nFind the value of cos \\(\\frac{\\pi}{3}\\) – sin \\(\\frac{\\pi}{6}\\) – tan3<\/sup> \\(\\frac{\\pi}{4}\\)
\nAnswer:
\n\"1st<\/p>\n

Question 2.
\nFind value of 3 tan2<\/sup>30\u00b0 + \\(\\frac{4}{3}\\)sin2<\/sup>60\u00b0 – \\(\\frac{1}{2}\\)cosec2<\/sup>30\u00b0 – \\(\\frac{1}{3}\\)cos2<\/sup> 45\u00b0
\nAnswer:
\n\"1st<\/p>\n

Question 3.
\nFind x, if \\(\\frac{x \\csc ^{2} 30^{\\circ} \\cdot \\sec ^{2} 45^{\\circ}}{6 \\cos ^{2} 45^{\\circ} \\sin 30^{\\circ}}\\) = tan2<\/sup> 45\u00b0 – tan2<\/sup> 60\u00b0
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 4.
\nShow that cos2<\/sup> \\(\\frac{\\pi}{4}\\) – cos4<\/sup> \\(\\frac{\\pi}{6}\\) + sin4<\/sup> \\(\\frac{\\pi}{6}\\) + sin4<\/sup> \\(\\frac{\\pi}{3}\\) = \\(\\frac{9}{16}\\)
\nAnswer:
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 5.
\n\"1st
\nAnswer:
\n\"1st<\/p>\n

Heights and Distances<\/span><\/p>\n

Question 1.
\nA person is at the top of tower 75 feet high from there he observes a vertical pole and finds the angles of depression of the top and the bottom of the pole which are 30\u00b0 and 60\u00b0 respectively. Find the height of the pole.
\nAnswer:
\nLet AB be the tower = 75 feet,
\nCD be the pole = x
\nAX is \u2225lel<\/sup> to MC
\n\u2234 A\u0108M = X\u00c2C = 30\u00b0
\nAX \u2225lel<\/sup> BD
\nHence AD\u0302B = X\u00c2D = 60\u00b0
\n\"1st
\nNow from the right triangle ADB, we have
\ntan 60\u00b0 = \\(\\frac{\\mathrm{AB}}{\\mathrm{BD}}\\)
\n\u21d2 BD = \\(\\frac{\\mathrm{AB}}{\\tan 60^{\\circ}}\\) = BD = \\(\\frac{75}{\\sqrt{3}}\\)
\ntan 30\u00b0 = \\(\\frac{\\mathrm{AM}}{\\mathrm{MC}}\\)
\n\u21d2 AM = MC . tan 30\u00b0
\n\u21d2 AM = BD \\(\\frac{1}{\\sqrt{3}}=\\frac{75}{\\sqrt{3}} \\cdot \\frac{1}{\\sqrt{3}}\\) = 25
\nNow x = CD = MB
\n\u21d2 x = AB – AM = 75 – 25 = 50 feet<\/p>\n

Question 2.
\nA person standing on the bank of a river observe that the angle of elevation of the top of a tree on the opposite bank is 60\u00b0, when he returns 60 feet from the bank, he finds the angle to be 30\u00b0. Find the height of the tree and breadth of the river.
\nAnswer:
\nPQ represent tree A and B be the points of observation. BQ is the breadth of the river. From the figure
\n\"1st
\ntan 60\u00b0 = \\(\\frac{P Q}{B Q}\\)
\n\u21d2 PQ = BQ . \u221a3
\nAgain from the figure we have
\ntan 30\u00b0 = \\(\\frac{P Q}{A Q}\\)
\n\u21d2 AQ = \\(\\frac{P Q}{\\tan 30^{\\circ}}\\) = AQ = \u221a3PQ
\n\u21d2 AQ = \u221a3 . (\u221a3BQ) = 3BQ. Now BQ \u21d2 AQ – AB
\n\u21d2 BQ = 3BQ – AB
\n\u21d2 2BQ = 60
\n\u2234 BQ = 30\u00b0 is the breadth of the river
\n\u2234 height of the tree = PQ = \u221a3.BQ = \u221a3.30 = 30\u221a3<\/p>\n

\"1st<\/p>\n

Question 3.
\nThe angle of elevation of a stationary cloud from a point 3000 cm above the lake is 30\u00b0 and the angle of depression of its reflection in the lake is 60\u00b0. What is the height of the cloud above the lake.
\nAnswer:
\nD – the image of the cloud C.
\nAB the surface of the lake
\nAlso CB = BD
\nLet P be the position of observation.
\nThus by data AP = 3000 cm
\nBy data, EP\u0302C = 30\u00b0, EP\u0302D = 60\u00b0
\n\"1st
\n\"1st
\n\u21d2 PE = \u221a3 . CE ….. (1)
\n\"1st
\n\u21d2 PE = \\(\\frac{\\mathrm{ED}}{\\sqrt{3}}\\) ….. (2)
\nFrom (1) and (2) we get
\n\u21d2 \u221a3CE = \\(\\frac{1}{\\sqrt{3}}\\)ED = 3CE = ED
\n\u21d2 3(BC – BE) = EB + BD
\n\u21d2 3(BC – 3000) = 3000 + BC (\u2235 BD = BC and BE PA = 3000)
\n\u21d2 3BC – 9000 = 3000 + BC
\n\u21d2 2BC = 12,000 = BC = 6000 cm
\n\u2234 height of the cloud = 600 cms<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Chapter 13 Angles and Trigonometric Ratios Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Basic Maths Question Bank Chapter 13 Angles and Trigonometric Ratios Question 1. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22480"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=22480"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22480\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=22480"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=22480"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=22480"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}