{"id":47073,"date":"2023-12-21T06:55:27","date_gmt":"2023-12-21T01:25:27","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=47073"},"modified":"2023-12-21T16:32:00","modified_gmt":"2023-12-21T11:02:00","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-18-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-18-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 18 Trigonometric Ratios and Standard Angles Chapter Test"},"content":{"rendered":"
Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question P.Q. (tan\u03b8 – 1) (cosec 3\u03b8 – 1) = 0 Question 7. Question P.Q. Question 8. Question 9. Question 10. Question 11. (ii) \u03b8 and (\u03b8 – 20\u00b0) are acute angles Question 12. ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 18 Trigonometric Ratios and Standard Angles Chapter Test Question 1. Find the values of : (i) sin2 60\u00b0 – cos2 45\u00b0 + 3tan2 30\u00b0 (ii) (iii) sec 30\u00b0 tan 60\u00b0 + sin 45\u00b0 cosec 45\u00b0 + cos 30\u00b0 cot 60\u00b0 Answer: (i) sin2 60\u00b0 – cos2\u00a045\u00b0 […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[],"yoast_head":"\n
\nFind the values of :
\n(i) sin2<\/sup> 60\u00b0 – cos2<\/sup> 45\u00b0 + 3tan2<\/sup> 30\u00b0
\n(ii) \\(\\frac{2 \\cos ^{2} 45^{\\circ}+3 \\tan ^{2} 30^{\\circ}}{\\sqrt{3} \\cos 30^{\\circ}+\\sin 30^{\\circ}}\\)
\n(iii) sec 30\u00b0 tan 60\u00b0 + sin 45\u00b0 cosec 45\u00b0 + cos 30\u00b0 cot 60\u00b0
\nAnswer:
\n(i) sin2<\/sup> 60\u00b0 – cos2<\/sup>\u00a045\u00b0 + 3tan2<\/sup> 30\u00b0
\n
\n
\n<\/p>\n
\nTaking A = 30\u00b0, verify that
\n(i) cos4<\/sup> A – sin4<\/sup> A = cos 2A
\n(ii) 4cos A cos (60\u00b0 – A) cos (60\u00b0 + A) = cos 3 A.
\nAnswer:
\n(i) cos4<\/sup> A – sin4<\/sup> A = cos 2A
\nTaking A = 30\u00b0
\nL.H.S.: cos4<\/sup> A – sin4<\/sup> A = cos4<\/sup> 30\u00b0 – sin4<\/sup> 30\u00b0
\n
\nHence, L.H.S. = R.H.S. result is proved.
\n(ii) 4 cos A cos (60\u00b0- A) cos (60\u00b0 + A) = cos 3 A
\nL.H.S. = 4cos A cos (60\u00b0 – A) cos (60\u00b0 + A)
\n= 4cos 30\u00b0 cos (60\u00b0 – 30\u00b0) cos (60\u00b0 + 30\u00b0) [\u2235 taking A = 30\u00b0]
\n= 4cos 30\u00b0 cos 30\u00b0 cos 90\u00b0 = 4 \u00d7 \\(\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2}\\) \u00d7 0 = 0
\nR.H.S.: cos 3A
\n= cos (3 \u00d7 30\u00b0) = cos 90\u00b0 = 0
\nHence, L.H.S. = R.H.S. result is proved.<\/p>\n
\nIf A = 45\u00b0 and B = 30\u00b0, verify that
\n
\nAnswer:
\n
\n<\/p>\n
\nTaking A = 60\u00b0 and B = 30\u00b0, verify that
\n
\nAnswer:
\n(i) A = 60\u00b0, B = 30\u00b0
\n
\n
\n<\/p>\n
\nIf \\( \\sqrt{{2}} \\) tan 2 \u03b8 = \\( \\sqrt{{6}} \\) and \u03b8\u00b0 < 2\u03b8 < 90\u00b0, find the value of sin \u03b8 + \\( \\sqrt{{3}} \\) cos \u03b8 – 2 tan2<\/sup> \u03b8.
\nAnswer:
\n<\/p>\n
\nIf 3\u03b8 is an acute angle, solve the following equations for \u03b8 :
\n(i) ( cosec 3\u03b8 – 2) (cot 2\u03b8 – 1) = 0<\/p>\n
\n(tan\u03b8 – 1) (cosec 3\u03b8 – 1) = 0
\nAnswer:
\n(i) (cosec 3\u03b8 – 2) (cot 2\u03b8 – 1) = 0
\nEither, cosec 3\u03b8 – 2 or cot 2\u03b8 – 1 = 0
\n\u21d2 cosec 3\u03b8 = 2 or cot 2\u03b8 = 1
\n\u21d2 cosec 3\u03b8 = cosec 3\u03b8\u00b0 or cot 2\u03b8 =cot 45\u00b0
\n\u21d2 3\u03b8 = 30\u00b0 or 2\u03b8 = 45\u00b0
\n<\/p>\n
\nEither tan\u03b8 – 1 = 0 or cosec 3\u03b8 – 1 = 0
\n\u21d2 tan\u03b8 = 1 or cosec 3\u03b8 = 1
\n\u21d2 tan\u03b8 = tan45\u00b0 or cosec 3 \u03b8 = cosec 90\u00b0
\n\u21d2 \u03b8 = 45\u00b0 or 3\u03b8 = 90\u00b0 or \u03b8 = \\(\\frac{90^{\\circ}}{3}\\)
\n= 30\u00b0
\nHence, \u03b8 = 45\u00b0 or 30\u00b0<\/p>\n
\nIf tan (A + B) = \\( \\sqrt{{3}} \\) and tan (A – B) = 1 and A, B (B < A) are acute angles, find the values of A and B.
\nAnswer:
\nGiven that tan (A + B) = \\( \\sqrt{{3}} \\)
\n\u21d2 tan ( A + B) = tan 60\u00b0 [tan 60\u00b0 = \\( \\sqrt{{3}} \\) ]
\n\u21d2 A + B = 60\u00b0 ………..(1)
\nAlso, tan (A – B) = 1
\n\u21d2 tan (A – B) = tan 45\u00b0 [tan 45\u00b0 = 1]
\n\u21d2 A – B = 45\u00b0 …………..(2)
\nFrom equation (1) and (2), we get
\n
\nSubstituting the value of A in equation (1), we get
\n<\/p>\n
\nFind the values of :
\n
\nAnswer:
\n
\n
\n<\/p>\n
\nWithout using trigonometrical tables, evaluate the following :
\n(i) sin2<\/sup> 28\u00b0 + sin2<\/sup> 62\u00b0 – tan2<\/sup> 45\u00b0
\n(ii) \\(2 \\frac{\\cos 27^{\\circ}}{\\sin 63^{\\circ}}+\\frac{\\tan 27^{\\circ}}{\\cot 63^{\\circ}}+\\cos 0^{\\circ}\\)
\n(iii) cos 18\u00b0 sin 72\u00b0 + sin 18\u00b0 cos 72\u00b0
\n(iv) 5 sin 50\u00b0 sec 40\u00b0 – 3 cos 59\u00b0 cosec 31\u00b0
\nAnswer:
\n(i) sin2<\/sup> 28\u00b0 + sin2<\/sup> 62\u00b0 – tan2<\/sup> 45\u00b0
\n= sin2<\/sup> 28\u00b0 + sin2<\/sup> (90\u00b0 – 28\u00b0) – tan2<\/sup> 45\u00b0
\n= sin2<\/sup> 28\u00b0 + cos2<\/sup> 28\u00b0 – tan2<\/sup> 45\u00b0
\n= 1 – (1)2<\/sup> = 1 – 1 = 0 (\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1 and tan 45\u00b0 = 1)
\n
\n= 2 \u00d7 1 + 1 + 1 = 2 + 1 + 1= 4
\n(iii) cos 18\u00b0 sin 12\u00b0 + sin 18\u00b0 cos 12\u00b0
\n= cos (90\u00b0 – 12\u00b0) sin 72\u00b0 + sin (90\u00b0 – 12\u00b0) cos 12\u00b0
\n= sin 72\u00b0.sin 12\u00b0 + cos 12\u00b0 cos 12\u00b0
\n= sin2<\/sup> 12\u00b0 + cos2<\/sup> 12\u00b0 = 1 (\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1)
\n(iv) 5 sin 50\u00b0 sec 40\u00b0 – 3 cos 59\u00b0 cosec 31\u00b0
\n
\n<\/p>\n
\nProve that:
\n
\nAnswer:
\n<\/p>\n
\nWhen 0\u00b0 < A < 90\u00b0, solve the following equations :
\n(i) sin 3A = cos 2A
\n(ii) tan 5A = cotA.
\nAnswer:
\n(i) sin 3A = cos 2A
\n\u21d2 sin 3A = sin (90\u00b0 – 2A)
\n\u2234 3A = 90\u00b0 – 2A
\n3 A + 2A = 90\u00b0 \u21d2 5A = 90\u00b0
\n\u2234 A = \\(\\frac{90^{\\circ}}{5}\\) = 18\u00b0
\n(ii) tan 5A = cot A
\ntan 5A = tan(90\u00b0 – A)
\n\u2234 5A = 90\u00b0- A \u21d2 5A + A = 90\u00b0
\n\u21d2 6A = 90\u00b0
\n\u2234 A = \\(\\frac{90^{\\circ}}{5}\\) = 15\u00b0<\/p>\n
\nFind the value of \u03b8 if
\n(i) sin (\u03b8 + 36\u00b0) = cos \u03b8, where \u03b8 and \u03b8 + 36\u00b0 are acute angles.
\n(ii) sec 4\u03b8 = cosec (\u03b8 – 20\u00b0), where 4\u03b8 and \u03b8 – 20\u00b0 are acute angles.
\nAnswer:
\n(i) \u03b8 and (\u03b8 + 36\u00b0) are acute angles
\nsin (\u03b8 + 36\u00b0) = cos \u03b8 = sin (90\u00b0 – \u03b8) {sin (90\u00b0 – \u03b8) = cos \u03b8}
\nComparing, we get
\n\u03b8 + 36\u00b0 = 90\u00b0 – \u03b8
\n\u03b8 + \u03b8 = 90\u00b0 – 36\u00b0
\n2\u03b8 = 54\u00b0 \u21d2 \u03b8 = \\(\\frac{54^{\\circ}}{2}\\) = 27\u00b0
\n\u2234 \u03b8 = 27\u00b0<\/p>\n
\nsec 4\u03b8 = cosec (\u03b8 – 20\u00b0)
\ncosec (90\u00b0 – 4\u03b8) = cosec (\u03b8 – 20\u00b0) {\u2235 cosec (90\u00b0 – \u03b8) = sec \u03b8}
\nComparing, we get
\n90\u00b0 – 4\u03b8 = \u03b8 – 20\u00b0
\n\u21d2 90\u00b0 + 20\u00b0 = \u03b8 + 4\u03b8
\n\u21d2 5\u03b8 = 110\u00b0
\n\u03b8 = \\(\\frac{110^{\\circ}}{5}\\) = 22\u00b0
\n\u03b8 = 22\u00b0<\/p>\n
\nIn the adjoining figure, ABC is right-angled triangle at B and ABD is right angled triangle at A. If BD \u22a5 AC and BC = 2\\( \\sqrt{{3}} \\) cm, find the length of AD.
\n
\nSolution:
\n\u2206ABC and \u2206ABD are right angled triangles
\nin which \\( \\sqrt{{B}} \\) = 90\u00b0 and \\( \\sqrt{{A}} \\) = 90\u00b0
\nBC = 2\\( \\sqrt{{3}} \\) cm AC and BD intersect each other at E at right angle and \u2220CAb = 30\u00b0.
\nNow in right \u2206ABC,
\ntan \u03b8 = \\(\\frac{B C}{A B}\\) \u21d2 tan 30\u00b0= \\(\\frac{2 \\sqrt{3}}{\\mathrm{AB}}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}=\\frac{2 \\sqrt{3}}{\\mathrm{AB}}\\)
\n\u21d2 AB = 2\\( \\sqrt{{3}} \\) \u00d7 \\( \\sqrt{{3}} \\) = 2 \u00d7 3 = 6 cm.
\nIn \u2206ABE, \u2220EAB = 30\u00b0
\nand \u2220EAB = 90\u00b0
\n\u2234 \u2220ABE or \u2220ABD = 180\u00b0 – 90\u00b0 – 30\u00b0
\n= 60\u00b0
\nNow in right \u2206ABD,
\ntan 60\u00b0 = \\(\\frac{\\mathrm{AD}}{\\mathrm{AB}}\\) \u21d2 \\( \\sqrt{{3}} \\) = \\(\\frac{\\mathrm{AD}}{6}\\)
\n\u21d2 AD = 6\\( \\sqrt{{3}} \\) cm.<\/p>\nML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"