{"id":30139,"date":"2022-03-09T15:47:11","date_gmt":"2022-03-09T10:17:11","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=30139"},"modified":"2022-03-10T13:27:30","modified_gmt":"2022-03-10T07:57:30","slug":"ap-board-10th-class-maths-solutions-chapter-11-exercise-11-3-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-10th-class-maths-solutions-chapter-11-exercise-11-3-in-telugu\/","title":{"rendered":"AP Board 10th Class Maths Solutions Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.3"},"content":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions<\/a> Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.3 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 10th Class Maths Solutions 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.3<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4a\u0c02\u0c21\u0c3f:
\n(i) \\(\\frac{\\tan 36^{\\circ}}{\\cot 54^{\\circ}}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\frac{\\tan 36^{\\circ}}{\\cot 54^{\\circ}}\\)
\n= \\(\\frac{\\tan 36^{\\circ}}{\\cot \\left(90^{\\circ}-36^{\\circ}\\right)}\\) [\u2235 cot(90 – \u03b8) = tan \u03b8]
\n= \\(\\frac{\\tan 36^{\\circ}}{\\tan 36^{\\circ}}\\) = 1<\/p>\n

(ii) cos 12\u00b0 – sin 78\u00b0
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos 12\u00b0 – sin 78\u00b0 = cos 12\u00b0 – sin (90\u00b0 – 12\u00b0) [\u2235 sin(90 – \u03b8) = cos \u03b8]
\n= cos 12\u00b0 – cos 12\u00b0 = 0.<\/p>\n

\"AP<\/p>\n

(iii) cosec 31\u00b0 – sec 59\u00b0
\n\u0c38\u0c3e\u0c27\u0c28.
\ncosec 31\u00b0 – sec 59\u00b0 = cosec 31\u00b0 – sec (90\u00b0 – 31\u00b0)
\n[\u2235 sec (90 – \u03b8) = cosec \u03b8]
\n= cosec 31\u00b0 – cosec 31\u00b0 = 0<\/p>\n

(iv) sin 15\u00b0 sec 75\u00b0
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 15\u00b0 sec 75\u00b0 = sin 15\u00b0 . sec (90\u00b0 – 15\u00b0)
\n= sin 15\u00b0. cosec 15\u00b0 [\u2235 sec (90 – \u03b8) = cosec \u03b8]
\n= sin 15\u00b0 . \\(\\frac{1}{\\sin 15^{\\circ}}\\) [\u2235 cosec 15\u00b0 = \\(\\frac{1}{\\sin 15^{\\circ}}\\)]<\/p>\n

(v) tan 26\u00b0 tan 640
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan 26\u00b0 tan 64\u00b0
\n= tan 26\u00b0 . tan (90\u00b0 – 26\u00b0)
\n= tan 26\u00b0. cot 26\u00b0 [\u2235 tan (90 – \u03b8) = cot \u03b8]
\n= tan 26\u00b0. \\(\\frac{1}{\\tan 26^{\\circ}}\\) [\u2235 cot \u03b8 = \\(\\left.\\frac{1}{\\tan \\theta}\\right]\\)]<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n(i) tan 48\u00b0 tan 16\u00b0 tan 42\u00b0 tan 74\u00b0 = 1
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S. = tan 48\u00b0 tan 16\u00b0 tan 42\u00b0 tan 74\u00b0
\n= tan 48o. tan 16\u00b0
\n= tan(90\u00b0 – 48\u00b0). tan(90\u00b0 – 16\u00b0)
\n= tan 48\u00b0. tan 16\u00b0. cot 48\u00b0. cot 16\u00b0 [\u2235 tan (90 – \u03b8) = cot \u03b8]
\n= 1 = R.H.S.
\n\u2234 L.H.S. = R.H.S.<\/p>\n

(ii) cos 36\u00b0 cos 54\u00b0 – sin 36\u00b0 sin 54\u00b0 = 0
\n\u0c38\u0c3e\u0c27\u0c28.
\nL.H.S.= cos36\u00b0 cos54o – sin36\u00b0 sin54\u00b0
\n= cos(90\u00b0 – 54\u00b0). cos(90\u00b0 – 36\u00b0) – sin 36\u00b0 . sin 54\u00b0 [\u2235 cos (90 – \u03b8) = sin \u03b8]
\n= sin 54\u00b0 . sin 36\u00b0 – sin 54\u00b0 . sin 36\u00b0
\n= 0 = R.H.S
\n\u2234 L.H.S. = R.H.S.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\ntan 2A = cot (A – 18\u00b0), 2A \u0c32\u0c18\u0c41\u0c15\u0c4b\u0c23\u0c02 \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4a\u0c02\u0c21\u0c3f
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41: tan 2A = cot (A – 18\u00b0)
\ncot (90\u00b0 – 2A) = cot (A – 18\u00b0) [\u2235 tan \u03b8 = cot (90 – \u03b8)]
\n\u21d2 90\u00b0 – 2A = A – 18\u00b0
\n\u21d2 108\u00b0 = 3A
\nA = \\(\\frac{108^{\\circ}}{3}\\) = 36\u00b0
\n\u2234 A \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 36\u00b0.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\nA, B \u0c32\u0c41 \u0c32\u0c18\u0c41\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan A = cot B . \u0c05\u0c2f\u0c3f\u0c28 A + B = 90\u00b0 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 : tan A = cot B
\n\u21d2 cot (90\u00b0 – A) = cot B [\u2235 tan \u03b8 = cot (90 – \u03b8]]
\n\u21d2 90\u00b0 – A = B
\n\u2234 A + B = 90\u00b0.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\nA, B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C \u0c32\u0c41 . \u2206ABC \u0c32\u0c4b\u0c28\u0c3f \u0c05\u0c02\u0c24\u0c30 \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c2f\u0c3f\u0c28 tan\\(\\left(\\frac{\\mathbf{A}+\\mathbf{B}}{2}\\right)\\) = cot \\(\\frac{C}{2}\\) = \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u2206ABC \u0c32\u0c4b A, B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C \u0c32\u0c41 \u0c05\u0c02\u0c24\u0c30 \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41
\n\u0c15\u0c3e\u0c35\u0c41\u0c28 A + B + C = 180\u00b0.
\n\u0c2a\u0c48 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02\u0c28\u0c41 ‘2’ \u0c1a\u0c47 \u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c17\u0c3e,
\n\\(\\frac{A+B}{2}+\\frac{C}{2}=\\frac{180^{\\circ}}{2}\\) = 90\u00b0
\n\\(\\frac{A+B}{2}\\) = 90\u00b0 – \\(\\frac{C}{2}\\)
\n\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32\u0c3e “tan” \u0c05\u0c28\u0c41 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f\u0c28\u0c3f \u0c24\u0c40\u0c38\u0c41\u0c15\u0c4a\u0c28\u0c17\u0c3e
\ntan\\(\\left(\\frac{\\mathbf{A}+\\mathbf{B}}{2}\\right)\\) = tan (90 – \\(\\frac{C}{2}\\))
\ntan\\(\\left(\\frac{\\mathbf{A}+\\mathbf{B}}{2}\\right)\\) = \\(\\frac{C}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\nsin 75\u00b0 + cos 65\u00b0 \u0c28\u0c41 0\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 45\u00b0 \u0c2e\u0c27\u0c4d\u0c2f\u0c17\u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 75\u00b0 + cos 65\u00b0 = sin (90\u00b0 – 15\u00b0) + cos (90\u00b0 – 25\u00b0)
\n= cos 15\u00b0 + sin 25\u00b0
\n[\u2235 sin (90 – \u03b8) = cos \u03b8 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cos (90 – \u03b8) = sin \u03b8]<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.3 Textbook Exercise Questions and Answers. AP State Syllabus 10th Class Maths Solutions 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.3 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4a\u0c02\u0c21\u0c3f: (i) \u0c38\u0c3e\u0c27\u0c28. = [\u2235 cot(90 – \u03b8) = tan \u03b8] = = 1 (ii) cos 12\u00b0 – sin 78\u00b0 \u0c38\u0c3e\u0c27\u0c28. cos 12\u00b0 …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/30139"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=30139"}],"version-history":[{"count":3,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/30139\/revisions"}],"predecessor-version":[{"id":30160,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/30139\/revisions\/30160"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=30139"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=30139"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=30139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}