{"id":30019,"date":"2022-03-08T16:32:09","date_gmt":"2022-03-08T11:02:09","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=30019"},"modified":"2022-03-08T17:27:21","modified_gmt":"2022-03-08T11:57:21","slug":"ap-board-10th-class-maths-solutions-chapter-11-exercise-11-1-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-10th-class-maths-solutions-chapter-11-exercise-11-1-in-telugu\/","title":{"rendered":"AP Board 10th Class Maths Solutions Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.1"},"content":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions<\/a> Chapter 11 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f Exercise 11.1 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 11.1<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c12\u0c15 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 ABC\u0c32\u0c4b \u0c2d\u0c41\u0c1c\u0c3e\u0c32\u0c41 AB, BC \u0c2e\u0c30\u0c3f\u0c2f\u0c41 CA \u0c32 \u0c2a\u0c4a\u0c21\u0c35\u0c41\u0c32\u0c41 \u0c35\u0c30\u0c41\u0c38\u0c17\u0c3e 8 \u0c38\u0c46\u0c02.\u0c2e\u0c40., 15 \u0c38\u0c46\u0c02.\u0c2e\u0c40 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 17 \u0c38\u0c46\u0c02.\u0c2e\u0c40 \u0c05\u0c2f\u0c3f\u0c28 sinA, cos A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan A \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u2206ABC \u0c12\u0c15 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41.
\n\u0c2d\u0c41\u0c1c\u0c3e\u0c32 \u0c15\u0c4a\u0c32\u0c24\u0c32\u0c41 AB = 8 \u0c38\u0c46\u0c02.\u0c2e\u0c40. BC = 15 \u0c38\u0c46\u0c02.\u0c2e\u0c40. \u0c2e\u0c30\u0c3f\u0c2f\u0c41 CA = 17 \u0c38\u0c46\u0c02.\u0c2e\u0c40.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c15\u0c4a\u0c32\u0c24\u0c32\u0c32\u0c4b \\(\\overline{\\mathrm{CA}}\\) \u0c2a\u0c4a\u0c21\u0c35\u0c48\u0c28 \u0c2d\u0c41\u0c1c\u0c2e\u0c41 \u0c15\u0c3e\u0c35\u0c41\u0c28 \u2206ABC \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c15\u0c30\u0c4d\u0c23\u0c2e\u0c41 CA \u0c05\u0c17\u0c41\u0c28\u0c41.<\/p>\n

\"AP<\/p>\n

\u2220A \u0c2a\u0c30\u0c02\u0c17\u0c3e \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c02 = BC = 15 \u0c38\u0c46\u0c02.\u0c2e\u0c40.
\n\u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c2e\u0c41 = AB = 8 \u0c38\u0c46\u0c02.\u0c2e\u0c40. \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c30\u0c4d\u0c23\u0c2e\u0c41 = AC = 17 \u0c38\u0c46\u0c02.\u0c2e\u0c40.
\nA \u0c15\u0c41 \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c2e\u0c41 = BC = 15<\/p>\n

sin A = \"AP =\\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{15}{17}\\)<\/p>\n

cos A = \"AP =\\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{8}{17}\\)<\/p>\n

tan A = \"AP =\\(\\frac{\\mathrm{BC}}{\\mathrm{AB}}=\\frac{15}{8}\\)
\n\u2234 sin A = \\(\\frac{15}{17}\\); cos A = \\(\\frac{8}{17}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan A = \\(\\frac{15}{8}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 POR \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c2d\u0c41\u0c1c\u0c3e\u0c32\u0c41 PQ = 7 \u0c38\u0c46\u0c02.\u0c2e\u0c40., QR = 25 \u0c38\u0c46\u0c02.\u0c2e\u0c40. \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220P = 90\u00b0 \u0c05\u0c2f\u0c3f\u0c28 tanQ – tan R \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 POR\u0c12\u0c15 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 PQ = 7 \u0c38\u0c46\u0c02.\u0c2e\u0c40, PR = 25 \u0c38\u0c46\u0c02.\u0c2e\u0c40, \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220Q = 90\u00b0.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c38\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c2e\u0c41 . PQ2<\/sup> + PR2<\/sup> = QR2<\/sup>
\n(7)2<\/sup> + PR2<\/sup> = (25)2<\/sup>
\nPR2<\/sup> = 625 – 49
\nPR2<\/sup> = 576
\n\u21d2 PR = \u221a576 = 24
\n\u2234 tan Q = \"AP
\n= \\(\\frac{P R}{P Q}=\\frac{24}{7}\\)<\/p>\n

tan R = \"AP
\n= \\(\\frac{P Q}{P R}=\\frac{7}{24}\\)<\/p>\n

tan Q – tan R = \\(\\frac{24}{7}-\\frac{7}{24}=\\frac{(24)^{2}-(7)^{2}}{24 \\times 7}\\)
\n= \\(\\frac{576-49}{168}=\\frac{527}{168}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\nB \u0c35\u0c26\u0c4d\u0c26 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23\u0c02 \u0c15\u0c32\u0c4d\u0c17\u0c3f\u0c28 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 ABC\u0c32\u0c4b ‘ a = 24 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41, b = 25 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220BAC = \u03b8 \u0c05\u0c2f\u0c3f\u0c28 cos 6 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan 6 \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 ABC \u0c32\u0c4b B \u0c35\u0c26\u0c4d\u0c26 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23\u0c02 \u0c15\u0c32\u0c26\u0c41. \u0c2e\u0c30\u0c3f\u0c2f\u0c41 a = BC = 24 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41, b = CA = 25 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220BAC = \u03b8 \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c3f\u0c28<\/p>\n

\"AP<\/p>\n

Bb 24 Ac \u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c38\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n(25)2<\/sup> = AB2<\/sup> + 242<\/sup>
\nAB2<\/sup> = 252<\/sup> – 242<\/sup>
\n= 625 – 576
\nAB2<\/sup> = 49
\nAB = \u221a49 = 7
\n\u2220BAC = \u03b8 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c2e\u0c41
\n\u03b8\u0c15\u0c41 \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c2e\u0c41 = BC = 24 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41
\n\u03b8\u0c15\u0c41 \u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c2e\u0c41 = AB = 7 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41
\n\u0c15\u0c30\u0c4d\u0c23\u0c2e\u0c41 = AC = 25 \u0c2f\u0c42\u0c28\u0c3f\u0c1f\u0c4d\u0c32\u0c41<\/p>\n

\"AP<\/p>\n

= \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{7^{\\mathrm{\\kappa}}}{25}\\)<\/p>\n

\"AP<\/p>\n

= \\(\\frac{\\mathrm{BC}}{\\mathrm{AB}}=\\frac{24^{\\mathrm{\\kappa}}}{7}\\)
\n\u2234 cos \u03b8 = \\(\\frac{7}{25}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan \u03b8 = \\(\\frac{24}{7}\\).<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\ncos A = \\(\\frac{12}{13}\\) \u0c05\u0c2f\u0c3f\u0c28 sin A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan A \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 Cos A = \\(\\frac{12}{13}\\)
\nCos A = \"AP = \\(\\frac{12}{13}\\)
\n\u2234\u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c2e\u0c41 : \u0c15\u0c30\u0c4d\u0c23\u0c2e\u0c41 = 12 : 13.<\/p>\n

\"AP<\/p>\n

\u0c15\u0c4b\u0c23\u0c02 ‘A’ \u0c2a\u0c30\u0c02\u0c17\u0c3e, \u0c06\u0c38\u0c28\u0c4d\u0c28\u0c2d\u0c41\u0c1c\u0c2e\u0c41 = AB = 12k \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c30\u0c4d\u0c23\u0c2e\u0c41 = AC = 13 k(‘k’ \u0c12\u0c15 \u0c27\u0c28 \u0c38\u0c02\u0c16\u0c4d\u0c2f)
\n\u2206ABC \u0c28\u0c41\u0c02\u0c21\u0c3f
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\u21d2 (13k)2<\/sup> = (12.k)2<\/sup> + BC2<\/sup>
\n\u21d2 169 k2<\/sup> = 144 k2<\/sup> + BC2<\/sup>
\n\u21d2 BC2<\/sup> = 169 k2<\/sup> – 144 k2<\/sup>
\n\u21d2 (169 – 144) k2<\/sup> = 25 k2<\/sup>
\n\u21d2 BC = \u221a(25k2<\/sup>)
\n\u21d2 5k = \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c2e\u0c41
\n\u0c07\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, \u0c2e\u0c3f\u0c17\u0c3f\u0c32\u0c3f\u0c28 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32\u0c41
\nsin A = \"AP
\n= \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{5 \\mathrm{k}}{13 \\mathrm{k}}=\\frac{5}{13}\\)<\/p>\n

tan A = \"AP
\n= \\(\\frac{B C}{A B}=\\frac{5 k}{12 k}=\\frac{5}{12}\\)
\n\u0c06 \u0c35\u0c3f\u0c27\u0c2e\u0c41\u0c17\u0c3e sinA = \\(\\frac{5}{13}\\), tanA = \\(\\frac{5}{12}\\) \u0c05\u0c17\u0c41\u0c28\u0c41.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n3 tan A = 4 \u0c05\u0c2f\u0c3f\u0c28 sin A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 Cos A \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 3 tan A = 4 \u21d2 tan A = \\(\\frac{4}{3}\\)
\n\u0c15\u0c3e\u0c28\u0c3f tan A = \"AP = \\(\\frac{4}{3}\\)
\n\u2220A \u0c15\u0c41 \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c2e\u0c41 = 4k \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220A \u0c15\u0c41 \u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c02 = 3k.
\n\u0c08 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c28\u0c41 \u2206ABC \u0c32\u0c4b \u0c1a\u0c42\u0c2a\u0c17\u0c3e<\/p>\n

\"AP<\/p>\n

\u2206ABC \u0c12\u0c15 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41 \u0c15\u0c3e\u0c35\u0c41\u0c28 AC2<\/sup> = AB2<\/sup> + BC2<\/sup> (\u2235 \u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c38\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02)
\n= (3k)2<\/sup> + (4k)2<\/sup> = 9k2<\/sup> + 16k2<\/sup>
\nAC2<\/sup> = 25k2<\/sup>
\nAC = \u221a(25 k2<\/sup>) = 5k
\nsin A = \\(\\frac{B C}{A C}=\\frac{4 k}{5 k}=\\frac{4}{5}\\)<\/p>\n

cos A = \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{3 \\mathrm{k}}{5 \\mathrm{k}}=\\frac{3}{5}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\u2206ABC, \u2206XYZ\u0c32\u0c32\u0c4b cos A = cos X \u0c05\u0c2f\u0c4d\u0c2f\u0c47\u0c1f\u0c1f\u0c4d\u0c32\u0c41 4A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220X\u0c32\u0c41 \u0c32\u0c18\u0c41 \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c2f\u0c3f\u0c28 \u2220A = \u2220X \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.<\/p>\n

\"AP<\/p>\n

\u2206ABC \u0c32\u0c4b
\ncos A = \"AP
\n= \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}\\)<\/p>\n

\"AP<\/p>\n

\u0c05\u0c32\u0c3e\u0c17\u0c47 \u2206XYZ \u0c32\u0c4b 2
\ncos X = \\(\\frac{\\mathrm{XY}}{\\mathrm{XZ}}\\)
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02, cosA = cos X
\n\u2234 \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{\\mathrm{XY}}{\\mathrm{XZ}}\\) = k \u0c05\u0c28\u0c41
\n\u21d2 \\(\\frac{\\mathrm{AB}}{\\mathrm{XY}}=\\frac{\\mathrm{AC}}{\\mathrm{XZ}}\\) ………………(1)
\n\u2234 AB = kAC \u0c2e\u0c30\u0c3f\u0c2f\u0c41 XY = kXZ
\n\u0c07\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41
\n\"AP
\n(1) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 (2) \u0c32 \u0c28\u0c41\u0c02\u0c21\u0c3f,
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{XY}}=\\frac{\\mathrm{AC}}{\\mathrm{XZ}}=\\frac{\\mathrm{BC}}{\\mathrm{YZ}}\\)
\n\u2234 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c \u0c2d\u0c41\u0c1c\u0c3e\u0c32\u0c15\u0c41 \u0c05\u0c28\u0c41\u0c2a\u0c3e\u0c24\u0c02\u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41.
\n\u2234 \u2206ABC ~ \u2206XYZ
\n\u21d2 \u2220A = \u2220X (\u2235 \u0c38\u0c30\u0c42\u0c2a \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c3e\u0c32\u0c32\u0c4b \u0c05\u0c28\u0c41\u0c30\u0c42\u0c2a \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c3e\u0c32\u0c41)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\ncot \u03b8 = \\(\\frac{7}{8}\\) \u0c05\u0c2f\u0c3f\u0c28
\n(i) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\n(ii) \\(\\frac{(1+\\sin \\theta)}{\\cos \\theta}\\) \u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 cot \u03b8 = \\(\\frac{7}{8}\\)
\n\"AP
\nAB = 7k \u0c2e\u0c30\u0c3f\u0c2f\u0c41 BC = 8k \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c3f\u0c28 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41\u0c32\u0c4b
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (\u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c38\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02)
\n= (7k)2<\/sup> + (8k)2<\/sup>
\nAC2<\/sup> = 49k2<\/sup> + 64k2<\/sup>
\nAC2<\/sup> = 113 k2<\/sup>
\n\u21d2 AC = \u221a113k
\n\u0c07\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, sin \u03b8 = \"AP
\n= \\(\\frac{8 k}{\\sqrt{113} k}=\\frac{8}{\\sqrt{113}}\\)
\ncos \u03b8 = \"AP
\n= \\(\\frac{7 k}{\\sqrt{113} k}=\\frac{7}{\\sqrt{113}}\\)<\/p>\n

(i) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\n= \\(\\frac{1^{2}-\\sin ^{2} \\theta}{1^{2}-\\cos ^{2} \\theta}\\) [\u2235 (a + b) (a – b) = a2<\/sup> – b<sup.2)<\/p>\n

= \\(\\frac{1-\\left(\\frac{8}{\\sqrt{113}}\\right)^{2}}{1-\\left(\\frac{7}{\\sqrt{113}}\\right)^{2}}=\\frac{1-\\frac{64}{113}}{1-\\frac{49}{113}}=\\frac{\\frac{113-64}{113}}{\\frac{113-49}{113}}=\\frac{49}{64}\\)<\/p>\n

(ii) \\(\\frac{(1+\\sin \\theta)}{\\cos \\theta}\\)
\n= \\(\\frac{1+\\frac{8}{\\sqrt{113}}}{\\frac{7}{\\sqrt{113}}}\\)<\/p>\n

= \\(\\frac{\\frac{\\sqrt{113}+8}{\\sqrt{113}}}{\\frac{7}{\\sqrt{113}}}=\\frac{\\sqrt{113}+8}{7}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\nB \u0c35\u0c26\u0c4d\u0c26 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23\u0c02 \u0c15\u0c32\u0c4d\u0c17\u0c3f\u0c28, \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 ABC \u0c32\u0c4b tan A = \u221a3 \u0c05\u0c2f\u0c3f\u0c28
\n(i) sin A cos.C + cos A sin C
\n(ii) cos A cos C-sin Asin C \u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 : tan A = \\(\\frac{\\sqrt{3}}{1}\\\\)
\n\u0c15\u0c3e\u0c28\u0c3f tan A = \"AP = \\(\\frac{\\sqrt{3}}{1}\\\\)
\n\u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c02 = \u221a3k \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c02 = 1k<\/p>\n

\"AP<\/p>\n

\u2206ABC \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41\u0c32\u0c4b
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (\u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c38\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c2e\u0c41 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02)
\n\u21d2 AC2<\/sup> = (1k)2<\/sup> + (J3k)2<\/sup>
\n\u21d2 AC2<\/sup> = 1k2<\/sup> + 3k2<\/sup>
\n\u21d2 AC2<\/sup> = 4k2<\/sup>
\n\u2234 AC = \u221a4k2<\/sup> = 2k
\nBC
\nsin A = \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{\\sqrt{3} \\mathrm{k}}{2 \\mathrm{k}}=\\frac{\\sqrt{3}}{2}\\)
\ncos A = \\(=\\frac{A B}{A C}=\\frac{1 k}{2 k}=\\frac{1}{2}\\)
\nsin C = \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{1 \\mathrm{k}}{2 \\mathrm{k}}=\\frac{1}{2}\\)
\ncos C = \\(\\frac{B C}{A C}=\\frac{\\sqrt{3} k}{2 k}=\\frac{\\sqrt{3}}{2}\\)<\/p>\n

(i) sin A . cos C + cos A . sin C
\n= \\(\\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{2}+\\frac{1}{2} \\cdot \\frac{1}{2}=\\frac{(\\sqrt{3})^{2}}{4}+\\frac{(1)^{2}}{4}\\)
\n= \\(\\frac{3}{4}+\\frac{1}{4}=\\frac{3+1}{4}=\\frac{4}{4}\\) = 1<\/p>\n

(ii) cos A. cos C-sin A. sin C
\n= \\(\\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{2}-\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{2}\\)
\n= \\(\\frac{\\sqrt{3}}{4}-\\frac{\\sqrt{3}}{4}\\) = 0<\/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 11.1 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 11.1 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c12\u0c15 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 ABC\u0c32\u0c4b \u0c2d\u0c41\u0c1c\u0c3e\u0c32\u0c41 AB, BC \u0c2e\u0c30\u0c3f\u0c2f\u0c41 CA \u0c32 \u0c2a\u0c4a\u0c21\u0c35\u0c41\u0c32\u0c41 \u0c35\u0c30\u0c41\u0c38\u0c17\u0c3e 8 \u0c38\u0c46\u0c02.\u0c2e\u0c40., 15 \u0c38\u0c46\u0c02.\u0c2e\u0c40 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 17 \u0c38\u0c46\u0c02.\u0c2e\u0c40 \u0c05\u0c2f\u0c3f\u0c28 sinA, cos A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 …<\/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\/30019"}],"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=30019"}],"version-history":[{"count":3,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/30019\/revisions"}],"predecessor-version":[{"id":30046,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/30019\/revisions\/30046"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=30019"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=30019"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=30019"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}