{"id":128024,"date":"2023-09-08T16:34:00","date_gmt":"2023-09-08T11:04:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=128024"},"modified":"2023-11-08T12:33:20","modified_gmt":"2023-11-08T07:03:20","slug":"rd-sharma-class-10-solutions-chapter-12-exercise-12-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 12- Some Applications of Trigonometry Exercise 12.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-128033\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-12-Exercise-12.1.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-12-Exercise-12.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-12-Exercise-12.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1:\u00a0<\/strong>Students can download the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-some-applications-of-trigonometry\/\"><strong>RD Sharma Class 10 Solutions Chapter 12<\/strong><\/a> Exercise 12.1\u00a0PDF to learn how to solve the questions in this exercise correctly. Students wishing to brush up on their concepts can check the\u00a0<a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a>.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d02b142c1a3\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d02b142c1a3\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#download-rd-sharma-class-10-solutions-chapter-12-exercise-121-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#access-answers-to-maths-rd-sharma-solutions-for-class-10-chapter-12-%e2%80%93-some-applications-of-trigonometry\" title=\"Access Answers to Maths RD Sharma Solutions for Class 10 Chapter 12 \u2013 Some Applications Of Trigonometry\">Access Answers to Maths RD Sharma Solutions for Class 10 Chapter 12 \u2013 Some Applications Of Trigonometry<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#exercise-121-page-no-1229\" title=\"Exercise 12.1 Page No: 12.29\">Exercise 12.1 Page No: 12.29<\/a><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#faqs-on-rd-sharma-class-10-solutions-chapter-12-exercise-121\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1\">FAQs on RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#where-can-i-get-rd-sharma-class-10-maths-solutions-chapter-12-exercise-121-free-pdf\" title=\"Where can I get RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 Free PDF?\">Where can I get RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 Free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#does-rd-sharma-class-10-maths-solutions-chapter-12-exercise-121-help-you-to-clear-board-exams\" title=\"Does RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 help you to clear board exams?\">Does RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 help you to clear board exams?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/#what-is-the-importance-of-studying-rd-sharma-solutions-for-class-10-maths-chapter-12-exercise-121\" title=\"What is the importance of studying RD Sharma Solutions for Class 10 Maths Chapter 12 Exercise 12.1?\">What is the importance of studying RD Sharma Solutions for Class 10 Maths Chapter 12 Exercise 12.1?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-12-exercise-121-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-12-compressed-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-12-compressed-1.pdf\">RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-maths-rd-sharma-solutions-for-class-10-chapter-12-%e2%80%93-some-applications-of-trigonometry\"><\/span>Access Answers to Maths RD Sharma Solutions for Class 10 Chapter 12 \u2013 Some Applications Of Trigonometry<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h4><span class=\"ez-toc-section\" id=\"exercise-121-page-no-1229\"><\/span>Exercise 12.1 Page No: 12.29<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>1. A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60\u00b0. What is the height of the tower?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-1.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/strong><\/p>\n<p>Given:<\/p>\n<p>Distance between the foot of the tower and point of observation = 20 m = BC<\/p>\n<p>Angle of elevation of the top of the tower = 60\u00b0 = \u03b8<\/p>\n<p>And, Height of tower (H) = AB<\/p>\n<p>Now, from fig. ABC<\/p>\n<p>\u0394ABC is a right-angle triangle,<\/p>\n<p>So,<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-2.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p><strong>2. The angle of elevation of a ladder against a wall is 60\u00b0 and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-3.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given:<\/p>\n<p>Distance between the wall and foot of the ladder = 9.5 m<\/p>\n<p>Angle of elevation (\u03b8) = 60\u00b0<\/p>\n<p>Length of the ladder = L = AC<\/p>\n<p>Now, from fig. ABC<\/p>\n<p>\u0394ABC is a right angle triangle,<\/p>\n<p>So,<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-4.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p><em>Thus, length of the ladder (L) = 19 m<\/em><\/p>\n<p><strong>3. A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of 60\u00b0 with the level of the ground. Determine the height of the wall.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-5.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>Distance between the wall and the foot of the ladder = 2m = BC<\/p>\n<p>Angle made by ladder with ground (\u03b8) = 60\u00b0<\/p>\n<p>Height of the wall (H) = AB<\/p>\n<p>Now, the fig. of ABC forms a right angle triangle.<\/p>\n<p>So,<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-6.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p><strong>4. An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45\u00b0 with the horizontal through the foot of the pole, find the length of the wire.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-7.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>Height of the electric pole = 10 m = AB<\/p>\n<p>The angle made by steel wire with the ground (horizontal) \u03b8 = 45\u00b0<\/p>\n<p>Let length of wire = L = AC<\/p>\n<p>So, from the figure formed we have ABC as a right triangle.<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-8.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p><strong>5. A kite is flying at a height of 75 meters from the ground level, attached to a string inclined at 60\u00b0 to the horizontal. Find the length of the string to the nearest meter.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-9.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>Height of kite flying from the ground level = 75 m = AB<\/p>\n<p>Angle of inclination of the string with the ground (\u03b8) = 60\u00b0<\/p>\n<p>Let the length of the string be L = AC<\/p>\n<p>So, from the figure formed we have ABC as a right triangle.<\/p>\n<p>Hence,<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-10.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p><strong>6. A ladder 15 metres long reaches the top of a vertical wall. If the ladder makes an angle of 60<sup>o<\/sup>\u00a0with the wall, find the height of the wall.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-11.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/strong><\/p>\n<p>Given,<\/p>\n<p>The length of the ladder = 15m = AO<\/p>\n<p>Angle made by the ladder with the wall = 60<sup>o<\/sup><\/p>\n<p>Let the height of the wall be h metres.<\/p>\n<p>And the horizontal ground taken as OX.<\/p>\n<p>Then from the fig. we have,<\/p>\n<p>In right \u0394ABO, using trigonometric ratios<\/p>\n<p>cos (60<sup>o<\/sup>) = AB\/AO<\/p>\n<p>1\/2 = h\/ 15<\/p>\n<p>h = 15\/2<\/p>\n<p>h = 7.5m<\/p>\n<p><em>Hence, the height of the wall is 7.5m<\/em><\/p>\n<p><strong>7. A vertical tower stands on a horizontal place and is surmounted by a vertical flag staff. At a point on the plane 70 meters away from the tower, an observer notices that the angles of elevation of the top and bottom of the flag-staff are respectively 60\u00b0 and 45\u00b0. Find the height of the flag staff and that of the tower.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-12.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>A vertical tower is surmounted by flag staff.<\/p>\n<p>Distance between observer and the tower = 70 m = DC<\/p>\n<p>Angle of elevation of bottom of the flag staff = 45\u00b0<\/p>\n<p>Angle of elevation of top of the flag staff = 60\u00b0<\/p>\n<p>Let the height of the flag staff = h = AD<\/p>\n<p>Height of tower = H = BC<\/p>\n<p>If we represent the above data in the figure then it forms right angle triangles \u0394ACD and \u0394BCD<\/p>\n<p>When \u03b8 is angle in right angle triangle we know that<\/p>\n<p>tan \u03b8 = opp. Side\/ Adj. side<\/p>\n<p>Now,<\/p>\n<p>tan 45<sup>o<\/sup>\u00a0= BC\/ DC<\/p>\n<p>1 = H\/ 70<\/p>\n<p>\u2234 H =70 m<\/p>\n<p>Again,<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-13.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>x = 70 (1.732-1)<\/p>\n<p><em>\u2234 x = 51.24 m<\/em><\/p>\n<p><em>Therefore, the height of tower = 70 m and the height of flag staff = 51.24 m<\/em><\/p>\n<p><strong>8. A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60\u00b0 with the ground. At what height from the ground did the tree break?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-14.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>The initial height of tree H = 15 m = AB + AC<\/p>\n<p>Let us assume that it is broken at point A.<\/p>\n<p>And, the angle made by broken part with the ground (\u03b8) = 60\u00b0<\/p>\n<p>Height from ground to broken points = h = AB<\/p>\n<p>So, we have<\/p>\n<p>H = AC + h<\/p>\n<p>\u27f9 AC = (H \u2013 h) m<\/p>\n<p>We get a right triangle formed by the above given data,<\/p>\n<p>So,<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-15.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Rationalizing denominator, we have<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-16.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p><em>Therefore, the height of broken point from the ground is 15(2\u221a3 \u2013 3)m<\/em><\/p>\n<p><strong>9. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are respectively 30\u00b0 and 60\u00b0. Find the height of the tower.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-17.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>Height of the flag staff = 5 m =AB<\/p>\n<p>Angle of elevation of the top of flag staff = 60\u00b0<\/p>\n<p>Angle of elevation of the bottom of the flagstaff = 30\u00b0<\/p>\n<p>Let height of tower be \u2018h\u2019 m = BC<\/p>\n<p>And, let the distance of the point from the base of the tower = x m<\/p>\n<p>In right angle triangle BCD, we have<\/p>\n<p>tan 30<sup>o<\/sup>\u00a0= BC\/DC<\/p>\n<p>1\/\u221a3 = h\/x<\/p>\n<p>x = h\u221a3 \u2026.. (i)<\/p>\n<p>Now, in \u0394ACD,<\/p>\n<p>tan 60<sup>o<\/sup>\u00a0= AC\/DC<\/p>\n<p>\u221a3 = (5 + h)\/ x<\/p>\n<p>\u221a3x = 5 + h<\/p>\n<p>\u221a3(h\u221a3) = 5 + h [using (i)]<\/p>\n<p>3h = 5 + h<\/p>\n<p>2h = 5<\/p>\n<p>h = 5\/2 = 2.5m<\/p>\n<p><em>Therefore, the height of the tower = 2.5 m<\/em><\/p>\n<p><strong>10<\/strong>.\u00a0<strong>A person observed the angle of elevation of a tower as 30\u00b0. He walked 50 m toward the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60\u00b0. Find the height of the tower.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-18.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Given,<\/p>\n<p>The angle of elevation of the tower before he started walking = 30<sup>o<\/sup><\/p>\n<p>Distance walked by the person towards the tower = 50m<\/p>\n<p>The angle of elevation of the tower after he walked = 60<sup>o<\/sup><\/p>\n<p>Let height of the tower (AB) = h m<\/p>\n<p>Let the distance BC = x m<\/p>\n<p>From the fig. in \u0394ABC,<\/p>\n<p>tan 60<sup>o<\/sup>\u00a0= AB\/ BC<\/p>\n<p>\u221a3 = h\/x<\/p>\n<p>x = h\/\u221a3 \u2026.(i)<\/p>\n<p>Now, in \u0394ABD<\/p>\n<p>tan 30<sup>o<\/sup>\u00a0= AB\/ BD<\/p>\n<p>1\/\u221a3 = h\/ (50 + x)<\/p>\n<p>\u221a3h = 50 + x<\/p>\n<p>\u221a3h = 50 + (h\/\u221a3) [using (i)]<\/p>\n<p>3h = 50\u221a3 + h<\/p>\n<p>2h = 50\u221a3<\/p>\n<p>h = 25\u221a3 = 25(1.73) =\u00a0<em>43.25m<\/em><\/p>\n<p><em>Therefore, the height of the tower = 43.25m<\/em><\/p>\n<p><strong>11. The shadow of a tower, when the angle of elevation of the sun is 45\u00b0, is found to be 10 m longer than when it was 60\u00b0. Find the height of the tower.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-19.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>Let the height of the tower(AB) = h m<\/p>\n<p>Let the length of the shorter shadow be x m<\/p>\n<p>Then, the longer shadow is (10 + x)m<\/p>\n<p>So, from fig. In \u0394ABC<\/p>\n<p>tan 60<sup>o<\/sup>\u00a0= AB\/BC<\/p>\n<p>\u221a3 = h\/x<\/p>\n<p>x = h\/\u221a3\u2026. (i)<\/p>\n<p>Next, in \u0394ABD<\/p>\n<p>tan 45<sup>o<\/sup>\u00a0= AB\/BD<\/p>\n<p>1 = h\/(10 + x)<\/p>\n<p>10 + x = h<\/p>\n<p>10 + (h\/\u221a3) = h [using (i)]<\/p>\n<p>10\u221a3 + h = \u221a3h<\/p>\n<p>h(\u221a3 -1) =10\u221a3<\/p>\n<p>h = 10\u221a3\/ (\u221a3 -1)<\/p>\n<p>After rationalising the denominator, we have<\/p>\n<p>h = [10\u221a3 x (\u221a3 + 1)]\/ (3 \u2013 1)<\/p>\n<p>h = 5\u221a3(\u221a3 + 1)<\/p>\n<p>h = 5(3 + \u221a3) = 23.66 [\u221a3 = 1.732]<\/p>\n<p><em>Therefore, the height of the tower is 23.66 m.<\/em><\/p>\n<p><strong>12. A parachute is descending vertically and makes angles of elevation of 45\u00b0 and 60\u00b0 at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-20.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/strong><\/p>\n<p>Let the parachute at highest point A and let C and D be points which are 100 m apart on ground where from then CD = 100 m<\/p>\n<p>Angle of elevation from point D = 45\u00b0 = \u03b1<\/p>\n<p>Angle of elevation from point C = 60\u00b0 = \u03b2<\/p>\n<p>Let B be the point just vertically down the parachute<\/p>\n<p>Let us draw figure according to above data then it forms the figure as shown in which ABC and ABD are two triangles<\/p>\n<p>Maximum height of the parachute from the ground<\/p>\n<p>AB = H m<\/p>\n<p>Distance of point where parachute falls to just nearest observation point = x m<\/p>\n<p>If in right angle triangle one of the included angles is \u03b8 then<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-21.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>From (a) and (b)<\/p>\n<p><img src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry-ex-12-1-22.png\" alt=\"RD Sharma Solutions For Class 10 Maths Chapter 12 Solutions\" \/><\/p>\n<p>x = 136.6 m in (b)<\/p>\n<p>H = \u221a3 \u00d7 136.6 = 236.6m<\/p>\n<p>Therefore,<\/p>\n<p>The maximum height of the parachute from the ground, H = 236.6m<\/p>\n<p><em>The distance between the two points where the parachute falls on the ground and just the observation is x = 136.6 m<\/em><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-12-exercise-121\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631703230838\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-maths-solutions-chapter-12-exercise-121-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631703310288\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"does-rd-sharma-class-10-maths-solutions-chapter-12-exercise-121-help-you-to-clear-board-exams\"><\/span>Does RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 help you to clear board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 of Class 10 Maths is an essential chapter. These answers are centered on understanding several Math shortcuts and strategies for quick and easy calculations.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631703641449\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-is-the-importance-of-studying-rd-sharma-solutions-for-class-10-maths-chapter-12-exercise-121\"><\/span>What is the importance of studying RD Sharma Solutions for Class 10 Maths Chapter 12 Exercise 12.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1 solutions assist students in obtaining a summary of the question paper format, which includes a range of questions, such as largely repeated questions, concise and long response type questions, multiple-choice questions, and marks, among other things. The more problems you solve, the more confident you will be in your ability to succeed.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1:\u00a0Students can download the RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1\u00a0PDF to learn how to solve the questions in this exercise correctly. Students wishing to brush up on their concepts can check the\u00a0RD Sharma Class 10 Solutions. Download RD Sharma Class 10 Solutions Chapter 12 &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 12- Some Applications of Trigonometry Exercise 12.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-12-exercise-12-1\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 12- Some Applications of Trigonometry Exercise 12.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":128033,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128024"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=128024"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128024\/revisions"}],"predecessor-version":[{"id":504313,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128024\/revisions\/504313"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/128033"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=128024"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=128024"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=128024"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}