{"id":25179,"date":"2020-03-05T15:46:35","date_gmt":"2020-03-05T10:16:35","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25179"},"modified":"2021-07-02T15:21:20","modified_gmt":"2021-07-02T09:51:20","slug":"kseeb-solutions-for-class-10-maths-chapter-12-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-12-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 12 Some Applications of Trigonometry Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 12 Some Applications of Trigonometry Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 12 Some Applications of Trigonometry Additional Questions<\/h2>\n

I. Multiple Choice Questions:<\/span><\/p>\n

Question 1.
\nThe angle of elevation of the top of a tower from a point at a distance of 100 m from the base of the tower is found to be 45\u00b0 then height of the tower is
\na. 50 m
\nb. 100m
\nc. 50\\(\\sqrt{2}\\) m
\nd. 50\\(\\sqrt{2}\\) m
\nAnswer:
\nb. 100m<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the angle of elevation of the top of a tower, whose height is 100 m, at a point whose distance from the base of the tower is 100 m.
\na. 30\u00b0
\nb. 60\u00b0
\nc. 45\u00b0
\nd. 90\u00b0
\nAnswer:
\nc. 45\u00b0<\/p>\n

Question 3.
\nThe angle of elevation of the top of tree from a point at a distance of 200 m from its base is 60\u00b0 the height of the tree is
\na. 50\\(\\sqrt{3}\\)m
\nb. 100\\(\\sqrt{3}\\) m
\nc. 200\\(\\sqrt{3}\\)m
\nd. \\(\\frac{200}{\\sqrt{3}}\\) m
\nAnswer:
\nc. 200\\(\\sqrt{3}\\)m<\/p>\n

Question 4.
\nFind the length of the shadow of 10 m high tree of the angle of elevation of the sun is 30\u00b0
\na. 10 m
\nb. \\(\\frac{10}{\\sqrt{3}}\\) m
\nc. 10\\(\\sqrt{3}\\)m
\nd. 20 m
\nAnswer:
\nd. 20 m<\/p>\n

Question 5.
\nIf the shadow of 10 m high tree is 10\\(\\sqrt{3}\\)m then find the angle of elevation of the sun
\na. 60\u00b0
\nb. 90\u00b0
\nc. 45\u00b0
\nd. 30\u00b0
\nAnswer:
\nc. 45\u00b0<\/p>\n

Question 6.
\nThe ratio of the length of a tree and its shadow is 1 : \\(\\frac{1}{\\sqrt{3}}\\). The angle of the sun’s elevation is
\na. 30\u00b0
\nb. 45\u00b0
\nc. 60\u00b0
\nd. 90\u00b0
\nAnswer:
\nc. 60\u00b0<\/p>\n

Question 7.
\nThe ratio of the length of a rod to its shadow is 1: \\(\\sqrt{3}\\). The angle of elevation of the sun is
\na. 30\u00b0
\nb. 60\u00b0
\nc. 45\u00b0
\nd. 90\u00b0
\nAnswer:
\na. 30\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIf the angle of elevation of the sun is 45\u00b0, then find the length of the shadow of a tower whose height is ‘h’m
\na. \\(\\frac{h}{2}\\) m
\nb. h m
\nc. 2h m
\nd. h\\(\\sqrt{3}\\) m
\nAnswer:
\nb. h m<\/p>\n

Question 9.
\nThe angle of elevation of the sun is 45\u00b0. Then, the length of the shadow of a 12 m high tree is
\na. 6\\(\\sqrt{3}\\) m
\nb. 12\\(\\sqrt{3}\\) m
\nc. \\(\\frac{12}{\\sqrt{3}}\\) m
\nd. 12 m
\nAnswer:
\nd. 12 m<\/p>\n

Question 10.
\nFrom a bridge, 25 m high, the angle of depression of a boat is 45\u00b0. Find the horizontal distance of the boat from the bridge.
\na. 25 m
\nb. \\(\\frac{25}{2}\\)
\nc. 50 m
\nd. 25\\(\\sqrt{3}\\) m
\nAnswer:
\na. 25 m<\/p>\n

Question 11.
\nIf the angle of elevation of the sun is top 60\u00b0, then find the ratio of the height of a tree with its shadow.
\na. \\(\\sqrt{3}\\) : 1
\nb. 1 : \\(\\sqrt{3}\\)
\nc. 3 : 1
\nd. 1 : 3
\nAnswer:
\na. \\(\\sqrt{3}\\) : 1<\/p>\n

II. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nThe angle of elevation of an aeroplane from a point on the ground is 45\u00b0. After a flight of 15 seconds horizontally, the angle of elevation changes to 30\u00b0 if the aeroplane is flying at a constant height of 2500 m then find the speed of the plane.
\nAnswer:
\n\"KSEEB
\n2500 + BD = 2500\\(\\sqrt{3}\\)
\nBD = 2500\\(\\sqrt{3}\\) – 2500
\nBD = 2500(\\(\\sqrt{3}\\) – 1)m
\nBut BD = CE = 2500 (\\(\\sqrt{3}\\) – 1) m
\n\"KSEEB
\nSpeed = 118.33.m \/ sec<\/p>\n

Question 2.
\nFrom a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of a tower is 30\u00b0 the angle of elevation of the top of water tank on top of the tower is 45\u00b0. Find
\n(i) height of the tower
\n(ii) depth of the tank
\nAnswer:
\n\"KSEEB
\nAD = 40 m
\n\u2234 CD = AD – AC = 40 – \\(\\frac{40}{\\sqrt{3}}\\)
\nCD = 40\\(\\left(1-\\frac{1}{\\sqrt{3}}\\right)\\) m<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nA tall building casts a shadow of 300 m long when the sun’s altitude elevation is 30\u00b0. Find the height of the tower.
\nAnswer:
\n\"KSEEB
\n= 100\\(\\sqrt{3}\\) m
\nHeight of the tower is 100\\(\\sqrt{3}\\) m<\/p>\n

Question 4.
\nA tree is broken over by the wind forms a right angled triangle with the ground. If the broken parts makes an angle of 60\u00b0 with the ground and the top of the tree is now 20 m from its base, how tall was the tree.
\nAnswer:
\n\"KSEEB
\nAB = 20\\(\\sqrt{3}\\) m.
\nHeight of tree = AD = AB + BC
\n= 40 + 20\\(\\sqrt{3}\\)
\n= 20(2 + \\(\\sqrt{3}\\))<\/p>\n

Question 5.
\nThe angle of elevation of the top of a flag post from a point on a horizontal ground is found to be 30\u00b0. On walking 6 m towards the post, the elevation increases by 15\u00b0.
\nFind the height of the flag post.
\nAnswer:
\n\"KSEEB
\nIn \u2206 ABC \u03b8 = 30\u00b0.
\n\"KSEEB
\nY + 6 = x\\(\\sqrt{3}\\) \u2192 (1)
\nIn \u2206 ABD, \u03b8 = 45\u00b0
\n\"KSEEB
\nx = y \u2192 (2)
\nPut equation (2) in equation (1)
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nFrom the top of a building 16 m high, the angular elevation of the top of a hill is 60\u00b0 and the angular depression of the foot of the hill is 30\u00b0. Find the height of the hill.
\nAnswer:
\n\"KSEEB
\nIn \u2206 AED, \u03b8 = 30\u00b0
\n\"KSEEB
\nIn \u2206 BCE, \u03b8 = 60\u00b0
\n\"KSEEB
\nFrom (1) and (2)
\n\"KSEEB
\n16 \u00d7 3 = x – 16
\nx = 48 + 16 = 64
\n\u2234 The height of the hill is 64 m.<\/p>\n

Question 7.
\nTwo windmills of height 50 m and 40 m are on either side of the field. A person observes the top of the windmills from a point in between the towers. The angle of elevation was found to be 45\u00b0 in both the cases. Find the distance between the windmills.
\nAnswer:
\n\"KSEEB
\nLet AB be a tower of height = 50 m
\nCD be another tower of height = 40 m
\nThe observer is at P and angle of elevation.
\n\"KSEEB
\nPD = 40 m
\n\u2234 BD = BP + PD = 50 + 40 = 90 m
\n\u2234 The distance between the wind mills on either side of field is 90 m.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 12 Some Applications of Trigonometry Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 12 Some Applications of Trigonometry Additional Questions I. Multiple Choice …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25179"}],"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=25179"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25179\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25179"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25179"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}