{"id":20703,"date":"2020-01-22T11:59:43","date_gmt":"2020-01-22T06:29:43","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20703"},"modified":"2021-07-02T15:27:59","modified_gmt":"2021-07-02T09:57:59","slug":"2nd-puc-basic-maths-question-bank-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-13-ex-13-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 13 Heights And Distances Ex 13.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 13.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 13 Heights and Distances Ex 13.1<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Heights and Distances Ex 13.1 Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nThe angle of elevation of the top of a tower at a distance 500 metres from its foot is 30\u00b0. Find the height of the tower.
\nSol.
\nFrom the triangle ABC
\nB be the position of the observer & AC = height of the pole h.
\n\"2nd
\ntan 30\u00b0 = \\(\\frac{\\mathrm{AC}}{\\mathrm{AB}}\\); \\(\\frac{1}{\\sqrt{3}}=\\frac{h}{500}\\)
\n\u2234 h = \\(\\frac{500}{\\sqrt{3}}=500 \\sqrt{3}\\) Height of the tower m \\(\\frac{500 \\sqrt{3}}{2} \\mathrm{mts}\\)<\/p>\n

Question 2.
\nThe angle of elevation of the top of a chimney at a distance of 100 metres from a foot is 30\u00b0. Find its height.
\nAnswer:
\nin triangle ABC we have B as the position of observer, AC height of tan 30\u00b0 = \\(\\frac{A C}{A B}\\) the chimney = h
\n\"2nd
\n\\(\\frac{1}{\\sqrt{3}}=\\frac{h}{100}\\)
\n\\(\\mathrm{h}=\\frac{100}{\\sqrt{3}}=\\frac{100 \\sqrt{3}}{3} \\mathrm{m}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFrom a ship a mast head 40 meters high the angle of depression of a boat is observed to be 45\u00b0. Find its distance from the ship.
\nAnswer:
\nIn triangle ABC, we have AC as mast head, AB is distance from the ship
\n\"2nd
\ntan 45\u00b0 = \\(\\frac{A C}{A B}\\)
\n\\(1=\\frac{40}{x}\\)
\n\u21d2 x = 40
\n\u2234 Distance from the ship is 40 mts<\/p>\n

Question 4.
\nWhat is the angle of elevation of the sun when the length of the shadow of a pole is \\(\\frac{1}{\\sqrt{3}}\\) times the height of the pole?
\nAnswer:
\nLet the height of the pole be AC = h & length of the shadow = AB = \\(\\frac{1}{\\sqrt{3}}\\) h
\n\"2nd
\nFrom triangle ABC we have tan
\ntan \u03b8 = \\(\\frac{A C}{A B}\\) tan \u03b8 = \\(\\frac{\\mathrm{h}}{\\frac{1}{\\sqrt{3}}}=\\sqrt{3}\\) = tan 60\u00b0
\n\u21d2 \u03b8 = 30\u00b0<\/p>\n

Question 5.
\nFind the angle of elevation of the sun when the shadow of a tower 75 meters high is meters \\(25\\sqrt{3}\\) long.
\nAnswer:
\nAC = height of the tower = h = 75 mts AB = Shadow of a tower = \\(25\\sqrt{3}\\) from triangle ABC, we have
\n\"2nd
\ntan \u03b8 = \\(\\frac{A C}{A B}=\\frac{75}{25 \\sqrt{3}}=\\frac{3}{\\sqrt{3}}=\\sqrt{3}\\) = tan 60\u00b0
\n\u21d2 \u03b8 = 60\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA kite flying at a height of h is tied to a thread which is 500m long. Assuming that there is no kink in the thread and it makes an angle of 30\u00b0 with the ground. Find the height of the kite.
\nAnswer:
\nLet AC = height of the kite BC = 500 mts From \u2206 ABC we have
\n\"2nd
\nsin 30\u00b0 = \\(\\frac{A C}{B C}=\\frac{h}{500}\\)
\n\\(\\frac{1}{2}=\\frac{h}{500} \\Rightarrow h=\\frac{500}{2}=250 \\mathrm{mts}\\)
\n\u2234 the heights of the kite = 250 mts.<\/p>\n

Question 7.
\nA ladder leaning against a wall makes an angle of 60\u00b0 with the ground. The foot of the ladder is 6m away from the wall. Find the length of the ladder.
\nAnswer:
\n\"2nd
\nLet AC be the ladder from triangle ABC we have
\ncos 60\u00b0 = \\(\\frac{A B}{B C}=\\frac{6}{B C}\\)
\n\\(\\frac{1}{2}=\\frac{6}{\\mathrm{BC}} \\Rightarrow \\mathrm{BC}=12 \\mathrm{mts}\\)
\n\u2234 The length of the ladder = 12 mts.<\/p>\n

Question 8.
\nFind the angle of elevation of the sun’s rays from a point on the ground at a distance of \\(3 \\sqrt{3}m\\) , from the foot of tower 3m high.
\nAnswer:
\n\"2nd
\nLet AC be the height of the pole AC = 3mts & AB = \\(3 \\sqrt{3}m\\) mts From the triangle ABC we have
\ntan \u03b8 = \\(\\frac{A C}{A B}=\\frac{3}{3 \\sqrt{3}}=\\frac{1}{\\sqrt{3}}\\) = tan 30 \u21d2 = 30\u00b0
\n\u2234 The angle of elevation is 30\u00b0<\/p>\n

\"KSEEB<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Heights and Distances Ex 13.1 Four or Five marks questions and answers<\/strong><\/p>\n

Question 1.
\nThe angles of elevation of the top of a tower from the base and the top of a building are 60\u00b0 and 45\u00b0 The building is 20 meters high. Find the height of the tower.
\nAnswer:
\nLet AD be the tower & CE be the building. Let the distance between the tower and building be x in \u2206 ABC,
\ntan 45\u00b0 = \\(\\frac{A B}{B C}=\\frac{h}{x}\\)
\n\\(1=\\frac{\\mathrm{h}}{\\mathrm{x}} \\Rightarrow \\mathrm{x}=\\mathrm{h}-(1)\\)
\n\"2nd
\n\"2nd<\/p>\n

Question 2.
\nThe shadow of a tower standing on a level plane is found to be 50 meters longer when sun’s altitude is 30\u00b0. Than when it is 60\u00b0. Find the height of the tower.
\nAnswer:
\n\"2nd
\nLet AB be the tower In triangle ABC, tan 60\u00b0 = \\(\\frac { h }{ x }\\)
\n\\(\\sqrt{3}=\\frac{h}{x}\\)
\n\\(h=\\sqrt{3 x}\\)
\nIn triangle ABD, tan 30\u00b0 = \\(\\frac{h}{50+x}\\)
\n50 + x = \\(\\sqrt{3}\\)
\n50 + x = \\(\\sqrt{3} \\cdot \\sqrt{3} x\\) \u2235 h = \\(\\sqrt{3} x\\)
\n50 = 3x – x \u21d2 2x = 50 \u21d2 x = 25 mts
\n\u2234 Height of the tower h = 25\\(\\sqrt{3}\\) mts<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nAn Aeroplane when flying at a heights of 2000 meters passes vertically above another plane at an instant, when their angles of elevation from the same point of observation are 60\u00b0 and 45\u00b0 respectively. Find the distance between the Aeroplanes.
\nAnswer:
\n\"2nd
\nLet A & B are Aeroplanes Let h be the distance between the aeroplanes
\nFrom triangle ACD
\n\"2nd
\nFrom triangle BCD
\n\"2nd<\/p>\n

Question 4.
\nFrom a point on the line joining the feet of two poles of equal heights, the angles of elevation of the tops of the poles are observed to be 30\u00b0 and 60\u00b0. If the distance between the poles is a Find (i) the height of the poles (ii) the position of the point of observation.
\nAnswer:
\n\"2nd
\nLet AB & DE are two poles
\n\"2nd
\n\u2234 Height of the poles is \\(\\frac{\\sqrt{3}}{4} a\\)
\nPosition of c from B is x = \\(\\sqrt{3 h}\\)
\n\u2234 position of the point of observation = \\(\\frac{\\sqrt{3} \\cdot \\sqrt{3} a}{4}=\\frac{3 a}{4}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nThe angles of elevation of the top of a tower from two points distant a and b (a < b) from its foot and the same straigth line from it are 30\u00b0 and 60\u00b0. Show that the height of the tower is \\(\\sqrt{a b}\\).
\nAnswer:
\n\"2nd
\nLet CD is the tower from rt angled triangle ACD
\nwe have tan 30 = \\(\\frac { h }{ a }\\) \u21d2 h = \\(\\frac{a}{\\sqrt{3}}\\) …(1)
\nFrom rt angled triangle BCD
\ntan 60\u00b0 = \\(\\frac { h }{ b }\\)
\n\\(\\sqrt{3}=\\frac{h}{b} \\Rightarrow h=\\sqrt{3} b-(2)\\)
\nFrom 1 & 2 we have
\n\"2nd<\/p>\n

Question 6.
\nA flag staff stands upon the top of a building. At a distance of 20 meters the angles of elevation of the top of the flag staff and building are 60\u00b0 and 60\u00b0 respectively. Find the height of the flag staff.
\nAnswer:
\n\"2nd
\nLet BC = Building
\nAB = flagstaff
\nFrom triangle ACD we have
\ntan 60\u00b0 = \\(\\frac{A C}{C D}=\\frac{A B+B C}{20}=\\sqrt{3}\\)
\nAB + BC = 20 \\(\\sqrt{3}\\) …. (1)
\nIn triangle BCD, tan 30\u00b0 = \\(\\frac{B C}{C D} ; \\quad \\frac{1}{\\sqrt{3}}=\\frac{B C}{20}\\)
\nBC = \\(\\frac{20}{\\sqrt{3}}\\) …. (2)<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFrom the top of a cliff, the angles of depression of two boats in the same vertical plane as the observer are 30\u00b0 and 45\u00b0. If the distance between the boats is 100 meters, find the height of the cliff.
\nAnswer:
\n\"2nd
\nLet the height of the cliff be h A & B are two boats
\nFrom triangle DCA
\ntan 45\u00b0 = \\(\\frac{D C}{A C}=\\frac{h}{x}\\)
\n\\(1=\\frac{\\mathrm{h}}{\\mathrm{x}} \\Rightarrow \\mathrm{h}=\\mathrm{x}\\) ….. (1)
\nFrom triangle BCD
\n\"2nd<\/p>\n

Question 8.
\nFrom a point A due north of the tower, the elevation of the top of the tower is 60\u00b0. From a point B due south, the elevantion is 45\u00b0, if AB = 100 metres. Show that the height of the tower is 50 \\(\\sqrt{3}(\\sqrt{3}-1)\\) meters.
\nAnswer:
\n\"2nd
\nA and B are positions of observation AB = 100 mts and CD is the height of tower and CD = h
\nFrom rt angled triangle ACD
\ntan 60\u00b0 = \\(\\frac{\\mathrm{CD}}{\\mathrm{AD}}\\)
\n\\(\\sqrt{3}\\) = \\(\\frac{h}{A D} \\Rightarrow A D=\\frac{h}{\\sqrt{3}}\\) …..(1)<\/p>\n

From rt angled triangle BCD
\ntan 45\u00b0 = \\(\\frac{\\mathrm{CD}}{\\mathrm{BD}}\\)
\n\\(1=\\frac{h}{B D} \\Rightarrow B D=h\\) …. (2)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nA person at the top of a hill observes that the angles of depression of two consecutive kilometers stones on a road leading to the foot of the hill and in the same vertical plane containing the position of the observer are 30\u00b0 and 60\u00b0. Find the height of the hill.
\nAnswer:
\nLet CD be the hill and CD = h mts
\nIn \u2206 ADC, we have tan 60\u00b0 = \\(\\frac{C D}{A D} \\Rightarrow \\sqrt{3}=\\frac{h}{x}\\)
\n\\(\\Rightarrow \\quad x=\\frac{h}{\\sqrt{3}} \\Rightarrow h=\\sqrt{3} x\\) …. (1)
\nFrom 1 and 2 we get
\n1 + x = \\(\\sqrt{3} \\cdot \\sqrt{3} x\\) (h = \\(\\sqrt{3} x\\))
\n1 + x = 3x
\n1 = 2x \u21d2 x = 5 \\(\\frac{1}{2}\\)
\nThe height of the hill = h = \\(\\sqrt{3} x=\\sqrt{3} \\cdot \\frac{1}{2}=\\frac{\\sqrt{3}}{2} \\mathrm{mts}\\) mts
\n\"2nd<\/p>\n

Question 10.
\nThe angles of elevation of the top of a tower from the base the top of a building are 60\u00b0 & 45\u00b0 the building is 32 meters high find the height of the tower
\nAnswer:
\n\"2nd
\nLet CD is the tower
\nAB is the house = 32 meters
\nFrom the rt angled triangle ACE
\ntan 45\u00b0 = \\(\\frac{C E}{A E}=\\frac{C E}{B D}=1\\)
\n\u21d2 CE = BD …. (1)
\nAgain from rt angled triangle ABD
\ntan 30\u00b0 = \\(\\frac{A B}{B D}=\\frac{32}{B D}=\\frac{1}{\\sqrt{3}}\\)
\n\u21d2 BD = \\(32 \\sqrt{3}\\)
\nFrom 1 and 2 we get CE = \\(32 \\sqrt{3}\\) mts
\n\u2234 Height of the tower = h = DE + CE = 32 + \\(32 \\sqrt{3}\\) = -32(1 + \\(\\sqrt{3}\\))mts<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nThe angle of elevation of a tower from a point on the ground is 30\u00b0. At a point on the horizontal line passing through the foot of the tower and 100 metres nearer it, the angle of elevation is found to be 60\u00b0. Find the height of the pole.
\nAnswer:
\n\"2nd
\nLet CD = Tower
\nFrom a rt angled triangle ACD
\ntan 30\u00b0 = \\(\\frac{\\mathrm{CD}}{\\mathrm{AC}}\\)
\n\\frac{1}{\\sqrt{3}}=\\(\\frac{h}{100+x}\\)
\n100+ x = \\(\\mathrm{h} \\sqrt{3}\\) …. (1)<\/p>\n

From triangle BCD
\ntan 60\u00b0 = \\(\\frac{\\mathrm{CD}}{\\mathrm{CB}}\\)
\n\\(\\sqrt{3}=\\frac{h}{x} \\Rightarrow h=\\sqrt{3} x\\) …. (2)<\/p>\n

From 1 & 2 we get
\n100 + x = \\(\\sqrt{3} \\cdot \\sqrt{3} x\\)
\n100 = 3x – x \u21d2 2x = 100 \u21d2 x = 50 :
\n\u2234 Height of the tower = h = \\(\\sqrt{3 x}\\) = \\(50\\sqrt{3}\\) mts
\nHence Distance of the first point from the tower \u20b9 100 + x = 100 + 50 = 150 mts<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nA Person is at the top of a tower 75 feet high from there he observes a vertical pole and finds the angles :: of depressions of the top and the bottom of the pole which are 30\u00b0 and 60\u00b0 respectively. Find the height of the pole.
\nAnswer:
\n\"2nd
\nLet AB = Tower, CD = Pole = h
\nFrom triangle ABC we get
\ntan 60\u00b0 = \\(\\frac{A B}{A C} \\Rightarrow \\frac{75}{A C}=\\sqrt{3}\\)
\n\u21d2 AC = \\(\\frac{75}{\\sqrt{3}} ….. (1)\\) …. (1)<\/p>\n

From triangle BDE
\n\"2nd
\n\u21d2 AC = \\(\\sqrt{3}(75-h)\\)<\/p>\n

From 1 and 2 we get
\n\\(\\frac{75}{\\sqrt{3}}=\\sqrt{3}(75-h) \\Rightarrow 75=(\\sqrt{3})^{2}(75-h)\\)
\n75 = 225 – 3h 3h \u21d2 150 \u21d2 h = 50
\n\u2234 Height of the pole -50ft<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 13.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 13 Heights and Distances Ex 13.1 Part – A 2nd …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20703"}],"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=20703"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20703\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20703"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20703"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20703"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}