{"id":3199,"date":"2023-05-16T10:00:25","date_gmt":"2023-05-16T04:30:25","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=3199"},"modified":"2023-05-17T17:38:36","modified_gmt":"2023-05-17T12:08:36","slug":"ap-ssc-10th-class-maths-solutions-chapter-12-ex-12-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-ssc-10th-class-maths-solutions-chapter-12-ex-12-2\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2<\/h2>\n

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers<\/h3>\n

Question 1.
\nA TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60\u00b0. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30\u00b0. Find the height of the tower and the width of the road.
\nAnswer:
\n\"AP
\nLet the height of the tower = h mts say
\nWidth of the road be = x m.
\nDistance between two points of observation = 10 cm.
\nAngles of elevation from the two points = 60\u00b0 and 30\u00b0.
\nFrom the figure
\ntan 60\u00b0 = \\(\\frac{h}{x}\\)
\n\u221a3 = \\(\\frac{h}{x}\\)
\n\u21d2 h = \u221a3x …….(1)
\nAlso tan 30\u00b0 = \\(\\frac{h}{10+x}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{h}{10+x}\\)
\n\u21d2 h = \\(\\frac{10+x}{\\sqrt{3}}\\) ………(2)
\nFrom equations (1) and (2) h
\nh = \u221a3x = \\(\\frac{10+x}{\\sqrt{3}}\\)
\n\u2234 \u221a3x = \\(\\frac{10+x}{\\sqrt{3}}\\)
\n\u221a3 \u00d7 \u221a3x = 10 + x
\n\u21d2 3x – x = 10
\n\u21d2 2x = 10
\n\u21d2 x = \\(\\frac{10}{2}\\) = 5m
\n\u2234 Width of the road = 5 m
\nNow Height of the tower = \u221a3x = 5\u221a3 m.<\/p>\n

\"AP<\/p>\n

Question 2.
\nA 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30\u00b0 to 60\u00b0 as he walks towards the temple. Find the distance he walked towards the temple.
\nAnswer:
\n\"AP
\nHeight of the temple = 30 m
\nHeight of the man = 1.5 m
\nInitial distance between the man and temple = d m. say
\nLet the distance walked = x m.
\nFrom the figure
\ntan 30\u00b0 = \\(\\frac{30-1.5}{d}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{28.5}{d}\\)
\n\u2234 d = 28.5 \u00d7 \u221a3m ………(1)
\nAlso tan 60\u00b0 = \\(\\frac{28.5}{d-x}\\)
\n\u21d2 \u221a3 = \\(\\frac{28.5}{d-x}\\)
\n\u21d2 \u221a3(d-x) = 28.5
\n\u21d2 \u221a3(28.5 \u00d7 \u221a3-x) = 28.5
\n\u21d2 28.5 \u00d7 3 – \u221a3x = 28.5
\n\u21d2 \u221a3x = 3 \u00d7 28.5-28.5
\n\u21d2 \u221a3x = 2 \u00d7 28.5 = 57
\n\u2234 x = \\(\\frac{57}{\\sqrt{3}}=\\frac{19 \\times 3}{\\sqrt{3}}\\) = 19\u221a3
\n= 19 \u00d7 1.732
\n= 32.908 m.
\n\u2234 Distance walked = 32.908 m.<\/p>\n

Question 3.
\nA statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60\u00b0 and from the same point, the angle of elevation of the top of the pedestal is 45\u00b0. Find the height of the statue.
\nAnswer:
\n\"AP
\nHeight of the pedestal = 2 m.
\nLet the height of the statue = h m. Angle of elevation of top of the statue = 60\u00b0.
\nAngle of elevation of top of the pedestal = 45\u00b0.
\nLet the distance between the point of observation and foot of the pedestal = x m.
\nFrom the figure
\ntan 45\u00b0 = \\(\\frac{2}{x}\\)
\n1 = \\(\\frac{2}{x}\\)
\n\u2234 x = 2 m.
\nAlso tan 60\u00b0 = \\(\\frac{2+h}{x}\\)
\n\u21d2 \u221a3 = \\(\\frac{2+h}{x}\\)
\n\u21d2 2\u221a3 = 2 + h
\n\u21d2 h = 2\u221a3 – 2
\n= 2(\u221a3-1)
\n= 2(1.732 – 1)
\n= 2 \u00d7 0.732
\n= 1.464 m.<\/p>\n

\"AP<\/p>\n

Question 4.
\nFrom the top of a building, the angle of elevation of the top of a cell tower is 60\u00b0 and the angle of depression to its foot is 45\u00b0. If distance of the building from the tower is 7 m, then find the height of the tower.
\nAnswer:
\n\"AP
\nAngle of elevation of the top of the tower = 60\u00b0.
\nAngle of depression to the foot of the tower = 45\u00b0.
\nDistance between tower and building = 7 m.
\nLet the height of the building = x m and tower = y m.
\nFrom the figure
\ntan 45\u00b0 = \\(\\frac{x}{7}\\)
\n1 = \\(\\frac{x}{7}\\)
\n\u2234 x = 7 m.
\nAlso tan 60\u00b0 = \\(\\frac{y-x}{7}\\)
\n\u21d2 \u221a3 = \\(\\frac{y-x}{7}\\)
\n\u21d2 7\u221a3 = y – 7
\n\u2234 y = 7 + 7\u221a3
\n= 7 (\u221a3 + 1)
\n= 7(1.732 + 1)
\n= 2.732 \u00d7 7
\n= 19.124 m.<\/p>\n

Question 5.
\nA wire of length 18 m had been tied with electric pole at an angle of eleva\u00action 30\u00b0 with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60\u00b0 with the ground. How much length of the wire was cut ?
\nAnswer:
\n\"AP
\nLength of the wire = 18 m
\nLet the length of the wire removed = x
\nHeight of the pole be = h
\nFrom the figure
\nsin 30\u00b0 = \\(\\frac{h}{18}\\)
\n\u21d2 \\(\\frac{1}{2}\\) = \\(\\frac{h}{18}\\)
\n\u21d2 h = \\(\\frac{18}{2}\\) = 9 m
\nAlso sin 60\u00b0 = \\(\\frac{h}{18-x}\\)
\n\\(\\frac{\\sqrt{3}}{2}\\) = \\(\\frac{9}{18-x}\\)
\n\u221a3(18-x) = 9 \u00d7 2
\n18\u221a3 – \u221a3x = 18
\n\u221a3x = 18\u221a3 – 18
\n\u221a3x = 18(\u221a3-1)
\nx = \\(\\frac{18(\\sqrt{3}-1)}{\\sqrt{3}}\\)
\n= \\(\\frac{6 \\times 3(\\sqrt{3}-1)}{\\sqrt{3}}\\)
\n= 6\u221a3(\u221a3-1)
\n= 6(3-\u221a3)
\n= 18 – 6\u221a3
\n= 18 – 10.392
\n= 7.608 m.<\/p>\n

\"AP<\/p>\n

Question 6.
\nThe angle of elevation of the top of a building from the foot of the tower is 30\u00b0 and the angle of elevation of the top of the tower from the foot of the building is 60\u00b0. If the tower is 30 m high, find the height of the building.
\nAnswer:
\n\"AP
\nHeight of the tower = 30 m
\nAngle of elevation of the top of the tower = 60\u00b0.
\nAngle of elevation of the top of the building = 30\u00b0.
\nLet the distance between the foot of the tower and foot of the building be d m and height of the building be x m.
\nFrom the figure
\ntan 60\u00b0 = \\(\\frac{30}{d}\\)
\n\u221a3 = \\(\\frac{30}{d}\\)
\n\u21d2 d = \\(\\frac{30}{\\sqrt{3}}=\\frac{10 \\times 3}{\\sqrt{3}}\\) = 10\u221a3m
\nAlso tan 30\u00b0 = \\(\\frac{x}{d}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}=\\frac{x}{10 \\sqrt{3}}\\)
\n\u21d2 x = \\(\\frac{10 \\sqrt{3}}{\\sqrt{3}}\\) = 10 m
\n\u2234 Height of the building = 10 m<\/p>\n

Question 7.
\nTwo poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60\u00b0 and 30\u00b0 respectively. Find the height of the poles and the distances of the point from the poles.
\nAnswer:
\n\"AP
\nWidth of the road = 120 f.
\nAngle of elevation of the top of the 1st tower = 60\u00b0.
\nAngle of elevation of the top of the 2 tower = 30\u00b0.
\nLet the distance of the point from the 1st pole = x.
\nThen the distance of the point from
\nthe 2nd pole = 120 – x.
\nand height of each pole = h say.
\nFrom the figure
\ntan 60\u00b0 = \\(\\frac{h}{x}\\)
\n\u21d2 \u221a3 = \\(\\frac{h}{x}\\)
\n\u21d2 h = \u221a3x ……..(1)
\nAlso tan 30\u00b0 = \\(\\frac{\\mathrm{h}}{120-\\mathrm{x}}\\)
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}=\\frac{\\mathrm{h}}{120-\\mathrm{x}}\\)
\n\u21d2 h = \\(\\frac{120-x}{\\sqrt{3}}\\)
\nFrom (1) and (2)
\n\u221a3x = \\(\\frac{120-x}{\\sqrt{3}}\\)
\n\u21d2 \u221a3.\u221a3x = 120-x
\n\u21d2 3x = 120 – x
\n\u21d2 3x + x = 120
\n\u21d2 4x = 120
\n\u21d2 x = \\(\\frac{120}{4}\\) = 30 ft
\nNow h = \u221a3x = \u221a3 \u00d7 30 = 1.732 x 30 = 51.960 feet
\n\u2234 Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.
\nHeight of each pole = 51.96 ft.<\/p>\n

\"AP<\/p>\n

Question 8.
\nThe angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.
\nAnswer:
\n\"AP
\nLet the height of the tower = h m.
\nAngles of elevation of the top of the tower from two points = x\u00b0 and (90\u00b0 – x)
\nFrom the figure
\ntan x = \\(\\frac{h}{4}\\) ……. (1)
\nAlso tan (90\u00b0 – x) = \\(\\frac{h}{9}\\)
\n\u21d2 cot x = \\(\\frac{h}{9}\\)
\n\u21d2 \\(\\frac{1}{\\tan x}\\) = \\(\\frac{h}{9}\\)
\n\u2234 tan x = \\(\\frac{9}{h}\\) …….. (2)
\nFrom (1) and (2)
\ntan x = \\(\\frac{h}{4}\\) = \\(\\frac{9}{h}\\)
\n\u2234 \\(\\frac{h}{4}\\) = \\(\\frac{9}{h}\\)
\nh \u00d7 h = 9 \u00d7 4
\n\u21d2 h2<\/sup> = 36
\n\u21d2 h = 6 m<\/p>\n

Question 9.
\nThe angle of elevation of a jet plane from a point A on the ground is 60\u00b0. After 4 flight of 15 seconds, the angle of elevation changes to 30\u00b0. If the jet plane is flying at a constant height of 1500\u221a3 meter, find the speed of the jet plane. (\u221a3 = 1.732)
\nAnswer:
\n\"AP
\nHeight of the plane from the ground PM = RN = 1500\u221a3 m.
\nAngle of elevation are 30\u00b0 and 60\u00b0.
\nFrom the figure
\ntan 60\u00b0 = \\(\\frac{PM}{QM}\\)
\n\u221a3 = \\(\\frac{1500 \\sqrt{3}}{\\mathrm{QM}}\\)
\nQM = \\(\\frac{1500 \\sqrt{3}}{\\sqrt{3}}\\) = 1500 m
\nAlso tan 30\u00b0 = \\(\\frac{RN}{QN}\\)
\n\\(\\frac{1}{\\sqrt{3}}=\\frac{1500 \\sqrt{3}}{\\mathrm{QM}+\\mathrm{MN}}\\)
\nQM + MN = 1500\u221a3 \u00d7 \u221a3
\n1500 + MN = 1500 \u00d7 3
\nMN = 4500 – 1500
\nMN = 3000 mts.
\n\u2234 Distance travelled in 15 seconds = 3000 mts.
\n\u2234 Speed of the jet plane = \\(\\frac{\\text { distance }}{\\text { time }}=\\frac{3000}{15}\\) = 200 m\/s
\n= 200 \u00d7 \\(\\frac{18}{5}\\) kmph
\n= 720 kmph
\nSpeed = 200 m\/sec. or 720 kmph.<\/p>\n

AP SSC 10th Class Maths Textbook Solutions<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2 10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers Question 1. A TV tower … Read more<\/a><\/p>\n","protected":false},"author":5,"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.in\/wp-json\/wp\/v2\/posts\/3199"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=3199"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/3199\/revisions"}],"predecessor-version":[{"id":40529,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/3199\/revisions\/40529"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=3199"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=3199"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=3199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}