{"id":119436,"date":"2022-05-04T15:57:27","date_gmt":"2022-05-04T10:27:27","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119436"},"modified":"2022-05-04T15:57:27","modified_gmt":"2022-05-04T10:27:27","slug":"chapter-9-some-applications-of-trigonometry-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-9-some-applications-of-trigonometry-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 9 – Some Applications of Trigonometry Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 9.1<\/h2>\n

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30\u00b0. (see fig. 9.11)<\/h2>\n

\"\"
\nSolution:
\nLength of the rope is 20 m and angle made by the rope with the ground level is 30\u00b0.
\nGiven:\u00a0AC = 20 m and angle C = 30\u00b0
\nTo Find: Height of the pole
\nLet AB be the vertical pole
\nIn right \u0394ABC, using sine formula
\nsin 30\u00b0 = AB\/AC
\nUsing value of sin 30 degrees is \u00bd, we have
\n1\/2 = AB\/20
\nAB = 20\/2
\nAB = 10
\nTherefore, the height of the pole is 10 m.<\/h3>\n

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30\u00b0 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.<\/h2>\n

Solution:
\nUsing given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30\u00b0
\nBC = 8 m
\nTo Find: Height of the tree, which is AB<\/h3>\n

\"\"
\nFrom figure: Total height of the tree is the sum of AB and AC i.e. AB+AC
\nIn right \u0394ABC,
\nUsing Cosine and tangent angles,
\ncos 30\u00b0 = BC\/AC
\nWe know that, cos 30\u00b0 = \u221a3\/2
\n\u221a3\/2 = 8\/AC
\nAC = 16\/\u221a3 \u2026(1)
\nAlso,
\ntan 30\u00b0 = AB\/BC
\n1\/\u221a3 = AB\/8
\nAB = 8\/\u221a3 \u2026.(2)
\nTherefore, total height of the tree = AB + AC = 16\/\u221a3 + 8\/\u221a3 = 24\/\u221a3 = 8\u221a3 m.<\/h3>\n

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30\u00b0 to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60\u00b0 to the ground. What should be the length of the slide in each case?<\/h2>\n

Solution:
\nAs per contractor\u2019s plan,<\/h3>\n

\"\"<\/p>\n

\"\"
\nLet, ABC is the slide inclined at 30\u00b0 with length AC and PQR is the slide inclined at
\n60\u00b0 with length PR.
\nTo Find: AC and PR
\nIn right \u0394ABC,
\nsin 30\u00b0 = AB\/AC
\n1\/2 = 1.5\/AC
\nAC = 3
\nAlso,
\nIn right \u0394PQR,
\nsin 60\u00b0 = PQ\/PR
\n\u21d2 \u221a3\/2 = 3\/PR
\n\u21d2 PR = 2\u221a3
\nHence, length of the slide for below 5 = 3 m and
\nLength of the slide for elders children = 2\u221a3 m<\/h3>\n

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30\u00b0. Find the height of the tower.<\/h2>\n

Solution:
\nLet AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.<\/h3>\n

\"\"
\nTo Find: AB (height of the tower)
\nIn right ABC
\ntan 30\u00b0 = AB\/BC
\n1\/\u221a3 = AB\/30
\n\u21d2 AB = 10\u221a3
\nThus, the height of the tower is 10\u221a3 m.<\/h3>\n

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60\u00b0. Find the length of the string, assuming that there is no slack in the string.<\/h2>\n

Solution:
\nDraw a figure, based on given instruction,<\/h3>\n

\"\"
\nLet BC = Height of the kite from the ground, BC = 60 m
\nAC = Inclined length of the string from the ground and
\nA is the point where string of the kite is tied.
\nTo Find: Length of the string from the ground i.e. the value of AC
\nFrom the above figure,
\nsin 60\u00b0 = BC\/AC
\n\u21d2 \u221a3\/2 = 60\/AC
\n\u21d2 AC = 40\u221a3 m
\nThus, the length of the string from the ground is 40\u221a3 m.<\/h3>\n

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30\u00b0 to 60\u00b0 as he walks towards the building. Find the distance he walked towards the building.<\/h2>\n

Solution:
\nLet the boy initially stand at point Y with inclination 30\u00b0 and then he approaches the building to the point X with inclination 60\u00b0.<\/h3>\n

\"\"
\nTo Find: The distance boy walked towards the building i.e. XY
\nFrom figure,
\nXY = CD.
\nHeight of the building = AZ = 30 m.
\nAB = AZ \u2013 BZ = 30 \u2013 1.5 = 28.5
\nMeasure of AB is 28.5 m
\nIn right \u0394ABD,
\ntan 30\u00b0 = AB\/BD
\n1\/\u221a3 = 28.5\/BD
\nBD = 28.5\u221a3 m
\nAgain,
\nIn right \u0394ABC,
\ntan 60\u00b0 = AB\/BC
\n\u221a3 = 28.5\/BC
\nBC = 28.5\/\u221a3 = 28.5\u221a3\/3
\nTherefore, the length of BC is 28.5\u221a3\/3 m.
\nXY = CD = BD \u2013 BC = (28.5\u221a3-28.5\u221a3\/3) = 28.5\u221a3(1-1\/3) = 28.5\u221a3 \u00d7 2\/3 = 57\/\u221a3 = 19\u221a3 \u00a0m.
\nThus, the distance boy walked towards the building is 19\u221a3 m.<\/h3>\n

7. From a point on the ground, the angles of elevation of the bottom and the top of a
\ntransmission tower fixed at the top of a 20 m high building are 45\u00b0 and 60\u00b0 respectively. Find the height of the tower.<\/h2>\n

Solution:
\nLet BC be the 20 m high building.
\nD is the point on the ground from where the elevation is taken.
\nHeight of transmission tower = AB = AC \u2013 BC<\/h3>\n

\"\"
\nTo Find: AB, Height of the tower
\nFrom figure, In right \u0394BCD,
\ntan 45\u00b0 = BC\/CD
\n1 = 20\/CD
\nCD = 20
\nAgain,
\nIn right \u0394ACD,
\ntan 60\u00b0 = AC\/CD
\n\u221a3 = AC\/20
\nAC = 20\u221a3
\nNow, AB = AC \u2013 BC = (20\u221a3-20) = 20(\u221a3-1)
\nHeight of transmission tower = 20(\u221a3 \u2013 1) m.<\/h3>\n

8. A statue, 1.6 m tall, stands on the top of a 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 pedestal.<\/h2>\n

Solution:
\nLet AB be the height of statue.
\nD is the point on the ground from where the elevation is taken.
\nTo Find: Height of pedestal = BC = AC-AB<\/h3>\n

\"\"
\nFrom figure,
\nIn right triangle BCD,
\ntan 45\u00b0 = BC\/CD
\n1 = BC\/CD
\nBC = CD \u2026..(1)<\/h3>\n

Again,
\nIn right \u0394ACD,
\ntan 60\u00b0 = AC\/CD
\n\u221a3 = ( AB+BC)\/CD
\n\u221a3CD = 1.6 + BC
\n\u221a3BC = 1.6 + BC (using equation (1)
\n\u221a3BC \u2013 BC = 1.6
\nBC(\u221a3-1) = 1.6
\nBC = [(1.6)(\u221a3+1)]\/[(\u221a3-1)(\u221a3+1)]
\nBC = [1.6(\u221a3+1)]\/(2) m
\nBC = 0.8(\u221a3+1)
\nThus, the height of the pedestal is 0.8(\u221a3+1) m.<\/h3>\n

9. The 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 50 m high, find the height of the building.<\/h2>\n

Solution:
\nLet CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building respectively.<\/h3>\n

\"\"
\nIn right \u0394BCD,
\ntan 60\u00b0 = CD\/BC
\n\u221a3 = 50\/BC
\nBC = 50\/\u221a3 \u2026(1)
\nAgain,
\nIn right \u0394ABC,
\ntan 30\u00b0 = AB\/BC
\n\u21d2 1\/\u221a3 = AB\/BC
\nUse result obtained in equation (1)
\nAB = 50\/3
\nThus, the height of the building is 50\/3 m.<\/h3>\n

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m 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.<\/h2>\n

Solution:
\nLet AB and CD be the poles of equal height.
\nO is the point between them from where the height of elevation taken. BD is the distance between the poles.<\/h3>\n

\"\"
\nAs per above figure, AB = CD,
\nOB + OD = 80 m
\nNow,
\nIn right \u0394CDO,
\ntan 30\u00b0 = CD\/OD
\n1\/\u221a3 = CD\/OD
\nCD = OD\/\u221a3 \u2026 (1)
\nAgain,
\nIn right \u0394ABO,
\ntan 60\u00b0 = AB\/OB
\n\u221a3 = AB\/(80-OD)
\nAB = \u221a3(80-OD)
\nAB = CD (Given)
\n\u221a3(80-OD) = OD\/\u221a3 (Using equation (1))
\n3(80-OD) = OD
\n240 \u2013 3 OD = OD
\n4 OD = 240
\nOD = 60
\nPutting the value of OD in equation (1)
\nCD = OD\/\u221a3
\nCD = 60\/\u221a3
\nCD = 20\u221a3 m
\nAlso,
\nOB + OD = 80 m
\n\u21d2 OB = (80-60) m = 20 m
\nThus, the height of the poles are 20\u221a3 m and distance from the point of elevation are 20 m and
\n60 m respectively.<\/h3>\n

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60\u00b0. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30\u00b0 (see Fig. 9.12). Find the height of the tower and the width of the canal.<\/h2>\n

\"\"
\nSolution: Given, AB is the height of the tower.
\nDC = 20 m (given)
\nAs per given diagram, In right \u0394ABD,
\ntan 30\u00b0 = AB\/BD
\n1\/\u221a3 = AB\/(20+BC)
\nAB = (20+BC)\/\u221a3 \u2026 (i)
\nAgain,
\nIn right \u0394ABC,
\ntan 60\u00b0 = AB\/BC
\n\u221a3 = AB\/BC
\nAB = \u221a3 BC \u2026 (ii)
\nFrom equation (i) and (ii)
\n\u221a3 BC = (20+BC)\/\u221a3
\n3 BC = 20 + BC
\n2 BC = 20
\nBC = 10
\nPutting the value of BC in equation (ii)
\nAB = 10\u221a3
\nThis implies, the height of the tower is 10\u221a3 m and the width of the canal is 10 m.<\/h3>\n

12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60\u00b0 and the angle of depression of its foot is 45\u00b0. Determine the height of the tower.<\/h2>\n

Solution:
\nLet AB be the building of height 7 m and EC be the height of the tower.
\nA is the point from where elevation of tower is 60\u00b0 and the angle of depression of its foot is 45\u00b0.
\nEC = DE + CD
\nAlso, CD = AB = 7 m. and BC = AD
\nTo Find: EC = Height of the tower
\nDesign a figure based on given instructions:<\/h3>\n

\"\"
\nIn right \u0394ABC,
\ntan 45\u00b0 = AB\/BC
\n1= 7\/BC
\nBC = 7
\nSince BC = AD
\nSo AD = 7
\nAgain, from right triangle ADE,
\ntan 60\u00b0 = DE\/AD
\n\u221a3 = DE\/7
\n\u21d2 DE = 7\u221a3 m
\nNow: EC = DE + CD
\n= (7\u221a3 + 7) = 7(\u221a3+1)
\nTherefore, Height of the tower is 7(\u221a3+1) m. Answer!<\/h3>\n

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30\u00b0 and 45\u00b0. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.<\/h2>\n

Solution:
\nLet AB be the lighthouse of height 75 m. Let C and D be the positions of the ships.
\n30\u00b0 and 45\u00b0 are the angles of depression from the lighthouse.
\nDraw a figure based on given instructions:<\/h3>\n

\"\"
\nTo Find: CD = distance between two ships<\/h3>\n

Step 1: From right triangle ABC,
\ntan 45\u00b0 = AB\/BC
\n1= 75\/BC
\nBC = 75 m<\/h3>\n

Step 2: Form right triangle ABD,
\ntan 30\u00b0 = AB\/BD
\n1\/\u221a3 = 75\/BD
\nBD = 75\u221a3<\/h3>\n

Step 3: To find measure of CD, use results obtained in step 1 and step 2.
\nCD = BD \u2013 BC = (75\u221a3 \u2013 75) = 75(\u221a3-1)
\nThe distance between the two ships is 75(\u221a3-1) m. Answer!<\/h3>\n

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60\u00b0. After some time, the angle of elevation reduces to 30\u00b0 (see Fig. 9.13). Find the distance travelled by the balloon during the interval.<\/h2>\n

\"\"<\/p>\n

Solution:
\nLet the initial position of the balloon be A and final position be B.
\nHeight of balloon above the girl height = 88.2 m \u2013 1.2 m = 87 m.
\nTo Find: Distance travelled by the balloon = DE = CE \u2013 CD
\nLet us redesign the given figure as per our convenient<\/h3>\n

\"\"
\nStep 1: In right \u0394BEC,
\ntan 30\u00b0 = BE\/CE
\n1\/\u221a3= 87\/CE
\nCE = 87\u221a3<\/h3>\n

Step 2:
\nIn right \u0394ADC,
\ntan 60\u00b0 = AD\/CD
\n\u221a3= 87\/CD
\nCD = 87\/\u221a3 = 29\u221a3<\/h3>\n

Step 3:
\nDE = CE \u2013 CD = (87\u221a3 \u2013 29\u221a3) = 29\u221a3(3 \u2013 1) = 58\u221a3
\nDistance travelled by the balloon = 58\u221a3 m.<\/h3>\n

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30\u00b0, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60\u00b0. Find the time taken by the car to reach the foot of the tower from this point.<\/h2>\n

Solution:
\nLet AB be the tower.
\nD is the initial and C is the final position of the car respectively.
\nSince man is standing at the top of the tower so, Angles of depression are measured from A.
\nBC is the distance from the foot of the tower to the car.<\/h3>\n

\"\"
\nStep 1: In right \u0394ABC,
\ntan 60\u00b0 = AB\/BC
\n\u221a3 = AB\/BC
\nBC = AB\/\u221a3
\nAB = \u221a3 BC<\/h3>\n

Step 2:
\nIn right \u0394ABD,
\ntan 30\u00b0 = AB\/BD
\n1\/\u221a3 = AB\/BD
\nAB = BD\/\u221a3<\/h3>\n

Step 3: Form step 1 and Step 2, we have
\n\u221a3 BC = BD\/\u221a3 (Since LHS are same, so RHS are also same)
\n3 BC = BD
\n3 BC = BC + CD
\n2BC = CD
\nor BC = CD\/2
\nHere, distance of BC is half of CD. Thus, the time taken is also half.
\nTime taken by car to travel distance CD = 6 sec. Time taken by car to travel BC = 6\/2 = 3 sec.<\/h3>\n

16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.<\/h2>\n

Solution:
\nLet AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As per question,<\/h3>\n

\"\"
\nIn right \u0394ABC,
\ntan x = AB\/BC
\ntan x = AB\/4
\nAB = 4 tan x \u2026 (i)
\nAgain, from right \u0394ABD,
\ntan (90\u00b0-x) = AB\/BD
\ncot x = AB\/9
\nAB = 9 cot x \u2026 (ii)
\nMultiplying equation (i) and (ii)
\nAB2\u00a0= 9 cot x \u00d7 4 tan x
\n\u21d2 AB2\u00a0= 36 (because cot x = 1\/tan x
\n\u21d2 AB = \u00b1 6
\nSince height cannot be negative. Therefore, the height of the tower is 6 m.
\nHence Proved.<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Mathematics NCERT book solutions for Chapter 9 – Some Applications of Trigonometry Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119436","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119436","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=119436"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119436\/revisions"}],"predecessor-version":[{"id":119476,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119436\/revisions\/119476"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119436"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119436"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}