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Height And Distance

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Published in: Mathematics
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This study note will help you to understand about height and distance and their application in trigonometry. this notes has also some important problem based on solved examples

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  1. An Invitation to Science and Mathematics The Institute of Mathematical sciences An application of Trigonometry - Heights and Distances 1 Revision Let us first recall the things we have studied about trigonometry : • Trigonometry is the study of measurements of triangles. • An angle is a portion of the 2-dimensional plane which resides between two different directed line segments. The starting position of the angle is known as the initial side and the ending position of the angle is known as the terminal side. The point from which both of the directed line segments originate is known as the vertex of the angle. p Figure 1: Here Q is the vertex, QR is the initial side and QP is the terminal side. Angle is denoted by ZPQR • A right-angled triangle is a special type of triangle where one of its interior angle is 900. Side opposite to 900 is called the Hypotenuse. • Pythagoras Theorem : In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. 1
  2. Perpendicular sine = Hypotenuse Base cose = Hypotenuse perpendicular Hypotenuse c Base Angles in Degrees sin cos tan csc Perpendicular tan B = cscB = Base Hypotenuse Perpendicular Hypotenuse sece Base Base cote = Figure 2: 00 0 1 0 Not Trigonometric ratios defined 1 sec Not cot defined 300 1 2 €3 2 €3 3 2 2V3 3 €3 450 2 2 1 €2 1 600 €3 1 2V3 3 2 €3 3 perpendicular 900 1 Not defined 1 Not defined Figure 3: Values of ratios for standard angles 2 Application . Now let us see some real life applications of Trigonometry. Trignometry is interesting and has many useful applications in the field of astronomy, geography etc. One of the useful applications of trignometry is to calculate the height of a tower or a peak or the distance of a ship sailing in the sea. we do not need measuring scales. All we need to know is the angle of elevation or angle of depression. 2
  3. Theodolite is an instrument which is used in measuring the angle be- tween an object and the eye of the observer. A theodolite consists of two graduated wheels placed at right angles to each other and a telescope. The wheels are used for the measurement of horizontal and vertical angles. The angle to the desired point is measured by positioning the telescope towards that point. The angle can be read on the telescopic scale. Figure 4: Theodolite 3 Definitions . Let us define a few terminologies so that we can understand and implement the concept of Trigonometry to find the heights and distances. • Line of sight is a straight line from our eye to the object. • If an object is below the horizontal line from the eye, we have to lower our head to view the object. In this process our eyes moves through an angle. This angle is called the angle of depression. The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal line, when the object is below the horizontal line. Similarly, if an object is above the horizontal line from our eyes we have to raise our head to view the object. In this process our eyes move through an angle formed by the line of sight and horizontal line which is called the angle of elevation. Few things to remember before we start solving problems . • Read the statements of the question carefully. • Construct a diagram for the problem for better understanding and visualization of the problem. 3
  4. Figure 5: xo is the angle of elevation and yo is the angle of depression Identify the trigonometrical ratio that Will be problem. useful for solving the Real world problems Will not have standard angles always. So use the below table whenever and wherever necessary. TABLE OF TRIGONOMETRIC RATIOS 40 o_00D 0.01,' 0_035 0-052 o_07D 0-087 o. 122 o. Iag o. 156 0_174 0208 0225 0-259 0.276 o_C2 0-309 0_325 0-342 0-358 n.375 0.40,' o_aa3 o asg 0-485 o_soo 0.515 o.eao 0-545 0±59 0-574 0.616 o.S29 o.E43 o.E-55 o.eg5 0.707 1.000 1.000 0.9gg D.9gg osga D.9gs 0.99' D.9g3 o.ggo D.988 D.97e 0.974 0.970 0.966 D.961 0.95S D.951 D.940 D.934 D.927 0.921 D.914 o.gos o.ggg D.ggl o.gag D.875 D.8S7 o.e4e D.83g o.gag D.819 0.7ge D.78e 0.777 o. 755 D.743 0.731 D.719 D.707 0 000 0017 0035 0052 0070 0087 0.105 0123 0158 0176 021a 0,231 0,249 0,268 0287 030s 0325 0,334 0384 O, 445 0,183 0,510 0532 os77 0,601 o S75 0,700 0727 0754 0,781 0,810 0.8-29 0900 osga 0,966 1.000 65 69 75 0.707 0.71 g 0.731 0.74? 0755 n.766 0.77,' 0,788 0.7gg n.809 0.81 g 0.82g 0839 0848 o.S57 0.875 0.891 0.914 n.921 0.92,' n.934 0846 0.951 0±56 0±70 0.974 0,978 o.gs2 o„g85 n.988 osga o.gas o.ws o.ws cl.geg o.ggg 1.0D0 707 o, 695 D 682 o.B69 0.64-3 0616 o. 588 0,545 c. sao 0,515 c. soo o„i3E 0,407 o.3g1 0,375 o. 358 D 3112 0 326 0309 0276 0,242 o. 225 o. 208 Digl 0,174 0.122 0,105 o. 087 0,070 o, DS2 035 0,017 o. DOO noo .036 1.111 1.235 .280 227 .376 42a 433 1 .soo 1.ss4 1.8D4 1.gB3 2.050 2.145 2.246 2.356 2.475 2.605 2.747 3.078 3.271 3.487 3.752 4.011 4.705 5.145 5.671 6.314 8.144 9.514 11.430 14.301 19.081 57.290 Figure 6: Ratio table
  5. 4 Solved Problems 1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 300 and 450 respectively. If the lighthouse is 100m high. What is the distance between the two ships? Solution : Let AB be the lighthouse and C and D be the positions of the ships. 300 c Here, we need to find CD 450 D AB — 100m, ZACB = 300andZADB = 450 1 — tan 300 —AB x 100 3m Now, — tan 450 — (100Vä + 100)m — (100 x 2.73)m — 273m 5
  6. 2. The angle of elevation of an aeroplane from a point A on the ground is 600 . After a flight of 15 seconds horizontally, the angle of elevation changes to 300 If the aeroplane is flying at a speed of 200m/ s, then find the constant height at which the aeroplane is flying. Solution : Let A be the point of observation. Let E and D be positions of the aeroplane initially and after 15 seconds respectively. Let BE and CD denote the constant height at which the aeroplane is flying. Given that Z DAC — 300 and ZEAB = 600. 300 60 x Let BE = CD = h metres. Let AB = x metres. The distance covered in 15 seconds, ED — Thus, BC — 3000m In right-angled A ADC tan 300 CD = x tan 300 Thus, h — (x + 3000) x In right-angled AAEB tan 600 BE AB — AB x tan 600 Thus, From (1) and (2), we have 1 — (x + 3000) x 6 — 200 x 15 — 3000m 1 (1) (2)
  7. a; + 3000 Thus, from (2), h = 1500v/ä m 5 Practice problems 1. The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be 300 and 600 If the distance between the earth stations is 4000km, find the distance between the satellite and earth. (O — 1.732) 2. From the top and foot of a 40 m high tower, the angles of elevation of the top of a lighthouse are found to be 300 and 600 respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower. 3. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 300 with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree. 4. A simple pendulum of length 40 cm subtends 600 at the vertex in one full oscillation. What will be the shortest distance between the initial position and the final position of the bob? (between the extreme ends) 5. Two crows A and B are sitting at a height of 15 m and 10 m in two different trees vertically opposite to each other. They view a vadai (an eatable) on the ground at an angle of depression 45c and 60crespectively. They start at the same time and fly at the same speed along the shortest path to pick up the vadai. Which bird will succeed in it? 7

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