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Dr. Mark Sullivan Analytical methods Part II BASIC TRIGONOMETRY Imagine two sticks, the bottom one being 1 meter long. As you open them up, two things happen; 1) The angle they make at the join will increase. Height 2) The height measured from the top stick to the end of the meter long base one increases. Angle There is a direct correlation between the angle and the height. A table of the heights is below. 1m Question: A) If the bottom stick was 2 meters long, how would the table opposite change? Answer; Angle Height 0 10 20 30 40 50 60 70 80 0.00 m 0.18 m 0.36 m 0.58 m 0.84 m 1.19 m 1.73 m 2.75 m 5.67 m From your answer to the last question you can see that we only need to multiply the height that we get from the table above by the length of the base (or distance to the object) to find the height of something. This saves a lot of shinning up towers! Height Base Height derived from angle Oaklands college Page 1 of 29 Dr. Mark Sullivan Analytical methods The Height of the tree is found by; h = d tan (This last symbol is a Greek letter pronounced Theeta.) SAQ Set A Set B A tower block is 125 m distant. The angle of elevation of the top of the tower is 38. How high is the tower? A tower block is 143 m distant. The angle of elevation of the top of the tower is 32. How high is the tower? A flag pole is 35 m distant. The angle of elevation of the top of the tower is 48. How high is the tower? A flag pole is 43 m distant. The angle of elevation of the top of the tower is 42. How high is the tower? Oaklands college Page 2 of 29 Dr. Mark Sullivan Analytical methods A Statue is 224 m distant. The angle of elevation of the top of the tower is 18. How high is the tower? A Statue is 231 m distant. The angle of elevation of the top of the tower is 20. How high is the tower? A Hilltop is 1.45 km distant. The angle of elevation of the top of the tower is 8. How high is the tower? A Hilltop is 16.8 km distant. The angle of elevation of the top of the tower is 3. How high is the tower? A Monument is 35 m distant. The angle of elevation of the top of the tower is 58. How high is the tower? A Monument is 83 m distant. The angle of elevation of the top of the tower is 42. How high is the tower? Formulae like Height Base Height derived from angle can be swapped around as long as you remember one golden rule; if you move something to the other side of the = sign, it must do the opposite job. For example, take the quantity ‘Base’ in the above formula. Its job, so to speak, is to multiply the ‘Height derived from angle’ quantity. When we move it to the other side it must do the opposite of multiply which is divide. So re-arranging gives; Height Base Height derived from angle This method of dividing the height (or opposite) by the base (or adjacent) is called ‘taking the tangent’ and is very useful in surveying techniques. Oaklands college Page 3 of 29 Dr. Mark Sullivan Analytical methods For any right-angled triangle, i.e. not one necessarily sitting on the ground, we can it label like this; In this case the tangent (tan) can be written as; Hypotenuse Opposite the angle tan Opp Adj Angle Adjacent to the angle This shouldn’t be too much of a surprise, after all we derived the tangent from the vertical measurement of height above a horizontal 1 m stick. In other words; the Opposite and the Adjacent. With similar logic, two other arrangements can be written. These are Sine and Cosine or sin and cos. Pronounced ‘Sign’ and ‘Koz’ sin Opp Hyp cos Adj Hyp Imagine our two sticks again. As you open them up, two things happen; Slope 1) The angle they make at the join will increase. 2) The distance measured from the top of the bob back along the top stick to the angle increases. There is a direct correlation between the angle and the slope. A table of the slope lengths is below. Table of the Sine of angles Oaklands college Angle 1m Angle 0 10 20 30 40 50 60 70 80 Height 0.00 m 0.17 m 0.34 m 0.50 m 0.64 m 0.77 m 0.87 m 0.94 m 0.98 m Page 4 of 29 Dr. Mark Sullivan Analytical methods Write down the following (first bit is already done) .3 5 4 Sin = = 3 5 0.6 Find the angle which gives this value _____________________________ Cos = ____ = Find the angle which gives this value Tan = = ____ Find the angle which gives this value _____________________________ _____________________________ SAQ Find the angles marked below, the triangles being right-angled. Oaklands college Page 5 of 29 Dr. Mark Sullivan Analytical methods 5 4 3.5 1.8 51.58 42.36 In ABC, C = 90, B = 23.3 and AC = 11.2 cm. Find AB. In ABC, B = 90, A = 67.5 and AC = 0.86 m. Find BC. Oaklands college Page 6 of 29 Dr. Mark Sullivan Analytical methods SINE AND COSINE RULES The Sin, Cos and Tan functions only work for right-angled triangles. This isn’t too helpful for surveying as land seldom has nice tidy right-angles in it. Much more useful for us are two rules which apply to any triangle weather it has a right-angle or not. They’re very imaginatively called the Sine rule and the Cosine rule. a b c sin A sin B sin C Sine rule 1) Where to use it Use when you have a) A side and its opposite angle b) Any other side or angle 2) How to use it Write down a fraction with a known side on the top and the sin of the known angle opposite that side on the bottom. Write down an ‘=’ sign and then another fraction with a side and opposite angle as you did in the first step except this time one must be known and the other must be what you’re trying to find. Rearrange the equation so that the quantity you want to find is isolated and work out its value. Oaklands college Page 7 of 29 Dr. Mark Sullivan Analytical methods Worked example; b Next step; Write down a fraction with a known side on the top and the sin of the known angle opposite that side on the bottom. a 30 40 10 sin 110 10m The span of a roof is 10 m with c pitches of 30 and 40. Find the length of the slope a. Next, write down an ‘=’ sign and then First step; Do we have two angles? Yes, then find the third. No, then skip the step. must be what you’re trying to find; To find the missing angle use 'sum of angles'. another fraction with a side and opposite angle as you did in the first step except this time one must be known and the other 10 a sin 110 sin 30 Finally, rearrange the equation so that the quantity you want to find is isolated a 30 + 40 + C = 180 C = 180 - (30 + 40) 10 sin 30 sin 110 a C = 110 10 0.5 0.9397 = 5.231 m = 5.230 m (to the nearest mm) . Try these (SAQ) Set A Set B The span of a roof is 12 m with pitches of 35 and 45. Find the length of the slope a. The span of a roof is 14 m with pitches of 33 and 47. Find the length of the slope a. Oaklands college Page 8 of 29 Dr. Mark Sullivan Analytical methods The span of a roof is 8 m with pitches of 20 and 50. Find the length of the slope a. The span of a roof is 9 m with pitches of 25 and 40. Find the length of the slope a. In triangle ABC, a = 15cm, b = 25cm, and angle A = 47. What is angle B (angle opposite side b). In triangle ABC, a = 12cm, b = 20cm, and angle A = 35. What is angle B (angle opposite side b). Oaklands college Page 9 of 29 Dr. Mark Sullivan Analytical methods c 2 a 2 b 2 2ab cos c Cosine rule 3) Where to use it Use when you have a) Two sides and the angle in-between 4) How to use it This can be used in two forms; if you want to find the length of a side then this formula is just a matter of plugging numbers into the calculator. Here is a re-wording of the Cosine rule that might be more useful. Side First side 2 Second side 2 2 First side Second side Cos of angle between if, however, you want to find an angle, then this version of the Cosine rule is the one to use; a2 b2 c2 Angle Cos 1 2 ab Worked example C 25 100m A Using the cosine rule: 130m B Two points a and b lie on opposite sides of a gully. A point c is 100 m from a and 130 m from b and angle C = 25. How far horizontally is a from b? Oaklands college c2 = a2 + b2 - 2 a x b x cos C c2 = 1002 + 1302 - 2 x 100 x 130 x cos 25 c = 57,76 m Page 10 of 29 Dr. Mark Sullivan Analytical methods Try these (SAQ) Set A Set B Two points a and b lie on opposite sides of a gully. A point c is 80 m from a and 90 m from b and angle C = 35. How far is a from b? Two points a and b lie on opposite sides of a gully. A point c is 70 m from a and 100 m from b and angle C = 45. How far is a from b? Two points a and b lie on opposite sides of a road. A point c is 180 m from a and 130 m from b and angle C = 46. How far is a from b? Two points a and b lie on opposite sides of a road. A point c is 140 m from a and 110 m from b and angle C = 54. How far is a from b? Two points a and b lie on opposite sides of a chip shop. A point c is 18.345 m from a and 13.042 m from b and angle C = 46 34’ 40”. How far is a from b? Two points a and b lie on opposite sides of a chip shop. A point c is 12.382 m from a and 10.012 m from b and angle C = 36 44’ 50”. How far is a from b? Oaklands college Page 11 of 29 Dr. Mark Sullivan Analytical methods Sine and Cosine rules In triangle ABC, angle A = 55, a = 7.2, and c = 6. Find the angles B and C and the length of the side b. In triangle ABC, angle A = 40, a = 6.2, and c = 5. Find the angles B and C and the length of the side b. In triangle ABC, c = 8, angle A = 30, and angle B = 42. Find the length of the sides b and a. In triangle ABC, c = 12, angle A = 40, and angle B = 52. Find the length of the sides b and a. Two points of interest; c & d are on opposite sides of a wood with no line of sight between them. A surveyor takes measurements at a & b and draws the following sketch (not to scale). Find the distance between c & d. a 15 m 57 58 b 48 61 c d Oaklands college Page 12 of 29 Dr. Mark Sullivan Analytical methods ERRORS. In all applications, numbers come in two parts. One is the value, the other is its accuracy or more correctly its probable error. With a standard ruler, Could you measure something accurately to the nearest centimetre? What about to the nearest millimetre? To the nearest tenth of a millimetre? Almost certainly. Very probably. No chance! A ruler is only accurate to about 1 mm, possibly 0.5 mm if you use it carefully but beyond that, confidence is lost. Correctly then, all measurements should be recorded with an appropriate error estimate, for example; 105 mm 0.5 mm With a tape measure on-site, do you think you would be able to measure a distance of about 100 m to an accuracy of 1 mm ? 1 cm? 1.5 cm? 1 m? SAQ Try this; A tower block is 625 m distant. The angle of elevation of the top of the tower is 18 34’ 10”. How high is the tower? Oaklands college Page 13 of 29 Dr. Mark Sullivan Analytical methods Your collegue does the same exercise at the same distace and finds an angle of elevation greater than yours. What is the difference in your calculated heights if his measurments are a) 1 larger b) 1’ larger c) 10” larger d) 1” larger What would be an acceptable level of error and what accurassy will you have to achieve with your measurments to be of professional standard? (One or two sentences justifying your value(s)) Oaklands college Page 14 of 29 Dr. Mark Sullivan Analytical methods SIX - FIGURE GRID REFERENCES A six figure grid reference is normally the one you would choose to use to locate a point (say a building). To quote a six figure reference, first locate the square that you are interested in. Then imagine dividing that square into 10 equally spaced intervals along and up the square. This would form an imaginary grid of 100 squares within the grid square. Then quote a further reference using these numbers. For example, an object in the exact centre of a square would have a reference along the lines of ‘something something 5 something something 5’. The first two numbers would be the eastings of the grid square, the second two 'somethings' would be the northings of the grid square. The 5 and 5 are there because it would be 5 in and 5 up within the blue grid square (i.e. in the centre). For ease of use, when we write this, we write all the eastings together, then all the northings together. Have a look at the example below and see if it helps. Here we are trying to find the six figure grid reference of the church with a tower in Bishop's Tachbrook. The large cross represent a distance of 5 tenths of the way into the square (in other words halfway) horizontally and vertically. Eastings First (Vertical Lines) – between 31 & 32 so 31something. Estimate (or measure for accuracy) the number of tenths of a square it is along. Think of it as 31.4 across Northings Second (Horizontal Lines) – between 61 & 61 so 62something. Estimate (or measure for accuracy) the number of tenths of a square it is up. Think of it as 61.4 up The reference is 314614. Further examples are: Mill Mound 314617 New House Fm 304621 Make sure you can locate these on the map. Oaklands college Page 15 of 29 Dr. Mark Sullivan Analytical methods NAVIGATING FROM ONE POINT TO ANOTHER To find the distance between two points for which you have map references we need to reacquaint ourselves with good old Pythagoras. You remember him, the ancient Greek, about 500BC who did stuff with triangles. Pythagoras said that if you square the small sides of a right-angled triangle, their sum total will be eqal to the square of the long side; c a c2 = a2 + b2 b Using the basic rule of algebra, you can swap things about as long as they do the opposite job on the other side of the equals, I can move the squaring 2 from the a to the other side where it does the opposite job and becomes a root. c a 2 b2 Now consider our triangle on a grid map with its points at positions A & B. 58 If we want to know the distance between point A & B its like finding the hypotenuse of a right-angled triangle. 57 All we need are the distances East-West between them and North-South. B N 56 55 21 A 22 23 24 25 Consider points A (220560) and B (250570) The first 3 digits refer to the didtance East-West and the last 3 to the distance North-South of each point from a datum point. The difference between them is the length of the base and height of our triangle. Oaklands college Page 16 of 29 Dr. Mark Sullivan Analytical methods E-W N-S B= 250 570 A= 220 560 30 10 Difference = Now using Pythagoras’ formula in the form c a 2 b 2 we can calculate the length of c, the distance between A & B. Using the calculator; ( 30 x2 + 10 x2 ) =. . 31.6227766 Remember that the map references were given to a tenth of a 1 km square accuracy, which is an accuracy of a tenth of a km. If that’s what the references were measured to, then that is the units of the answer. To swap from tenths of a km to a more recognisable unit say km, then divide by ten (move the decimal one place to the left). . 3.16227766 km. Presented here in km, the first number after the decimal is 1 and represents tenths of a km. The next number after the decimal, 6, represents hundredths of a km but the references were given only to a tenth of a km. Any numbers after the first, tenths, place are therefore irrelevant The distance from A to B is = 3.1 km 100 m SAQ Find the distance between the points used on the last map; Mill Mound 314617 New House Fm 304621 Oaklands college Page 17 of 29 Dr. Mark Sullivan Analytical methods Baring of B from A Consider again our triangle on a grid map with its points at positions A & B (reproduced here). If we want to know the baring of point B from A we can use the tangent function. We know the distances EastWest between them and NorthSouth. tan 58 B 57 N 56 A 55 North South East West 21 22 23 24 25 Again using the rule of algebra, I will move the tan function to the other side where it swaps to anti-tan or inverse tan; North South East West tan 1 in the case of the example in the diagram, we already know the E-W & N-S distances so we can use the calculator again; SHIFT tan ( 10 30 ) =. . 18.43494882 this is the value of the angle in decimal degrees. To swap this to degrees, minutes and seconds; press; SHIFT ‘ “ . 18 26’ 5.82” The final step is to convert this angle to one reading from North. This is angle in the diagram (pronounced Theeta). We know that it’s a right-angle between West and North i.e. 90 so we can just subtract from 90 to find the whole circle baring. = 90 - = 90 - 18 26’ 5.82” = 71 33’ 54.18” Oaklands college Page 18 of 29 Dr. Mark Sullivan Analytical methods SAQ Set A Set B Find the distance and baring of point B from point A for the following; Find the distance and baring of point Z from point Y for the following; A = 258369 B = 266381 Y = 258369 Z = 366381 A = 369478 B = 464481 Y = 269255 Z = 465281 A = 470585 B = 567681 Y = 270364 Z = 364387 A = 581692 B = 661381 Y = 281471 Z = 299381 A = 681392 B = 661381 Y = 381471 Z = 299381 Oaklands college Page 19 of 29 Dr. Mark Sullivan Analytical methods Finding a position from a baring Point B lies 2.8 km on a baring of 030 from an observer located at 220560. what is the position of point B? Now consider our triangle on the grid map again with its points at positions A & B. 58 we know the distance between point A & B is 2.8 km in this example. 57 We also know that angle is 030 (whole circle barings are always quoted with three digits even if the first is a zero) 56 B N A 55 21 If = 030 or 30 then angle = 90 - 30 = 60 22 23 24 25 Northings; Eastings; The Northing or height of the triangle is opposite the angle that is known. The Easting or base of the triangle is adjacent to the angle that is known. The distance between the points A & B is the length of the hypotenuse. The distance between the points A & B is the length of the hypotenuse. The sine function is defined as being equal to the opposite side divided by the hypotenuse; The cosine function is defined as being equal to the adjacent side divided by the hypotenuse; sin opposite Hypotenuse sin Rearranging this we get; adjacent Hypotenuse Rearranging this we get; Opposite = Hypotenuse sin Adjacent = Hypotenuse cos = 2.8 km sin 60 = 2.8 km cos 60 = 2.42 km = 1.4 km = 2.4 km 0.1 km = 1.4 km 0.1 km Now add on Northings and Eastings; Position of A = 220560 This is; 220 plus +2.4 +1.4 Position of B; 244 Oaklands college 560 574 244574 Page 20 of 29 Dr. Mark Sullivan Analytical methods SAQ Set A Set B Find the position of point B given its baring and distance from A; Find the position of point Z given its baring and distance from Y; 2.3 km baring 030 2.1 km baring 020 12.4 km baring 040 2.9 km baring 030 24.7 km baring 056 24’ 7.5 km baring 075 43’ 3.1 km baring 189 4.7 km baring 230 5.6 km baring 280 62.4 km baring 330 Oaklands college Page 21 of 29 Dr. Mark Sullivan Analytical methods VECTORS Breaking a direction down into North and East directions is called ‘resolving into vectors’ and can be used for a whole load more than just working out where you are. Before we can proceed further we need to explore the idea of a vector and for that matter scalars also. All quantities, it turns out, fall into one of two categories; scalars or vectors. A scalar has only its numeric value for example, 1kg of sugar, 3 mins 20 secs delay or 34 C. A vector, on the other hand, is a quantity which expresses both magnitude and direction for example, a velocity of 45 km/s in a Northerly direction, a force of 300 kN at 45 to the vertical or a fresh Easterly breeze. Graphically we represent a vector as an arrow. In typeset notation a vector is represented by a boldface character, while in handwriting an arrow is drawn over the character representing the vector. Ax = A sin Ay = A cos A The angle θ represents the whole circle orientation of a two dimensional vector which determines the quantities Ax and Ay through the cos and sin functions. The quantities Ax, & Ay, represent the Cartesian components of the vectors in the figure opposite. A vector can be represented either by its Cartesian components, which are just the projections of the vector onto the map grid, or by its direction and magnitude (polar co-ordinates). The direction of a vector in two dimensions is best represented by the clockwise angle of the vector relative to North or the y axis. Use this space to derive expressions for the Cartesian quantities Ax and Ay from the polar quantities A and . Oaklands college Page 22 of 29 Dr. Mark Sullivan Analytical methods Method of employing trigonometric functions to determine components of a vector: 1. Draw a sketch (no scale needed) of the vector in the indicated direction; label its magnitude and the angle which it makes with the horizontal. 2. draw a rectangle about the vector so that the vector is the diagonal of the rectangle; beginning at the tail of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the head of the vector; the sketched lines will meet to form a parallelogram. 3. Draw the components of the vector; the components are the sides of the rectangle; be sure to place arrowheads on these components to indicate their direction (up, down, left, right). 4. Meaningfully label the components of the vectors with symbols to indicate which component is being represented by which side. 5. Determine the length of the side opposite the indicated angle using trig functions; substitute the magnitude of the vector for the length of the hypotenuse; use some algebra to solve the equation for the length of the side opposite the indicated angle. 6. Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle. The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tensional force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions. If we use vectors to represent force instead of movement across the ground, we can use them to work out stresses on a structure. Consider the following example; Oaklands college Page 23 of 29 Dr. Mark Sullivan Analytical methods 1. A 3m uniform beam of mass 5 kg/m is supported horizontally above the ground by a hanger at one end attached to a wall and by a rope at the other. The other end of the rope is fixed to the wall vertically above the beam. a) Derive an expression for the tension in the rope. b) Calculate the tension for a rope fixed 60 cm, 1m, 2m and 3m above the beam. c) Comment on how the tension varies with height ____________________________________________ First let’s draw a sketch of the structure. T cos T T sin ½Mg ½ Mg The tension in the rope, T, is shown in its component form as a vertical and a horizontal force T sin and T cos . The load is the weight of the beam whose mass, M, is acted upon by gravity resulting in a downward force of Mg (mass times the force due to gravity, g). the beam is, however, supported at both ends so this force is supported equally at each end; ½Mg. It is clear from the diagram that the upward force due to the tension in the rope must be exactly equal to the downward force produced by gravity acting on the beam; ½Mg = T sin The question asks for an expression for tension, T, so we must re-arrange the equation for T; 1 Mg 2 T sin By convention we write it the other way round; Oaklands college Mg T 2 sin T Mg 2 sin Page 24 of 29 Dr. Mark Sullivan Analytical methods b) Calculate the tension for a rope fixed 60 cm, 1m, 2m and 3m above the beam. Look first at the expression we just worked out; T Mg and identify which bits the question 2 sin is asking us to do repeatedly. In this case the only bit that will change as we put the different lengths in will be , the angle of the rope. All other parts we only have to work out once. T Mg 2 sin where M = 5 kg/m 3 m = 5 kg m kg 3 m 15 15 kg . m m G = 9.81 m/s2 So Mg = 15 kg 9.81 m/s2 = 15 kg 9.81 kg m m 147.15 2 2 s s kg m/s2 is the fundamental units of force. We call these units a Newton, N. Now do the working out for the different heights given. This is best done in a table; Height Length Height = tan Length 60 cm 3m 0.6/3 = 0.2 1m 3m 1/3 2m 3m 3m 3m sin Mg/(2 sin ) 11 18’ 35.76” 0.1961 375.19 N = 0.333’ 18 26’ 5.82” 0.3162 232.69 N 2/3 = 0.666’ 33 41’ 24.24” 0.5547 132.64 N 3/3 = 1.0 45 0.7071 104.05 N The tensions for the different height fixings are; Oaklands college 60 cm 375.19 N 1m 232.69 N 2m 132.64 N 3m 104.05 N Page 25 of 29 Dr. Mark Sullivan Analytical methods c) Comment on how the tension varies with height In order to say something about the results obtained in part b, it would be useful to ‘see’ them The clue is in the question, we were asked to find tensions for different heights so plot height versus tension. The heights do not all go in regular steps so a scatter plot is the right one here. On an x-y axis we plot the values of height along the x-axis and tension up the y-axis Plot of rope tension versus height of fixing. 400 Tension in rope /N 375.160361 300 232.664579 200 132.639218 104.050763 100 0 0 1 2 3 4 Height of fixing /m notice that the line connecting the points is curved. This indicates that at small values a small change in fixing height leads to a rapid change of tension. At large heights, the tension is low and not very sensitive to change. SAQ Try some for yourselves; 1. A 5 m uniform beam of mass 12 kg/m is supported horizontally above the ground by a hanger at one end attached to a wall and by a rope at the other. The other end of the rope is fixed to the wall vertically above the beam. a) Derive an expression for the tension in the rope. b) Calculate the tension for the rope fixed 60 cm, 1m, 2m and 3m above the beam. c) Comment on how the tension varies with height Set A Set B Your beam is 8 m long & has a mass of 12 kg/m. Your beam is 6 m long & has a mass of 18 kg/m. Oaklands college Page 26 of 29 Dr. Mark Sullivan Analytical methods Answer;- Oaklands college Page 27 of 29 Dr. Mark Sullivan Analytical methods 2. A rope is laid across two parallel rails. Three equal masses are suspended, one at either end and one in the middle between the two rails. Calculate the angle that the string makes with the vertical between the central mass and one of the rails. Answer;- Oaklands college Page 28 of 29 Dr. Mark Sullivan Analytical methods 3. A telegraph pole has three wires attached to it. Their bearings from the pole are 315, 270 and 210 and the forces in the wires are 1000N, 1100N and 1050N respectfully. Calculate the bearing a support stay should be fixed at in order that there are no unbalanced lateral forces acting on the pole. Answer;- Oaklands college Page 29 of 29