MATHEMATICS FOR ENGINEERING TRIGONOMETRY

Transcription

1 MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL SOME MORE RULES OF TRIGONOMETRY This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals. The tutorial contains the following. Revise the Radian Revise sine, cosine and tangent Two Dimensional Coordinate systems Cartesian Polar Sine Rule Cosine Rule The area of a triangle D.J.Dunn 1

2 1. RADIAN In Engineering and Science, we use the radian to measure angle as well as degrees. This is defined as the angle created by placing a line of length 1 radius around the edge of the circle as shown. In mathematical words it is the angle subtended by an arc of length one radius. This angle is called the RADIAN. The circumference of a circle is πr. It follows that the number of radians that make a complete circle is π R or π. R There are π radians in one revolution so 360 o =π radian 1 radian = 360/π = o. TWO DIMENSIONAL COORDINATE SYSTEMS CARTESIAN In a two dimensional system the vertical direction is usually y (positive up) and the horizontal is direction is x (positive to the right). Other letters may be used to designate an axis and they don t have to be vertical and horizontal. The origin o is where the axis cross at x = 0 and y = 0 A point p on this plane has coordinates x, y and this is usually written as p (x,y) POLAR If a line is drawn from the origin to point p it is a radius R and forms an angle θ with the x axis. The angle is positive measured from the x axis in a counter clockwise direction. A vector with polar coordinates is denoted R θ CONVERSION The two systems are clearly linked as we can convert from one to the other using trigonometry and Pythagoras theorem. y = R sin θ x = R cos θ y/x = tan θ R = (x + y ) ½ WORKED EXAMPLE No. 1 The x, y coordinates of a point is 4, and 6. Calculate the polar coordinates. R = (4 + 6 ) 1/ = 7.11 θ = tan -1 (6/4) = o D.J.Dunn

3 SELF ASSESSMENT EXERCISE No Convert 60 o to radian. (1.047 rad). Convert π/6 radian into degrees. (30 o ) 3. The x, y Cartesian coordinates of a vector are and 7.Express the vector in Polar coordinates. ( o ) 4. A vector with Cartesian coordinates 10, 0 is added to a vector with coordinates -0,10. What are the polar coordinates of the resulting vector? ( o ) 3. TRIGONOMETRIC RATIOS The ratios of the lengths of the sides of a right angle triangle are always the same for any given angle θ. These ratios are very important because they allow us to calculate lots of things to do with triangles. In the following the notation above is used with the corners denoted AB and C SINE The ratio Opposite Hypotenuse CB = is called the sine of the angle A. (note we usually drop the e on sine) AB Before the use of calculators, the values of the sine of angles were placed in tables but all you have to do is enter the angle into your calculator and press the button shown as sin. For example if you enter 60 into your calculator in degree mode and press sin you get If you enter 0. into your calculator in radian mode and press sin you get Note that sin(θ) = -sin (-θ) and sin (180 θ) = sin(θ) COSINE Adjacent AC The ratio = is called the cosine of the angle A. Hypotenuse AB On your calculator the button is labelled cos. For example enter 60 into your calculator in degree mode and press the cos button. You should obtain 0.5 If you enter 0. into your calculator in radian mode and press cos you get Note that cos(θ) = cos (-θ) and cos(180 θ) = -cos(θ) TANGENT The ratio Opposite BC = is called the tangent of the angle A. Adjacent AC On your calculator the button is labelled tan. For example enter 60 into your calculator in degree mode and press the tan button and you should obtain If you enter 0. into your calculator in radian mode and press tan you get 0.07 Note that sin(θ)/cos(θ) = tan(θ) and tan(90 θ) = 1/tan(θ) D.J.Dunn 3

4 4. SOLVING ANY TRIANGLE The following work enables us to solve triangles other than right angles triangles. SINE RULE Consider the diagram. h = b sin A = a sin B a b It follows that = sina sinb If we did the same for another perpendicular to side b or a we could show that a sina = b sinb = c sinc WORKED EXAMPLE No. Find the length of the two unknown side in the triangle shown. a = 50 mm A = 30 o B = 45 o C = 180 o - 30 o - 45 o = 105 o a b c 50 b c = = = = sina sinb sinc sin(30) sin(45) sin(105) 50 sin(45) 50 sin(105) b = = mm c = = mm sin (30) sin (30) WORKED EXAMPLE No. 3 A weight of 300 N is suspended on two ropes as shown. Calculate the length of the ropes Draw the vector diagram for the three forces in equilibrium. Calculate the forces in the ropes. The third internal angle is 110 o. 4/sin 110 = L 1 /sin 50 = L /sin 0 L 1 = 3.61 m and L = m Next draw the triangle of forces as shown. F 1 /sin 40 o = 300/sin 70 o F 1 = 05. N F /sin 70 o = 300/sin 70 o F 1 = 300 N D.J.Dunn 4

5 COSINE RULE Consider the diagram. Using Pythagoras we have: h = a (b x) and h = c x a (b + x bx) = c x a b - x + bx) = c x a = b + x bx + c x a = b + c bx substitute x = c cos(a) a = b + c bc cos(a) = c b + c a cos(a) = bc If we repeated the process with h drawn normal to the other sides we could show that : c + a b a + b c cos(b) = cos(c) = ca ab You can see a pattern for remembering the formulae. This is a useful formula for solving a triangle with three known sides or two known sides and the angle opposite the unknown side. WORKED EXAMPLE No. 4 Find the length of the unknown side in the triangle shown. Find the other internal angles. a + b c cos(c) = ab cos(60 o )()(60)(70) = = 8500 c c = 4300 c = mm o c cos(60 ) = ()(60(70) + 70 c b + c a cos(a) = = bc ()(70)(65.574) A = 5.4 o B = = 16.5 o = 0.61 WORKED EXAMPLE No. 5 Find the resultant of the two forces shown. The addition of the two force is done as shown. o R cos(135 ) = ()(100(100) o cos(135 )()(100)(100) = = 0000 R R = R = N R D.J.Dunn 5

6 SELF ASSESSMENT EXERCISE No. 1. Find the resultant of the two forces shown. (Answer 38.8 N). Vector A has polar coordinates 1 60 o and vector B has polar coordinates 5 0 Find the resultant in polar form. ( o ) 3. The diagram shows a weight suspended from two ropes. Calculate the angles of the ropes to the horizontal support. o (Answers 49 o and 59 o ) 4. A weight of 4 Tonne is suspended on two ropes as shown. Calculate the length of the ropes and the forces in them. (Answers m, m, 3.17 T and.98 T) D.J.Dunn 6

7 5. AREAS OF TRIANGLES The diagram shows that any triangle has half the area of a rectangle of dimensions B h where B is the base of the triangle and h is the perpendicular height It follows that the area of a triangle is A = Bh/ SELF ASSESSMENT EXERCISE No.3 1. A cone has a base diameter of 60 mm and a vertical height of 60 mm. Calculate the area of a vertical cross section through the apex. (Answer 1800 mm ). Calculate the area of the triangle shown. (Answer 3591 mm ) 3. Calculate the area of the triangle shown. (You should use the cosine rule to find one of the angles) (Answer 38 mm ) D.J.Dunn 7