Oblique Triangles and Vectors 15 Oblique Triangles and Vectors Click on the computer image at the bottom right for a direct web link to an interesting Wikipedia Math Site. Prepared by: Richard Mitchell Humber College

15.1 - Trigonometric Functions of Any Angle

15.1-DEFINITIONS Definition of the Trigonometric Functions 15.1-Trigonometric Functions of Any Angle. From any point P on the terminal side of the angle, we draw a perpendicular to the x axis forming a right triangle with legs x and y and with a hypotenuse r. The trigonometric ratios are generated as follows.

15.1-DEFINITIONS Definition of the Trigonometric Functions 15.1-Trigonometric Functions of Any Angle.

15.1-DEFINITIONS Definition of the Trigonometric Functions 15.1-Trigonometric Functions of Any Angle.

15.1-DEFINITIONS Definition of the Trigonometric Functions blank 15.1-Trigonometric Functions of Any Angle. blank

S+ A+ T+ C+ 15.1-DEFINITIONS CAST Rule blank 15.1-Trigonometric Functions of Any Angle. blank

15.1-EXAMPLE 1 A point on the terminal side of angle θ has the coordinates (-3, -5). Write the six trigonometric functions of θ to 3 significant digits. S+ A+ θ 15.1-Trigonometric Functions of Any Angle. T+ C+ (-3, -5) Terminal Side From the point P (-3, -5) on the terminal side of the angle, we draw a perpendicular to the x axis forming a right triangle with legs x = -3 and y = -5 and with a hypotenuse r = +5.83 as found by using the Pythagorean Theorem. The six trigonometric ratios are generated as follows.

15.1-EXAMPLES 3 to 6 Use your calculator to verify the following trigonometric ratios. S+ A+ T+ C+ 15.1-Trigonometric Functions of Any Angle. NOTE: There are infinitely many angles that have the exact same value of a trigonometric function. Usually, we only need two positive angles less than 3600 that have the required trigonometric function.

15.1-EXAMPLE 8 Evaluate the following expression to four significant digits. Note: Use DEGREE MODE 15.1-Trigonometric Functions of Any Angle.

15.1-DEFINITIONS Reference Angles (Examples 9 to 12) S+ A+ blank T+ C+ For an angle in standard position on the coordinate axis, the acute angle that its terminal side makes with the x axis is called the reference angle θ’. 15.1-Trigonometric Functions of Any Angle. blank

15.1-EXAMPLE 13 If sin θ = +0.6293, find two positive values of θ less than 3600 Work to the nearest tenth of a degree. S+ A+ T+ C+ NOTE: According to the CAST Rule, sine is positive (+0.6293) in the first and second quadrants. Thus, our reference angle of 39.00 is also found in the first and second quadrants. 15.1-Trigonometric Functions of Any Angle. NOTE: Since our reference angle of 39.00 is found in the first and second quadrants our two positive values of θ will also be found here. Thus our values are 39.00 and 141.00.

15.1-EXAMPLE 14 Find, to the nearest tenth of a degree, the two positive angles less than 3600 that have a tangent of -2.25. S+ A+ T+ C+ NOTE: According to the CAST Rule, tan is negative (-2,25) in the second and fourth quadrants. Thus, our reference angle of 66.00 is also found in the second and fourth quadrants. 15.1-Trigonometric Functions of Any Angle. NOTE: Since our reference angle of 66.00 is found in the second and fourth quadrants our two positive values of θ will also be found here. Thus our values are 114.00 and 294.00.

15.1-DEFINITIONS Special Angles 15.1-Trigonometric Functions of Any Angle. The angles 0°, 90°, 180°, and 360° are called quadrantal angles because the terminal side of each of them lies along one of the coordinate axes. Notice that the tangent is undefined for 90° and 270°, angles whose terminal side is on the y axis. The reason is that for any point (x, y) on the y axis, the value of x is zero. Since the tangent is equal to y/x, we have division by zero, which is not defined.

15.2 - Law of Sines 15.2-Law of Sines.

15.2-EXAMPLE 18 Solve triangle ABC where A = 32.50, B = 49.70 and a = 226. 97.80 = 321 15.2-Law of Sines.

15.2-EXAMPLE 18 Solve triangle ABC where A = 32.50, B = 49.70 and a = 226. = 417 97.80 = 321 15.2-Law of Sines.

15.2-EXAMPLE 20 Solve triangle ABC where A = 35.20, a = 525 and c = 412. 26.90 15.2-Law of Sines.

15.2-EXAMPLE 20 Solve triangle ABC where A = 35.20, a = 525 and c = 412. 26.90 = 805 117.90 15.2-Law of Sines.

15.2-DEFINITIONS The Ambiguous Case Another way to check for the ambiguous case is to make a sketch. 15.2-Law of Sines.

15.3 - Law of Cosines 15.3-Law of Cosines.

15.3-STRATEGY Law of Sines vs Law of Cosines NOTE: We use the law of sines when we have a known side opposite a known angle. We use the law of cosines only when the law of sines does not work, that is, for all other cases. In the figures shown, the heavy lines indicate the known information and might help in choosing the proper law. 15.3-Law of Cosines.

15.3-EXAMPLE 22 Solve triangle ABC where a = 184, b = 125 and C = 27.20. 38.10 = 92.6 114.70 BLANK 15.3-Law of Cosines. (Acute Angle)

15.3-EXAMPLE 24 Solve triangle ABC, where a = 128, b = 146 and c = 222. 108.10 33.20 (Largest Angle) 38.70 BLANK 15.3-Law of Cosines.

15.4 - Applications 15.4-Applications.

15.4-EXAMPLE 25 Find the area of the gusset shown below. 33.7 90.00 θ = 125.00 15.4-Applications.

15.4-EXAMPLE extra From a point on level ground between two power poles of the same height, cables are stretched to the top of each pole. One cable is 52.6 m long, the other is 67.5 m long, and the angle of intersection between the two poles is 1250 . Find the distance between the two poles. θ 15.4-Applications.

15.5 - Addition of Vectors 15.5-Addition of Vectors.

15.5-DEFINITIONS Vector Diagrams (Resultant Vector) 15.5-Addition of Vectors. If we draw two vectors tip to tail, the resultant R will be the vector that will complete the triangle when drawn from the tail of the first vector to the tip of the second vector. It does not matter whether vector A or vector B is drawn first; The same resultant will be obtained either way. The parallelogram method will give the same result. To add the same two vectors A and B as before, we first draw the given vectors tail to tail and complete a parallelogram by drawing lines from the tips of the given vectors, parallel to the given vectors. The resultant R is then the diagonal of the parallelogram drawn from the intersections of the tails of the original vectors.

15.5-STRATEGY Two vectors, A and B, make an angle of 47.20 with each other as shown in the figure below. If their magnitudes are A = 125 and B = 146, find the magnitude of the resultant R and the angle Ф1 that R makes with vector B. Φ2 is the angle between the smaller vector A and the resulting vector R determined by using the Law of Sines. Ф2 COMMON VECTOR ANGLE This is the angle that runs between vectors A and B . It is used to calculate Φ3 which is then used for the Law of Cosines. In this example, the common angle 47.20 is already given. 47.20 132.80 Ф3 15.5-Addition of Vectors. Φ3 is the angle needed to apply the Law of Cosines and is calculated by subtracting the common vector angle found between vector A and vector B from 1800 ( Φ3 = 1800 - 47.20 = 132.80 ) 125 Ф1 Φ1 is the angle between the larger vector B and the resulting vector R determined by using the Law of Sines. 146

15.5-EXAMPLE 27 Two vectors, A and B, make an angle of 47.20 with each other as shown in the figure below. If their magnitudes are A = 125 and B = 146, find the magnitude of the resultant R and the angle Ф1 that R makes with vector B. Ф2 248 Ф3 132.80 125 21.70 Ф1 47.20 146 15.5-Addition of Vectors.

15.5-STRATEGY Vector A has a magnitude of 125 at an angle of 77.20 and Vector B has a magnitude of 146 at an angle (Ф4 ) of 30.00 . Find the magnitude of the resultant and the angle (Ф5) that R makes with the x axis. Y-axis COMMON VECTOR ANGLE This is the angle that runs between vectors A and B . In this example, the common angle 47.20 is not given so we must calculate this angle by subtracting the angle that is given between vector B and the x-axis (300) from the angle that is given between vector A and the x-axis (77.20). The resulting angle (47.20) is then used to calculate Φ3 which is used for the Law of Cosines. Ф2 47.20 132.80 Ф3 15.5-Addition of Vectors. Φ3 is the angle needed to apply the Law of Cosines and is calculated by subtracting the common vector angle found between vector A and B from 1800 ( Φ3 = 1800 - 47.20 = 132.80 ) 125 Ф5 Φ5 is the resulting angle that we ultimately want and lies between the resulting vector R and the x-axis. This angle is determined by adding Φ4 (given) and Φ1 which is calculated from the Law of Sines. ( Φ5 = 300 + 21.70 = 51.70 ) Ф1 146 Ф4 Φ4 is given in this example as 300 and is the angle between vector B and the x axis. X-axis

15.5-EXAMPLE extra Vector A has a magnitude of 125 at an angle of 77.20 and Vector B has a magnitude of 146 at an angle (Ф4 ) of 30.00 . Find the magnitude of the resultant and the angle (Ф5) that R makes with the x axis. 248 Ф3 132.80 125 Ф5 Ф1 146 Ф4 15.5-Addition of Vectors.

15.5-STRATEGY Polar to Rectangular Conversions Vector A has a magnitude of 125 at an angle of 77.20 and Vector B has a magnitude of 146 at an angle of 30.00 . Find the magnitude of the resultant and the angle that R makes with the x axis. Polar to Rectangular Conversions Vector addition and subtraction are best done in Rectangular Form so we convert all of our Polar Coordinates into Rectangular Components using basic trigonometry or directly on our calculator. 15.5-Addition of Vectors. Add up the total x-components and total y-components. This represents our Resulting Vector in Rectangular Form. Convert your final answer back into Polar Form using basic trigonometry or directly on your calculator.

15.5-EXAMPLE 28 Find the resultant R of the four vectors shown in Fig. 15-39(a). Vector Rx Components Ry Components A 42.0 cos 58.0° = 22.3 42.0 sin 58.0° = 35.6 B 56.1 cos 148° = -47.6 56.1 sin 148° = 29.7 C 52.7 cos 232° = -32.4 52.7 sin 232° = -41.5 D + 45.3 cos 291° = 16.2 + 45.3 sin 291° = -42.3 R Rx = -41.5 Ry = -18.5 Step 1: Resolve each given vector into their Rx and Ry rectangular components using x = V cosθ and y = V sinθ. Then, add up the total Rx and Ry values giving the resulting vector R in rectangular form. 15.5-Addition of Vectors.

15.5-EXAMPLE 28 Find the resultant R of the four vectors shown in Fig. 15-39(a). Vector Rx Components Ry Components A 42.0 cos 58.0° = 22.3 42.0 sin 58.0° = 35.6 B 56.1 cos 148° = -47.6 56.1 sin 148° = 29.7 C 52.7 cos 232° = -32.4 52.7 sin 232° = -41.5 D + 45.3 cos 291° = 16.2 + 45.3 sin 291° = -42.3 R Rx = -41.5 Ry = -18.5 Horizontal Component Vertical Component Step 1: Resolve each given vector into their Rx and Ry rectangular components using x = V cosθ and y = V sinθ. Then, add up the total Rx and Ry values giving the resulting vector R in rectangular form. 15.5-Addition of Vectors. BLANK Step 2: Convert the Horizontal and Vertical rectangular components R = (-41.5, -18.5) into their equivalent polar form.

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