Engineering Mathematics 1 Angles and Arclengths PDF

Engineering Mathematics 1 Angles and Arclengths MPS Department/ FEU Institute of Technology  Objectives ▪ Define angles in terms of positive and negative rotations ▪ Determine the quadrant in which the terminal side of an angle in standard position lies ▪ Identif...

Engineering Mathematics 1 Angles and Arclengths MPS Department/ FEU Institute of Technology  Objectives ▪ Define angles in terms of positive and negative rotations ▪ Determine the quadrant in which the terminal side of an angle in standard position lies ▪ Identify the values of remaining trigonometric functions when one function is given With the development of Calculus and the physical sciences in the 17th Century, a different perspective arose – one that viewed the classic trigonometric relationships as functions with the set of real numbers as their domain. Consequently the applications expanded to include physical phenomena involving rotations and vibrations, including sound waves, light rays, planetary orbits, vibrating strings, pendulums, and orbits of atomic particles. We will explore both perspectives beginning with angles and their measures….. An angle is determined by rotating a ray about its endpoint. The starting position of called the initial side. The ending position is called the terminal side. An angle is at the standard position if its vertex is at the origin, and the 90  initial side is on the x-axis. II I 180  0  , 360  Initial Side III IV 270  Positive Angles are generated by counterclockwise rotation. Negative Angles are generated by clockwise rotation. go in a counter-clockwise rotation 90∘ 45∘ 180∘ 0∘ , 360o 270∘ go in a clockwise rotation − 270  − 180  0  , − 360  − 45  − 90  Angles that have the same initial and terminal side are called coterminal angles. See the illustrations below. 𝛼 𝛽 𝛼 𝛽 𝛼 and 𝛽 are coterminal angles The sum or difference of coterminal angles are positive or negative multiples of 360 𝛼 𝛽 𝛼 𝛽 Determine 2 coterminal angles, one positive and one negative. for a 60o angle 60  𝟔𝟎 + 𝟑𝟔𝟎 = 𝟒𝟐𝟎𝐨 𝟔𝟎 – 𝟑𝟔𝟎 = −𝟑𝟎𝟎𝐨 Determine 2 coterminal angles, one positive and one negative. for a 30o angle 30∘ 𝟑𝟎 + 𝟑𝟔𝟎 = 𝟑𝟗𝟎𝐨 𝟑𝟎 – 𝟑𝟔𝟎 = −𝟑𝟑𝟎𝐨 Determine 2 coterminal angles, one positive and one negative. for a 230o angle 𝟐𝟑𝟎 + 𝟑𝟔𝟎 = 𝟓𝟗𝟎𝐨 𝟐𝟑𝟎 – 𝟑𝟔𝟎 = −𝟏𝟑𝟎𝐨 Determine 2 coterminal angles, one positive and one negative. for a −20o angle −𝟐𝟎 + 𝟑𝟔𝟎 = 𝟑𝟒𝟎𝐨 −𝟐𝟎 – 𝟑𝟔𝟎 = −𝟑𝟖𝟎𝐨 Determine 2 coterminal angles, one positive and one negative. for a 460o angle Good but 𝟒𝟔𝟎 + 𝟑𝟔𝟎 = 𝟖𝟐𝟎𝐨 not best answer. 𝟒𝟔𝟎 – 𝟑𝟔𝟎 = 𝟏𝟎𝟎𝐨 𝟏𝟎𝟎 – 𝟑𝟔𝟎 = −𝟐𝟔𝟎𝐨 Two angles are Complementary if their sum is 90 Find the complement of each angles: 40 120 No Complement! 40 + x = 90 x = 50 degrees Two angles are Supplementary if their sum is 180 Find the supplement of each angles: 40 120 40 + x = 180 120 + x = 180 x = 140 degrees x = 60 degrees Coterminal Angles: 𝐴𝑛𝑔𝑙𝑒 ± 𝑘 360° , 𝑘 ∈ {,.. −2, −1,0,1,2, … } To find a Complementary Angle: 90 − Angle To find a Supplementary Angle: 180 − Angle One radian is the measure of the central angle,  , that intercepts an arc, s, that is equal in length to the radius r of the circle. C = 2 r C 2r 𝑟 𝑠=𝑟 = 𝜃 2 2 𝑟 C r= 2 So…1 revolution is equal to 2π radians Because the circumference of a circle is 2𝜋𝑟 units, is follows that the central angle of one complete revolution correspond to the arclength of 𝑠 = 2𝜋𝑟 Therefore, 2 radians = 360   radians = 180   radians = 90  2 Now, let’s add more…..  (1.57 rad ) Radians 1     2 2  3  radians radians 4 4   + 4 2 (3.14 rad ) (6.28 rad ) 5 7 radians radians 4 4 (4.71 rad ) Let’s take a look at more common angles that are found in the unit circle.  3  radians radians 4 4 (3.14 rad ) (6.28 rad ) 5 7 radians radians 4 4  3  radians radians 4 4 (3.14 rad ) (6.28 rad ) 5 7 radians radians 4 4  𝜋 𝜃= 2 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡 𝐼𝐼 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡 𝐼 𝜋 𝜋 sin 29° a true statement.  Two Special Triangles ◼ 30-60-90 Triangle ◼ 45-45-90 Triangle Can you reproduce these two triangles without looking at them? Try it now. It would be very handy for you later.  Function Values of Special Angles Remember the mnemonic sohcahtoa - “sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent.” 𝜃 sin 𝜃 cos 𝜃 tan 𝜃 cot 𝜃 sec 𝜃 csc 𝜃 30° 2 45° 1 1 60° 2 Now, try to use your two special triangles to check out these function values.  Function Values Using a Calculator ◼ Calculators are capable of finding trigonometric function values. ◼ When evaluating trigonometric functions of angles given in degrees, remember that the calculator must be set in degree mode. ◼ Remember that most calculator values of trigonometric functions are approximations.  Example ◼ a) ◼ b) cot 68.4832 ° ◼ Convert 38° to ◼ Use the identity decimal degrees. ◼ ◼ cot 68.4832 °.3942492  Angle Measures Using a Calculator ◼ Graphing calculators have three inverse functions. ◼ If x is an appropriate number, then gives the measure of an angle whose sine, cosine, or tangent is x. Note: Please go over page 15 of your Graphing Calculator Manual.  Example ◼ Use a calculator to find an angle in the interval that satisfies each condition. ◼ Using the degree mode and the inverse sine function, we find that an angle having sine value.8535508 is 58.6. We write the result as  Example (cont.) ◼ Use the identity Find the reciprocal of 2.48679 to get Now find using the inverse cosine function. The result is 66.289824  Significant Digits ◼ A significant digit is a digit obtained by actual measurement. ◼ Your answer is no more accurate then the least accurate number in your calculation. Number of Angle Measure to Nearest: Significant Digits 2 Degree 3 Ten minutes, or nearest tenth of a degree 4 Minute, or nearest hundredth of a degree 5 Tenth of a minute, or nearest thousandth of a degree  How to Solve a Right Triangle Given an Angle and a Side? ◼ Solve right triangle ABC, if A = 42° 30' and c = 18.4. ◼ B = 90 - 42° 30' B B = 47° 30' c = 18.4 42°30' a A sin A = C c b cos A = a c sin 42o30' = 18.4 b cos 42 30' = o a = 18.4sin 42o30' 18.4 a = 18.4(.675590207) b = 18.4cos 42o30' a  12.43 b  13.57  How to Solve a Right Triangle Given Two Sides? ◼ Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm. B c = 27.82 a = 11.47 C A ◼ B = 90 24.35° B = 65.65°  Solve the right triangle. Round decimal answers to GUIDED PRACTICE the nearest tenth. Example 1 A Find m∠ B by using the Triangle Sum Theorem. 42o 180o = 90o + 42o + m∠ B 70 48o = m∠ B 48o Approximate BC by using a tangent ratio. C B BC Approximate AB by using a cosine ratio. tan 42 = o 70 70 cos 42o = AB ANSWER 70 tan 42o = BC AB cos 42o = 70 70 0.9004 BC 70 The angle measures are 63.0 ≈ BC AB = cos 42o 42o, 48o, and 90o. The 70 side lengths are 70 feet, AB 0.7431 about 63.0 feet, and AB 94.2 about 94.2 feet.  Solve a right triangle that has a 40o angle and a 20 GUIDED PRACTICE inch hypotenuse. Example 2 Find m∠ X by using the X Triangle Sum Theorem. 180o = 90o + 40o + m∠ X 50o 50o = m∠ X 20 in Approximate YZ by using a sine ratio. XY sin 40o = 20 20 sin 40o = XY 40o 20 0.6428 ≈ XY Y Z 12.9 ≈ XY Approximate YZ by using a cosine ratio. YZ ANSWER cos 40o = 20 o 20 cos 40 = YZ The angle measures are 40o, 50o, 20 0.7660 ≈ YZ and 90o. The side lengths are 12.9 15.3 ≈ YZ in., about 15.3 in., and 20 in.  Solve the right triangle. Round to the nearest tenth. Example 3 P° + R° + Q° = 180° P° = 180-90-53 mQ = 53 P° = 37 mR = 90 mP = 37 PQ = 30° PR = 24.0 cos53 = p sin53 = q QR = 18.1 30 30 p  18.1 q  24.0  Solve the right triangle. Round decimals to the nearest tenth. Example 5 Example 4 mP = 90 − 37 = 53 mT = 90 − 24 = 66 sin37 = PQ cos37 = QR TR cos24 = 33 22 22 tan24 = AT 33 PQ = 13.2 QR = 17.6 TR = 14.7 AT = 36.1  Solving Right Triangles Example 6 Find the unknown measures. Round lengths to the nearest hundredth and angle measures to the nearest degree. Method 1: By the Pythagorean Theorem, Method 2: RT2 = RS2 + ST2 (5.7)2 = 52 + ST2 Since the acute angles of a right triangle are complementary, mT  90° – 29°  61°. , so ST = 5.7 sinR. Since the acute angles of a right triangle are complementary, mT  90° – 29°  61°.  Solve the right triangle. Round decimals the nearest tenth. Example 7 Use Pythagorean Theorem to find c… c 2 = 22 + 3 3 c = 3.6 Use an inverse trig function to AB = 3.6 find a missing acute angle… BC = 3 3 −1 AC = 2 mA = tan ( ) = 56.3 2 mA = 56.3 Use Triangle Sum Theorem to find the other acute angle… mB = °33.7 mC = °90 mB = 90 − 56.3 = 33.7 Example 8 Pythagorean Theorem A2 + b2 = c2 Where c is the hypotenuse. PN 2 = 112 + 182 PN = 21.9 11 −1 mN = tan ( ) = 31.4 18 mP = 90 − 31.4 = 58.6 Example 9 232 = TU 2 + 72 TU = 21.9 −17 mS = cos ( ) = 72.3 23 mU = 90 − 72.3 = 17.7 Ayres, Moyer, Schaum’s Outline of Theory and Problems of TRIGONOMETRY, Mc Grawhill Companies Richard N. Aufmann, Vernon C. Barker, Richard D. Nation, College Algebra and Trigonometry,Seventh Edition, © 2011, 2008 Brooks/Cole, Cengage Learning