MATH 120-Vectors, Law of Sinesw, Law of Cosines (20 )

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1 MATH 120-Vectors, Law of Sinesw, Law of Cosines (20 ) *Before we get into solving for oblique triangles, let's have a quick refresher on solving for right triangles' problems: Solving a Right Triangle I. If we are given three parts of a triangle (at least one is a side), we are able to find the other three parts. For consistenc, let s label the acute angles of a right triangle as A and B and label the right angle as C. The letter a, b, and c will denote the sides opposite these angles, respectivel (e. Side c is the hpoteneuse of the right triangle). sin A a c cos A b tan A a cot A b sec A c csc A c c b a b a sin B b cosb a tanb b cot B a secb c cscb c c c a b a b *Note: sin A cosb tan A cot B sec A csc B Conjunctions of acute complementar angles are equal. II. Procedures for Solving a Right Triangle A. Sketch a right triangle and label the known and unknown sides and angles. B. Epress each of the three unknown parts in terms of the known parts and solve for the unknown parts. C. Check the results. The sum of the angles should be 180. If onl one side is given, check the computed side with the Pthagorean Theorem. If two sides are given, check the angles and computed side b using appropriate trigonometric functions. 1

2 e. Finding side a (given A 50 and b 6. 7 ) and, knowing that tan A a, a b tan A 6. 7 tan b Finding side c (given A 50 and b 6. 7 )and. knowing that cos A b, c c b 6. 7 cos A cos Finding angle B (given a right triangle with an angle = 50 ), knowing that the sum of three angles would be 180,B or B Checking the angles: A B C A Checking the sides: (accepatable) slight variation due to rounding *General rule: the longest side is alwas opposite the largest angle, and the ****** shortest side is alwas opposite the smallest angle. In right triangle, the hpoteneuse is alwas the longest side. Applications of Right Triangles I. Angle of elevation: angle between the horizontal and the line of sight, when the object is above the horizontal. e. If the angle of elevation is 20 at a distance of 1000 ft from the base of a building, how high is this building? 2

3 h tan ft h tan 20 ( 1000 ft) 036. ( 1000 ft) 360 ft (Result is rounded off to 2 significant digits because the data is onl good to 2 sig. digits) II. Angle of depression: angle between the horizontal and the line of sight, when the object is below the horizontal. e. If a plane is 2500 ft above the ground (above a football field) and the angle of depression of the north goal line from the plane is How far is the observer in the plane from the goal line? 2500 cos 315. d d 2500 ft 2930 ft cos 315. *Remember the number of significant figures ****** Solving Oblique Triangles, Using the Law of Sines Oblique triangles: Triangles that do not contain a right angle. I. We need to know three parts and at least one of them a side, in order to solve a triangle. There are four possible combinations of parts: A. Two angles and one side. B. Two sides and the angle opposite one of them. C. Two sides and the included angle. D. Three sides. 3

4 II. Derivation of Law of Sines Let ABC be an oblique triangle with sides a, b, and c opposite angles A, B, and C, respectivel. B drawing a perpendicular h from B to side b, or its etension, we can see that: h csin A or h asin C h csin A or h asin( 180C ) asin C csin A asin C or a c sin A sinc B dropping a perpendicular from A to a, we also derive this result: csin B bsin C or b c sin B sinc For an triangle with sides a, b, and c, opposite angles A, B, and C, respectivel, we have the Law of Sines: a b c sin A sin B sinc 4

5 III. Eamples: A. Two angles and one side. Given: a 15, A 15, B 140 Find: b, c, C Solution: ( 1 ): C 180( ) 25 a b c ( 2 ): 15 sin A sin B sinc sin15 b sin140 c sin25 Since 15 15(sin 140 ) (sin 15 ) sin15 b b sin140 15(sin 140 ) b sin15 useb 37 ( 3 ):Since 15 (sin 15 ) 15(sin 25 ) sin15 c c sin25 15(sin 25 ) c sin15 use c 25 B. Two sides and an angle opposite one of them Given: a 5240., b , B Find: c, A, C a b Solution: ( 1 ): sin A sin B sin A sin

6 Since (sin A) (sin ) sin A sin A sin (. (sin. ) ) sin (. 8776) *Note: sin A A or ( 2 ): w / A : C 180 ( B A) 180 ( ) ( 3 ):w / A : c a c. sinc sin A sin sin Since c c(sin ) (sin 7 sin sin ) (sin ) c sin *Now, replace A , using the same steps, C , c ****** 6

7 Solving Oblique Triangles, Using the Law of Cosines I. Eamples a b c 2bccos A b a c 2accos B c a b 2ab cosc A. Two sides and the included angle Given: a 4530, b 924, C Find: A, B, c Solution: ( 1 ):using Law of Cosines in the form c 2 a 2 b 2 2abcosC c a b 2abcos C c ( 4530)( 924)cos 98.0 c ( ) c 4747 Or use c 4750 a c ( 2 ):using Law of Sines: sin A sinc 4530 sin A 4750 sin (sin 980. ) 4750(sin A) sin A (sin. ) 09444,. A sin 1 ( ) 708. b c ( 3 ): using Law of Sines: sin B sinc 924 sin B 4750 sin (sin 980. ) 4750(sin B) sin B (sin. ) , B sin (( )

8 B. Three sides Given: a , b , c Find: A, B, C Solution: ( 1 ):using Law of Cosines in the form a 2 b 2 c 2 2bccos A ( )( )cos A cos A cos A , A cos 1 ( ) a b ( 2 ): using Law of Sines: sin A sin B sin sin B (sin B) (sin ) sin B (sin ) B sin 1 ( ) ( 3 ):C 180 ( A B) 180 ( ) ***General Rule: a b c 1. Use Law of Sines ( ) for problems involving SSA or AAS. sin A sin B sinc *SSA: two sides and an angle opposite one of them. * The SSA case requires special consideration. If the side opposite the given angle is a. greater than the known adjacent side, there is onl one possible triangle. b. less than the known adjacent side but greater than the altitude, there are two possible triangles. c. less than the altitude, there is no possible triangle. *AAS: two angles and a side opposite one of them. 8

9 2. As a final check: a. alwas choose a given value over a calculated value for doing calculations. b. alwas check our results to see that the largest angle is opposite the largest side and that the smallest angle is opposite the smallest side a b c 2bccos A Use Law of Cosines: ( b a c 2accos B) for problems involving SAS or SSS c a b 2abcosC Introduction to Vectors I. Scalar quantit: magnitude of the quantit, number to represent amount of certain measurements. (e.: length, width, temperature, area, speed, etc.) Scalar quantit is represented b lightface A. II. Vector quantit: quantit that is described b both its magnitude and Direction. (e.: velocit, force, etc..), Vector is represented b A or A * speed ( 500 mi/h ) velocit ( 500 mi/h in a direction 30 north of west) III. Vector Additions A. Vector sum of A+B is the R. R is called the resultant, from initial point O to terminal point Q. Resultant is a single vector that is the vector sum of an number of other vectors.. 9

10 B. Polgon Method/Vector Triangle Method: Sum of A+B is R can be drawn from the tail of A to the head of B. C. Parallelogram Method: let two vectors being added be the sides of a Parallelogram (tail to tail). Resultant is the diagonal of the parallelogram. Initial point of the resultant is the common initial point of the vectors being added. D. Two vectors in different locations are same if the have the same magnitude and direction. E. Scalar Multiple of vector A, na, is a vector n times as long as A, but in the same direction. F. Consider A-B as A+(-B). IV. Displacement A. The distance from a reference point and the angle from a reference direction. B. Displacement is a vector quantit. C. If a traveler travels awa from the reference point for a given amount of distance and direction (angle) from the reference point and then returns to the reference point. Its displacement from the reference point would be zero. 10

11 V. Eamples: A. Which is a scalar or a vector: 1. A boat sailed 2 miles/min. 2. A boat sailed 10 miles/min toward northeast. 3. Joe is running awa from the school bull at mile/min northward heading home. B. Using parallelogram method to figure out the resultant. (Sum of 2 or 3 vectors ) C. Solve: A ship travels 20 km in a direction of 30 south of east and then turns due south for another 40 km. What is the ship s displacement from its initial position? D. Solve: Two forces that act on an airplane wing are called the lift and drag. Find the resultant of lift of 800 lbs. And drag of 300 lbs. 11

12 VI. Vector Components I. Components of the original vector: vectors, when added together, have a resultant equal to the given vectors. A. Initial points of these components are at the origin. Terminal points are located at the points where perpendicular lines from the terminal point of the given vector across the aes. B. Resolving the vector into its components: finding component vectors. C. Eample: Given vector A, A is related to A b: A cos A A Is related to A b: A A sin A = Acos A A sin D. Steps used in finding the - and - components: 1. Place vector A such that is in standard position. 2. Calculate A and A from A Acos and A Asin. We ma use the reference angle if we note the direction of the component. 3. Check the components to see if each is in the correct direction and has a magnitude that is proper for the reference angle. E. Tr to resolve each vector into its - and -components: 1. Vector A of magnitude 350 and direction Vector with magnitude 2.65 and direction

13 F. Eample problems: 1. At one point the Pioneer space probe was entering the gravitational field of Jupiter at an angle of 2.55 below the horizontal with a velocit of 18,550 mi/h. What were the components of its Velocit? 2. Two upward forces are acting on a bolt. One force of 60.5 lb acts at an angle of 80.0 above the horizontal, and the other force of 35.2 lb acts at an angle of 50.0 below the first force. What is the total upward force on the bolt? ****** Application of Vectors/Vector Addition b Components I. To add vectors, use the components of the vector, the Pthagorean theorem, and the tangent of the standard position angle of the resultant. *A vector is not completel specified unless both its components and its direction are given. E. If two vectors (.A = 15.1 and B = 8.25 ) are at right angles, we can find the resultant vector R b using Pthagorean theorem Magnitude of R: R = A B Direction of R: tan B A tan ( ) A 1 B 13

14 II. Procedure for Adding Vectors b Components 1. Resolve the given vectors into - and -components. 2. Add the -components to obtain R ;add the -components to obtain R. 3. Find the magnitude of the resultant R, using R R R Find the standard-position angle for the resultant R. First find the reference angle ref for the resultant R b using tan ref R R ref tan 1 R R III. Eamples: Find the resultant of two vectors: A = 1200, A B = 1550, and b Step 1.: A Acos cos B Bcos cos , A B Asin sin Bsin sin Step 2: R A B R A B Step 3: R R R ( 1299) Step 4: tan ref R 142 R 1299 ref tan ( )

15 Since R is negative and R is positive, is a second-quadrant angle IV. Tr to find the resultant of the three vectors given: A 6. 4, 126, B 5. 9, B 238, C 3. 2,C 72 A ***General Rule 1. Vector problems ma be solved: a. graphicall using i. parallelogram method or ii. vector triangle method/polgon method b. algebraicall using the law of sines and/or the law of cosines;or c. b the component method 2. Given v and angle, the horizontal and vertical components are found as: v v cos v v sin v v v 2 2 reference angle : tan v v *angle in standard position is determined from angle and the quadrant in which v lies. 3. Component method of adding vectors: To find the resultant vector R of two or more vectors using the component method: a. find the horizontal component, R, of vector R, b finding the algebraic sum of the horizontal components of each of the vectors being added. 15

16 b. find the vertical component, R, of vector R, b finding the algebraic sum of the vertical components of each of the vectors being added. c. find the length of R: R R R 2 2 d. find the angle, first find, the reference angle. tan R R 16