Section 8.3 The Law of Cosines

Transcription

1 147 Section 8.3 The Law of Cosines In this section, we will be solving SAS, SSS triangles. To help us do this, we will derive the Laws of Cosines. Objective 1: Derive the Laws of Cosines. To derive the Law of Cosines, we will need to consider a case for each type of oblique triangle. Let A,, and C be the angles of the oblique triangle with corresponding sides of length a, b, & c. Case 1: Acute Triangle Case : Obtuse Triangle c a a c A C A C b b Place AC so that the vertex of angle of A coincides with the origin and side b lies along the positive x-axis. Case 1: Acute Triangle Case : Obtuse Triangle (ccos(a), csin(a)) (ccos(a), csin(a) c a a c A C A C b (b, 0) b (b, 0) If we think of the side with length c as being the radius of a circle, then the coordinates of the vertex of angle is (ccos(a), csin(a)). Since side b lies along the positive x-axis, the vertex of angle C is (b, 0). We can use the distance formula to compute the length of side a: a = (x x 1 ) + (y y 1 ) = (b ccos(a)) + (0 csin(a)) (simplify) = (b ccos(a)) + ( csin(a)) (expand)

2 = b bccos(a) + c cos (A) + c sin (A) (factor out c ) = b bccos(a) + c [cos (A) + sin (A)] (cos (A) + sin (A) = 1) = b bccos(a) + c (rearrange) = b + c bccos(a) Hence, a = b + c bccos(a). y a similar argument, we can show that: b = a + c accos() and c = a + b cos(c). The Law of Cosines Theorem For an oblique triangle AC with opposites of length a, b, and c respectively, a = b + c bccos(a) b = a + c accos() c = a + b cos(c) "The first side squared equals the sum of the squares of the other two sides minus times the product the other two sides times the cosine of the angle opposite of the first side." Objective : Solve SAS Triangles (case 3) Solve the following triangle: Ex. 1 a = 5 m, c = 7 m, = 48 We will begin by sketching a picture = 48 of a triangle. Since we have two 7 m = c a = 5 m sides and an included angle, we will need to use the Law of Cosines. A C We know, so we will use the b form that will allow us to solve for side b: b = a + c accos() b = (5) + (7) (5)(7)cos(48 ) b = cos(48 ) b = b = ± = ± Reject the negative answer and round. b 5.1 m Now, let's use the Law of Sines to find A: 148

3 To find A: sin(a) a sin(a) 5 sin(a) = = sin() b sin(48 = o ) sin(48 o ) (plug in the known values, cross multiply and solve) (find the inverse sine) A = sin 1 5sin(48 ( o ) ) = sin 1 ( ) = In quadrant I, In quadrant II, A A = = The second angle, is not possible since the sum of and 48 exceeds 180. Finally, C = = Thus, b 5.1 m, A 45.48, and C 86.5 When we have to use the Law of Cosines to find an angle, it is easier to manipulate the Law of Cosines to solve for the cosine function first: c = a + b cos(c) (subtract a + b from both sides) c a b = cos(c) (divide by ) c a b cos(c) = a + b c Similarly, cos(a) = b + c a = cos(c) (simplify) bc and cos() = a + c b The Alternate Form For an oblique triangle AC with opposites of length a, b, and c respectively, cos(a) = b + c a bc cos() = a + c b ac cos(c) = a + b c Objective 3: Solve SSS Triangles (case 4) ac

4 Solve the following triangle: Ex. a = 9 m, b = 6 m, c = 11 m We will begin by sketching a picture of a triangle. Since we have all 11 m = c a = 9 m there sides, we will need to use the Law of Cosines. Let's pick A C alternate forms that will allow us b = 6 m to find A and : cos(a) = b + c a cos() = a + c b bc ac cos(a) = (6) + (11) (9) (6)(11) cos(a) = = cos() = (9) + (11) (6) (9)(11) cos() = = A = cos 1 ( ) = cos 1 ( ) A = = Now, C is: C = = Thus, the angles are A 54.85, 33.03, and C Objective 4: Solve Application Problems. Let's first re-examine an application problem we solved in the beginning of the chapter. Now, using the Law of Cosines Ex. 3 Two cities, San Antonio and Philadelphia, are 1494 miles apart. To avoid a bad storm, a pilot took a course that was 0 further south from San Antonio for 400 miles and then turned to fly towards Philadelphia. How much further did the pilot have to fly to reach Philadelphia? Round to the nearest mile. If we draw the picture, we realize that we have an SAS triangle. Thus, this problem is now easily solved using the Law of Cosines: SA 1494 miles miles Philadelphia?

5 151 Ex. 4 Ex. 5 a = b + c bccos(a) a = (400) + (1494) (400)(1494)cos(0 ) a = cos(0 ) a = a = ± a = ± (reject the negative answer and round) a Thus, the plane traveled = The plane then flew miles further. To approximate the length of a marsh from point A to point C, a surveyor walks 55 meters from point A to point. She then turns 70 counterclockwise and walks 400 meters to point C. What is the length of the marsh (AC)? Let's first draw a picture of the situation. Since AC and 70 form a straight line, then they are supplementary angles. Thus, = = 110. Since we have a SAS triangle, we will use the Law of Cosines to find AC. b = a + c accos() b = (400) + (55) (400)(55)cos(110 ) b = cos(110 ) b = b = ± b = ± b The marsh is meters long. (reject the negative answer and round) A ship leaves port in the Florida Keys with a bearing of S80 E with a speed of 18 knots. After two hours, the ship turns 75 toward the north. After 3 additional hours of maintaining a speed of 0 knots, what is the bearing of the ship and the distance of the ship from the port? C 400 m m A

6 15 We begin by drawing a diagram of the situation. The ship travels 18() or 36 nautical miles S80 E from the Florida Keys. it then turns 75 toward the north and travels 0(3) = nautical miles. If we connect a line from θ the Florida Keys to the ending point, we will form a oblique triangle. We will 36 lel the angle formed by the 36 nautical mile side and the line connecting the Florida Keys to the ending point as θ. The in the triangle adjacent to the 75 angle is equal to = 105 We can use the Law of Cosines to find the distance the ship is from the port. d = (36) + (60) (36)(60)cos(105 ) d = cos(105 ) d = d = ± = ± (reject answer & round) d = nautical miles Now that we know all three sides of the triangle, we can use the Alternate Form to find θ: cos(θ) = ( ) + (36) (60) = = ( )(36) θ = cos 1 ( ) = Since the 36 nautical mile side had a bearing of S80 E, the angle between that side and the positive x-axis is = 10. The angle between the positive x-axis and the line connecting the Florida Keys to the ending point is θ 10 = = Finally, the angle between the line connecting the Florida Keys to the ending point and the positive y-axis is = 51.6 Thus, the ship is miles from the port at a bearing of N51.6 E. For any AC, prove the following: Ex. 6a cos( C ) = s(s c) Ex. 6b sin( C ) = (s a)(s b) where s = 1 (a + b + c) where s = 1 (a + b + c)

7 153 a) We will first show that cos ( C ) = s(s c) cos ( C ) (cos ( C ) = 1+cos(C) ) = 1+cos(C) (but cos(c) = a + b c ) = 1+ a + b c = 1 + a + b c = +a + b c = a ++b c = (a+b) c = (a+b c)(a+b+c) = (a+b c)s = (s c)s = 4s(s c) (multiply the top & bottom by ) (regroup) (simplify) (but a + + b = (a + b) ) (factor (a + b) c = (a + b c)(a + b + c)) (since s = 1 (a + b + c), then s = a + b + c) (since s = a + b + c, then s c = a + b c) = s(s c) (reduce) = s(s c) Hence, cos ( C ) = s(s c) cos( C ) = ± s(s c) (factor out the ) (use the square root property) ut C is an angle of triangle, so C < 180 which means C < 90. Thus, cos( C Therefore, cos( C ) = ) has to be positive and we reject the negative root. s(s c) b) We will first show that sin ( C ) = (s a)(s b) sin ( C ) (sin ( C ) = 1 cos(c) ) = 1 cos(c) (but cos(c) = a + b c )

8 154 = 1 a + b c = 1 a + b c = a b +c = c a + b = c (a +b ) = c (a b) = [c (a b)][c+(a b)] = [ a+b+c][a b+c] = [s a][a b+c] = [s a][s b] = 4(s a)(s b) (multiply the top & bottom by ) (simplify and distribute the negative sign) (regroup) (factor out a 1 from a + b ) (but a + b = (a b) ) (factor c (a b) = [c (a b)][c + (a b)]) (simplify) (since s = a + b + c, then s a = a + b + c) (since s = a + b + c, then s b = a b + c) (factor out the 's) (reduce) = (s a)(s b) Hence, sin ( C ) = (s a)(s b) sin( C ) = ± (s a)(s b) (use the square root property) ut C is an angle of triangle, so C < 180 which means C < 90. Thus, sin( C Therefore, sin( C ) = ) has to be positive and we reject the negative root. (s a)(s b)