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137 Section 8.2 – The Law of Sines In this section and the next section, we will be solving triangles that are not right triangles or what are called oblique triangles. We will derive the laws of sines (section 8.2) and cosines (section 8.3) to help us do this. Definition An oblique triangle is a triangle where none of the angles are right angles. There are two types of oblique triangles; the first case is when all the angles are acute and the second case is when the two of the angles are acute and the third angle is obtuse. Objective 1: Derive the Laws of Sines. To derive the Law of Sines, we will need to consider a case for each type of oblique triangle. Let ∠A, ∠B, and ∠C be the angles of the oblique triangle with corresponding sides of length a, b, & c. Case 1: Acute Triangle Case 2: Obtuse Triangle B c B a a c A C A b C b Construct h as the height of the triangle and ∠P to be the supplementary angle of ∠A in the obtuse triangle: Case 1: Acute Triangle Case 2: B c Obtuse Triangle B a h A h C b a c P A C b 138 In both cases, looking at the right triangle formed by ∠C, sides a and h: sin(C) = opp hyp = h a which implies h = asin(C) In case 1, looking at the right triangle formed by ∠A, sides c and h: sin(A) = opp hyp = h c which implies h = csin(A) In case 2, looking at the right triangle formed by ∠P, sides c and h: sin(P) = opp hyp = h c which implies h = csin(P) But, P = 180˚ – A, so sin(P) = sin(180˚ – A) (use the difference of angles formula) = sin(180˚)cos(A) – cos(180˚)sin(A) (but sin(180˚) = 0 and cos(180˚) = – 1) = 0•cos(A) – (– 1)sin(A) = sin(A) which implies h = csin(P) = csin(A) Thus, in both cases, h = csin(A) This means that asin(C) = h = csin(A). asin(C) = csin(A) (divide by ac) sin(C) c = sin(A) a sin(C) sin(B) = . c b sin(C) for any oblique c Similarly, we can show that Thus, sin(A) a = sin(B) b = triangle. The Law of Sines Theorem For an oblique triangle ABC with opposites of length a, b, and c respectively, sin(A) a = sin(B) b = sin(C) c There are two cases were we will use the Law of Sines. The first case is when we know one side and two angles (ASA or SAA). The second case is when we know two sides and the one angle that is opposite of either the two known sides (SSA). As we solve for the various sides of an oblique triangle, it is important to always use the given information if possible rather than rounded value of something we have already found. Objective 2: Solve SAA or ASA Triangles (Case 1) 139 We will solve a couple examples where two angles and a side of the triangle are given. Solve the following triangle: Ex. 1 A = 50˚, C = 20˚, a = 3 m Solution: We will begin by sketching a picture B of a triangle. Since the sum of the c a=3m angles of a triangle is 180˚, we can find angle B: A = 50˚ C = 20˚ B = 180˚ – 50˚ – 20˚ = 110˚ b Since we know angle A and side a, (SAA) we will use those values on one side of the Laws of Sines: To find c: To find b: sin(A) a = sin(C) c sin(A) a = sin(B) b Plug in the known values, cross multiply and solve: sin(50o ) 3 = sin(20o ) c sin(50˚)c = 3sin(20˚) c= o 3sin(20 ) sin(50o ) sin(50o ) 3 = sin(110o ) b sin(50˚)b = 3sin(110˚) b= 3sin(110o ) sin(50o ) c ≈ 1.3394… ≈ 1.34 m b ≈ 3.6800… ≈ 3.68 m Thus, B = 110˚, c ≈ 1.34 m, and b ≈ 3.68 m Ex. 2 A = 78˚, B = 41˚, c = 11 in Solution: We will begin by sketching a picture B = 41˚ of a triangle. Since the sum of the c = 11 in a angles of a triangle is 180˚, we can find angle C: A = 78˚ C C = 180˚ – 78˚ – 41˚ = 61˚ b Since we know angle C and side c, (ASA) we will use those values on one side of the Laws of Sines: To find a: To find b: sin(A) a = sin(C) c sin(B) b = sin(C) c Plug in the known values, cross multiply and solve: 140 sin(78o ) a = sin(61o ) 11 sin(41o ) b sin(61˚)a = 11sin(78˚) a= = sin(61o ) 11 sin(61˚)b = 11sin(41˚) 11sin(78o ) b= sin(61o ) 11sin(41o ) sin(61o ) a ≈ 12.3020… ≈ 12.30 in b ≈ 8.2511… ≈ 8.25 in Hence, C = 61˚, a ≈ 12.3 in, and b ≈ 8.25 in. Objective 3: Solve SSA Triangles (Case 2). In solving oblique triangles that fit into second case when we know two sides and the one angle that is opposite of either the two known sides, there are several possible solutions. Consider ∆ABC where the sides a b a and b and ∠A are given. Let h = h height of the triangle (dashed line): A In examining the right triangle formed by ∠A, side b and h. Since sin(A) = opp hyp , then sin(A) = h b which implies that h = bsin(A). The length of side a dictates how many possible triangles we get for our solution. No Triangle Case If a < h = bsin(A), then a is not long enough to connect to the base (side c). Thus, no triangle can be formed: b h a A In this case, if we use the Law of Sines, we would get sin(B) > 1 which is impossible since the range of the sine function is [– 1, 1]. One Right Triangle Case If a = h = bsin(A), then a is just barely long enough to connect to the base (side c) at a right angle. Thus, one right triangle can be formed: b a=h A In this case, if we use the Law of Sines, we would get sin(B) = 1 which would tell us that B is a 90˚ angle. 141 Two Triangle Case If h = bsin(A) < a < b, then a is long enough to connect to the base (side c) at two different points. Thus, there two triangles that can be formed: b a One Oblique Triangle Case If a ≥ b then a is long enough to connect to the base (side c) at one point, but too long to connect to the base at another point. Thus, only one triangle can be formed: h a b A h a A In this case, when we use the Law of Sines, we will find an angle in the first quadrant and one in the second quadrant that will satisfy the triangle giving us two possible answers. In this case, when we use the Law of Sines, we will find an angle only in the first quadrant that will satisfy the triangle. The angle in the second quadrant will give us a sum of angles larger than 180˚. Solve the following triangle: Ex. 3 a = 2, c = 3, ∠A = 36˚ Solution: We will start by making a sketch of the triangle. Since we know ∠A and side a, we will use those values on one side of the Laws of Sines: To find ∠C: sin(A) sin(C) = (plug a c sin(36o ) sin(C) = 2 3 3sin(36o ) sin(C) = 2 3sin(36o ) = C = sin – 1 2 ( In quadrant I, C ≈ 61.85˚ ) c=3 B A = 36˚ a=2 C b in the known values, cross multiply and solve) (find the inverse sine) sin – 1(0.88167…) = 61.845432…˚ In quadrant II, C = 180˚ – 61.845432…˚ = 118.154567…˚ ≈ 118.15˚ Both angles are possible, so we have two possible triangles. 142 36˚ 61.85˚ Triangle 1: C ≈ 61.85˚ B = 180˚ – 36˚ – 61.845432…˚ = 82.154567… ≈ 82.15˚ Use the Laws of Sines to find b: sin(A) sin(B) = a b o sin(36 ) sin(82.154567...o ) = 2 b sin(36˚)b = 2sin(82.154567…˚) b= 2sin(82.154567...o ) sin(36o ) = 3.370754… ≈ 3.37 Hence, C = 61.85˚, B = 82.15˚ and b ≈ 3.37. Ex. 4 118.15˚ 36˚ Triangle 2: C ≈ 118.15˚ B = 180˚ – 36˚ – 118.154567…˚ = 25.845432…˚ ≈ 25.85˚ Use the Laws of Sines to find b: sin(A) sin(B) = a b o sin(36 ) sin(25.845432...o ) = 2 b sin(36˚)b = 2sin(25.845432…˚) b= 2sin(25.845432...o ) sin(36o ) = 1.483347… ≈ 1.48 Hence, C = 118.15˚, B = 25.85˚ and b ≈ 1.48. a = 9 m, b = 5 m, ∠A = 81˚ Solution: We will start by making a sketch of the triangle. Since we know ∠A and side a, we will use those values on one side of the Laws of Sines: To find ∠B: c B A = 81˚ C b = 5m sin(A) sin(B) = (plug in the known values, cross multiply a b sin(81o ) sin(B) = 9 5 5sin(81o ) (find the inverse sine) sin(B) = 9 5sin(81o ) = sin – 1(0.548715…) = 33.278952…˚ B = sin – 1 9 ( In quadrant I, B ≈ 33.28˚ a=9m and solve) ) In quadrant II, B = 180˚ – 33.278952…˚ = 146.721047…˚ ≈ 146.72˚ The second angle, 146.72˚ is not possible since the sum of 146.72˚ and 81˚ exceeds 180˚. 143 Triangle 1: B ≈ 33.28˚ C = 180˚ – 81˚ – 33.278952…˚ = 65.721047…˚ ≈ 65.72˚ Use the Laws of Sines to find c: sin(A) sin(C) = a c o sin(81 ) sin(65.721047...o ) = 9 c sin(81˚)c = 9sin(65.721047…˚) c= 9sin(65.721047...o ) sin(81o ) = 8.306253… ≈ 8.31 m Hence, B = 33.28˚, C = 65.72˚ and c ≈ 8.31 m. Ex. 5 a = 3 ft, c = 1.6 ft, ∠C = 54˚ Solution: We will start by making a sketch of the triangle. Since we know ∠C and side c, we will use those values on one side of the Laws of Sines: To find ∠A: sin(A) a sin(A) 3 c = 1.6 ft A B a = 3 ft 54˚ = C b sin(C) (plug in the known values, cross multiply and c sin(54o ) = 1.6 3sin(54o ) (find the inverse sine) sin(A) = 1.6 3sin(54o ) = sin – 1(1.516906…) which is undefined. A = sin – 1 1.6 = ( solve) ) Thus, there is no triangle with such measurements. Objective 4: Solve Applications. Solve the following: Ex. 5 An astronomer on Earth measures the angle between the Sun and Venus to be 22.4˚. If the Earth is approximately 93 million miles from the Sun and Venus is approximately 67 million miles from the sun, how far away is Venus from the Earth? Solution: We will first need to draw a picture of the situation: 144 Venus x V 67 million miles 22.4˚ 93 million miles S Earth Sun Since we know 22.4˚ and the opposite side (67 million miles), we will plug in those values on one side of the Law of Sines and plug in 93 million miles in on the other to find the angle on top: sin(22.4o ) 67 sin(V) 93 93sin(22.4o ) sin(V) = 67 93sin(22.4o ) V = sin – 1 67 = ( (cross multiply and solve) (find the inverse sine) ) = sin –1 (0.528948…) = 31.934432…˚ In quadrant I, V ≈ 31.93˚ In quadrant II, V = 180˚ – 31.934432…˚ = 148.06556…˚ ≈ 148.07˚ Both angles are possible, so we have two possible triangles. Triangle 1: V ≈ 31.93˚ Triangle 2: V ≈ 148.07˚ S = 180˚ – 22.4˚ – 31.934432…˚ S = 180˚ – 22.4˚ – 148.06556…˚ = 125.665567… ≈ 125.67˚ = 9.534432…˚ ≈ 9.53˚ Use the Laws of Sines to find ?: Use the Laws of Sines to find ?: sin(22.4o ) sin(125.665567...o ) = 67 x 67sin(125.665567...o ) x= sin(22.4o ) sin(22.4o ) sin(9.534432...o ) = 67 x o 67sin(9.534432... ) x= sin(22.4o ) = 142.84259… ≈ 142.84 = 29.122965… ≈ 29.12 Thus, Venus is either 142.84 million miles or 29.12 million miles from the Earth. Ex. 6 Lighthouse #1 is located 150 miles directly north of Lighthouse #2. Each lighthouse receives a distress signal from a ship out at sea. If the signal indicates that the ship has a bearing of S48˚W from lighthouse #1 and a bearing of N32˚W from lighthouse #2, how far is the ship from each lighthouse? Solution: 145 We will begin by draw a picture of the situation. We will first need to find angle B: C = 180˚ – 48˚ – 32˚ = 100˚ Since we know angle C and the opposite side (150 miles) , we can plug those values in on one side of the Laws of Sines: a 48˚ C 150 miles b To find a: sin(A) a = 32˚ To find b: sin(C) c sin(B) b = sin(C) c Plug in the known values, cross multiply and solve: sin(32o ) sin(100o ) = a 150 o 150sin(32 ) a= sin(48o ) sin(100o ) = b 150 o 150sin(48 ) b= sin(100o ) sin(100o ) a ≈ 80.7141… ≈ 80.71 miles b ≈ 113.19135… ≈ 113.19 miles The ship is ≈ 80.71 miles from lighthouse #1 and ≈ 113.19 miles from lighthouse #2. Prove the following: Ex. 7 Mollweide's Theorem: For any triangle, a+b c = 1 2 cos[ (A −B)] 1 2 sin( C) Solution: We will begin by using the Law of Sines to build the left side: a : c sin(A) sin(C) = a c a sin(A) = c sin(C) a b a+b add c + c = c : a+b sin(A) sin(B) = + c sin(C) sin(C) b : c sin(B) sin(C) = b c b sin(B) = c sin(C) Solve for Solve for Now, sin(A) + sin(B) sin(C) α +β α −β Use the formula (sin(α) + sin(β) = 2sin cos 2 2 1 numerator and rewrite C as 2( 2 C) in the denominator. = ( ) ( )) on the 146 A +B = A −B 2 sin( 2 ) cos( 2 ) 1 (use the double angle formula for sine) sin( 2( 2 C )) A +B = A −B 2 sin( 2 ) cos( 2 ) 1 1 (reduce) 2 sin( 2 C ) cos( 2 C ) A +B = A −B sin( 2 ) cos( 2 ) 1 1 (since A + B + C = 180˚, then A + B = 180˚ – C) sin( 2 C ) cos( 2 C ) = = = = 180o −C A −B ) cos( 2 ) 2 (simplify) 1 1 sin( 2 C ) cos( 2 C ) 1 A −B sin( 90 0 − 2 C ) cos( 2 ) (but, sin(90˚ 1 1 sin( 2 C ) cos( 2 C ) 1 A −B cos( 2 C ) cos( 2 ) (reduce) 1 1 sin( 2 C ) cos( 2 C ) 1 A−B cos[ (A − B)] cos( 2 ) 2 = . 1 1 C) sin( sin( 2 C ) 2 sin( – θ) = cos(θ))