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5.4 Solving Triangles and the Law of Cosines in This Section We Work out the Law of Cosines Using Our Earlier Identities

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5.4 Solving Triangles and the Law of Cosines in This Section We Work out the Law of Cosines Using Our Earlier Identities 5.4 Solving Triangles and the Law of Cosines In this section we work out the law of cosines using our earlier identities. We then practice applying this new identity. 5.4.1 The law of cosines and its proof Draw the triangle 4ABC on the Cartesian plane with the vertex C at the origin. (We will do the case where C is acute; the situation in which C is obtuse is similar.) See Figure 5, below. A(x; y) = A(b cos C; b sin C) b c y C(0; 0) x a − x D(x; 0) B(a; 0) Figure 5. y x Since, in Figure 5, sin C = and cos C = , we may relabel the x and y coordinates of A(x; y) as b b x = b cos C and y = b sin C: Now, as it turns out, we get some information if we compute c2: By the Pythagorean theorem, c2 = (y2) + (a − x)2 = (b sin C)2 + (a − b cos C)2 = b2 sin C2 + a2 − 2ab cos C + b2 cos2 C2: We can use the Pythagorean identity to simplify b2 sin C2 + b2 cos2 C2 = b2 and so c2 = a2 + b2 − 2ab cos C (25) This is the Law of Cosines. 5.4.2 One Angle and the Law of Cosines It is straightforward to use the law of cosines when we know one angle and its two adjacent sides. This is the Side-Angle-Side (SAS) case, in which we may label the angle C and its two sides a and b and so we can solve for the side c. Or, if we have the Side-Side-Side (SSS) situation, in which we know all three sides, we can label one angle C and solve for that angle in terms of the sides a; b and c, using the law of cosines. A Worked Problem. Solve the following triangle: C = 32◦; a = 100 feet; b = 150 feet: Solution. If C = 32◦; a = 100 feet; b = 150 feet then, by the law of cosines (c2 = a2 + b2 − 2ab cos C), c2 = (100)2 + (150)2 − 2(100)(150) cos(32◦) ≈ 7059: So p c ≈ 7059 ≈ 84:018 feet. 215 Now we apply the law of sines: sin 32◦ sin A sin B = = 84:018 100 150 This forces 100 sin 32◦ sin A = ≈ 0:6307 84:018 and 150 sin 32◦ sin B = ≈ 0:9461 84:018 If the sine of A is 0:6307 then either A ≈ sin−1(0:63007) ≈ 39:1◦ or A ≈ 180 − 39:1 = 140:9◦: The second answer is too big, so A = 39:1◦: Since sin B = 0:9461 then either B ≈ sin−1(0:94510) ≈ 71:1◦ or B = 180 − 71:1 = 108:9◦: The second answer, B = 108:9◦ , makes perfect sense because the angles of the triangle need to add up to 180 degrees. 5.4.3 The law of cosines and SSS Solve the following triangle: a = 30 feet; b = 20 feet; c = 15 feet: Solution. If a = 30 feet; b = 20 feet; c = 15 feet then the law of cosines tells us that 152 − 302 − 202 = cos C −2(30)(20) so −1075 43 = = cos C −1200 48 Therefore 43 C = cos−1( ) ≈ 26:38◦ : 48 By the law of sines sin 26:38◦ sin A sin B = = ; 15 30 20 so sin A = 2 sin 26:38◦: and so A = 62:7◦ or A = 180◦ − 62:7◦ = 117:3◦ Also 4 sin B = sin 26:38◦ 3 so B = 36:33◦ or 180◦ − 36:33◦ = 143:67◦ The second result here is too large. Our final answer, after checking that angles add up to 180◦, is A = 117:3◦;B = 36:3◦;C = 26:4◦: 216 Some Worked Problems. 1. A triangle has angle C = 20◦ and sides a = 10; b = 20. (a) Use the law of cosines to find the length of the side c. Solution. c2 = 102 + 202 − 2(10)(20) cos 20◦ = 500 − 400 cos 20◦ ≈ 124:1 p So c ≈ 124:1 ≈ 11:14. (b) Use the law of sines to find two possible values of the angle A (one in which A is acute and one in which A is obtuse.) Solution. The sine of A is 0:307 so A = 17:88◦ or 162:12◦: (c) Use the law of sines to find two possible values of the angle B (one in which B is acute and one in which B is obtuse.) Solution. The sine of B is 0:614 so B ≈ 37:88◦ or 142:12◦: (d) Solve the triangle. Solution. c ≈ 11:14;A ≈ 17:88◦ and B ≈ 142:12◦: 2. a = 15; b = 12; c = 10: Solution. By the Law of Cosines, C = 41:65◦: Now use the Law of Sines to get a = 15; b = 12; c = 10;A = 85:46◦;B = 52:89◦;C = 41:65◦. 3. a = 15; b = 12; c = 30: Solution. Since one side is longer than the sum of the other two sides, no such triangle is possible. (No calculations are necessary here.) 5.4.4 Heron's formula If we know the three sides a; b and c then in theory, since the triangle is fixed and we can compute the three angles, we should be able to compute the area of the triangle. A succinct formula for the area of a triangle, given the three sides, was worked out long ago by Heron of Alexadria. The perimeter of a polygon is the sum of the lengths of its sides. Thus the perimeter 1 of a triangle is represented by the sum a + b + c: Here we need the \semi-perimeter", s = (a + b + c), 2 half of the perimeter. Given the lengths a; b and c we can then compute the area K of the triangle as follows: K = ps(s − a)(s − b)(s − c) (26) This is Heron's formula. Proof of Heron's Formula. Heron's formula can be developed by using our earlier equation for the area of a triangle, found when we proved the Law of Sines: 1 K = ab sin C: 2 217 We combine this formula with the Law of Cosines and some algebra to get to Heron's formula. Here is how. First, let's square both sides and write 1 K2 = a2b2 sin2 C: 4 We replace sin2 C by 1 − cos2 C using a Pythagorean identity 1 K2 = a2b2(1 − cos2 C) 4 and multiply a2b2 through on the right side to achieve 1 1 K2 = a2b2 − a2b2 cos2 C (27) 4 4 Now, if we pause to look at the Law of Cosines and solve that equation for ab cos C we see that the Law of Cosines tells us that 1 ab cos C = (a2 + b2 − c2) (28) 2 and so 1 a2b2 cos2 C = (a2 + b2 − c2)2: 4 2 2 2 1 2 2 2 2 So we replace a b cos C by 4 (a + b − c ) to turn equation 27 into 1 1 1 K2 = a2b2 − (a2 + b2 − c2)2 = (4a2b2 − (a2 + b2 − c2)2) (29) 4 16 16 At this point a little algebra is necessary. Notice that 4a2b2 − (a2 + b2 − c2)2 is a difference of squares so 4a2b2 − (a2 + b2 − c2)2 = (2ab − a2 − b2 + c2)(2ab + a2 + b2 − c2): The right hand side can be rewritten to [c2 − (a − b)2][(a + b)2 − c2] and again we have difference of squares and so [c2 − (a − b)2][(a + b)2 − c2] = (c − (a − b))(c + (a − b))(a + b − c)(a + b + c): = (c − a + b)(c + a − b)(a + b − c)(a + b + c): If we return to equation 29 we have 1 K2 = (c − a + b)(c + a − b)(a + b − c)(a + b + c) (30) 16 and so, distributing the denominator 16 = 24 across the four terms, we can write this as −a + b + c a − b + c a + b − c a + b + c K2 = ( )( )( )( ) (31) 2 2 2 2 a+b+c At first glance this may seem a bit messy, but this can all be nicely expressed in terms of s = 2 : a+b+c Replacing 2 by s we have K2 = (s − a)(s − b)(s − c)s (32) Taking the square root of this equation gives us Heron's formula for the area of a triangle. K = p(s − a)(s − b)(s − c)s 218 5.4.5 Five Guys We have come across two very useful identities that are easy to remember: 1. The Pythagorean identity, cos2 θ + sin2 θ = 1 (and its two siblings), and sin A sin B sin C 2. The Law of Sines, = = . a b c There are five more identities that are very useful and if we know them (or have them handy) then the other trig identities follow easily from them. Our goal here is the understanding of mathematical concepts, not the memorization of \magical" formulas! But if one were to memorize any trig identities, in addition to the easy two above, I recommend the following five. 1. The formulas for the sine and cosine of the sum of two angles: cos(α + β) = cos α cos β − sin α sin β sin(α + β) = sin α cos β + cos α sin β 2. The power reduction formulas: 1 + cos(2α) cos2 α = 2 1 − cos(2α) sin2 α = 2 3. The law of cosines: c2 = a2 + b2 − 2ab cos C In my precalculus and trigonometry classes2, these five identities, boxed above, will be provided on quizzes and exams.
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