Solving Triangles and the Law of Cosines In this section we work out the law of cosines from our earlier identities and then practice applying this new identity. c2 = a2 + b2 − 2ab cos C: (1) Elementary Functions Draw the triangle 4ABC on the Cartesian plane with the vertex C at the Part 5, Trigonometry origin. Lecture 5.4a, The Law of Cosines Dr. Ken W. Smith Sam Houston State University 2013 In the drawing sin C = y and cos C = x : We may relabel the x and y Smith (SHSU) Elementary Functions 2013 1 / 22 Smith (SHSU) b Elementary Functionsb 2013 2 / 22 coordinates of A(x; y) as x = b cos C and y = b sin C: Solving Triangles and the Law of Cosines One Angle and the Law of Cosines We get information if we compute c2: By the Pythagorean theorem, c2 = (y2) + (a − x)2 = (b sin C)2 + (a − b cos C)2 = b2 sin2 C + a2 − 2ab cos C + b2 cos2 C: c2 = a2 + b2 − 2ab cos C: We use the Pythagorean identity to simplify b2 sin2 C + b2 cos2 C = b2 and so It is straightforward to use the law of cosines when we know one angle and c2 = a2 + b2 − 2ab cos C 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. Smith (SHSU) Elementary Functions 2013 3 / 22 Smith (SHSU) Elementary Functions 2013 4 / 22 A Worked Problem. A Worked Problem. Solve the triangle C = 32◦; a = 100 feet; b = 150 feet: ◦ Solution. If C = 32 ; a = 100 feet; b = 150 feet then, by the law of We solve the triangle C = 32◦; a = 100 feet; b = 150 feet: 2 2 2 cosines (c = a + b − 2ab cos C), We found c2 = (100)2 + (150)2 − 2(100)(150) cos(32◦) ≈ 7059: A ≈ sin−1(0:63007) ≈ 39:1◦ or A ≈ 180 − 39:1 = 140:9◦: So p c ≈ 7059 ≈ 84:018 feet. Now we apply the law of sines: The second answer is too big, so A = 39:1◦: sin 32◦ sin A sin B Since sin B = 0:9461 then either = = 84:018 100 150 B ≈ sin−1(0:94510) ≈ 71:1◦ or B = 180 − 71:1 = 108:9◦: This forces 100 sin 32◦ sin A = ≈ 0:6307 ◦ 84:018 The second answer, B = 108:9 , makes perfect sense because the angles and of the triangle need to add up to 180 degrees. 150 sin 32◦ sin B = ≈ 0:9461 84:018 If the sine of A is 0:6307 then either SmithA (SHSU)≈ sin−1(0:63007) ≈Elementary39:1◦ or FunctionsA ≈ 180 − 39:1 = 140:9◦:2013 5 / 22 Smith (SHSU) Elementary Functions 2013 6 / 22 The law of cosines and SSS The law of cosines and SSS Solve the triangle a = 30 feet; b = 20 feet; c = 15 feet: We solve the 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 We found tells us that A = 62:7◦ or A = 180◦ − 62:7◦ = 117:3◦ 152 − 302 − 202 = cos C −2(30)(20) Also 4 so sin B = sin 26:38◦ −1075 43 3 = = cos C −1200 48 so Therefore B = 36:33◦ or 180◦ − 36:33◦ = 143:67◦ 43 C = cos−1( ) ≈ 26:38◦ : 48 The second result here is too large. By the law of sines ◦ sin 26:38◦ sin A sin B Our final answer, after checking that angles add up to 180 , is = = ; 15 30 20 A = 117:3◦;B = 36:3◦;C = 26:4◦: so sin A = 2 sin 26:38◦: and so ◦ ◦ ◦ ◦ Smith (SHSU) A = 62:7 or AElementary= 180 Functions− 62:7 = 117:3 2013 7 / 22 Smith (SHSU) Elementary Functions 2013 8 / 22 Some Worked Problems. Some Worked Problems. A triangle has angle C = 20◦ and sides a = 10; b = 20. Use the law of cosines to find the length of the side c. A triangle has angle C = 20◦ and sides a = 10; b = 20. Use the law of Solution. sines to find two possible values of the angle B (one in which B is acute and one in which B is obtuse.) c2 = 102 + 202 − 2(10)(20) cos 20◦ = 500 − 400 cos 20◦ ≈ 124:1 p Solution. The sine of B is 0:614 so B ≈ 37:88◦ or 142:12◦: So c ≈ 124:1 ≈ 11:14. Solve the triangle. Use the law of sines to find two possible values of the angle A (one in Solution. c ≈ 11:14;A ≈ 17:88◦ and B ≈ 142:12◦: 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◦: Smith (SHSU) Elementary Functions 2013 9 / 22 Smith (SHSU) Elementary Functions 2013 10 / 22 Solving Triangles and the Law of Cosines Solving Triangles and the Law of Cosines Solve the triangle with sides a = 15; b = 12; c = 10: Solve the triangles with sides a = 15; b = 12; c = 30: Solution. By the Law of Cosines, C = 41:65◦: Now use the Law of Sines Solution. Since one side is longer than the sum of the other two sides, to get no such triangle is possible. a = 15; b = 12; c = 10;A = 85:46◦;B = 52:89◦;C = 41:65◦. (No calculations are necessary here. If we did a calculation, we would eventually take the arccosine of a number larger than 1, which is impossible.) Smith (SHSU) Elementary Functions 2013 11 / 22 Smith (SHSU) Elementary Functions 2013 12 / 22 The Law of Cosines Elementary Functions In the next presentation, we will look at Heron's formula and the \Five Part 5, Trigonometry Guys", (five trig identities I give to my classes.) Lecture 5.4b, Heron's Formula, Five Guys (End) Dr. Ken W. Smith Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 13 / 22 Smith (SHSU) Elementary Functions 2013 14 / 22 The area of an oblique triangle The area of an oblique triangle Since a triangle is half of a parallelogram, its area is one-half of the 1 product of its base and height. K = ab sin C (3) Let K represent the area of a triangle. Looking at the drawing below, we 2 see that K = 1 ch: We can summarize this by saying that the area of a triangle is one-half of 2 the product of the sine of an angle and its neighboring sides. But earlier, in our proof of the Law of Sines, we solved for h and we wrote h = b sin A and h = a sin B: So we can substitute for h and write the area as 1 1 K = 2 cb sin A and K = 2 ac sin B Or we could call the known angle C and just write 1 K = ab sin C (2) 2 Smith (SHSU) Elementary Functions 2013 15 / 22 Smith (SHSU) Elementary Functions 2013 16 / 22 Heron's formula Heron's formula Let's take the area equation 1 K = 2 ab sin C and square both sides. If we know the three sides a; b and c then in theory, since the triangle is 2 1 2 2 2 fixed and we can compute the three angles, we should be able to compute K = 4 a b sin C: the area of the triangle. By the Pythagorean identity replace sin2 C by 1 − cos2 C A first step is the formula we found when we proved the Law of Sines: 2 1 2 2 2 K = 4 a b (1 − cos C): 1 and distribute K = ab sin C 2 2 1 2 2 1 2 2 2 K = 4 a b − 4 a b cos C: A succinct formula for the area of a triangle, given the three sides, was Now use the Law of Cosines in the form worked out long ago by Heron of Alexandria. a2+b2−c2 cos C = 2ab to replace cos2 C: 2 1 2 2 1 2 2 a2+b2−c2 2 K = 4 a b − ( 4 a b )( 2ab ) and simplify 2 1 2 2 1 2 2 2 2 Smith (SHSU) Elementary Functions 2013 17 / 22 Smith (SHSU) K = 4 a bElementary− ( 16 Functions)(a + b − c ) 2013 18 / 22 Heron's formula Summary & Five Guys Using the law of sine and the law of cosines, we worked out a formula for the area: Summary of our identities K2 = 1 a2b2 − ( 1 )(a2 + b2 − c2)2 4 16 We have come across two very useful identities that are easy to remember: 2 1 2 2 2 2 2 2 K = 16 (4a b − (a + b − c ) ) 2 Now, with a little bit of algebra ... (skipping a bunch of steps!) ... we can 1 The Pythagorean identity, cos2 θ + sin θ = 1 (and its two siblings), get this into the form and sin A sin B sin C 2 a+b+c −a+b+c a−b+c a+b−c 2 K = ( 2 )( 2 )( 2 )( 2 ) The Law of Sines, a = b = c . The perimeter of the triangle is a + b + c: Half of the perimeter, written There are five more identities that are very useful and if we know them (or a+b+c s = 2 is called the semiperimeter.