Dr. Neal, WKU MATH 117 The Law of Cosines When we have a generic scalene triangle with no right angle, then the Pythagorean Theorem no longer applies. Moreover, we no longer have a hypotenuse, so the right triangle trig formulas do not apply either. In this case, we generally label the sides of the triangle as a, b, c and the angles as A, B, and C. By convention, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. abcCAB The Law of Cosines allows us to find one side in terms of the other two sides and the angle in between those sides (Side-Angle-Side). c2= a2+ b2! 2 ab cos C b2= a2+ c2! 2 a c cos B a2= b2+ c2! 2 b c cos A (given a, b and !C) (given a, c and !B) (given b, c and !A) Note: If C = 90º, then we have a right triangle. Then cosC = 0 and so c2= a2+ b2. To prove the Law of Cosines for side c, first drop a perpendicular to side b from angle B as shown below. abcC abcdefC We now can apply the Pythagorean Theorem to the resulting two triangles. First, d2+ e2= a2, so that e2= a2! d2. Also, f = b ! d. We then have c2= e2+ f2= (a2! d2) + (b ! d)2= a2! d2+ b2! 2b d + d2= a2+ b2! 2 bd . But then cosC =da; thus, d = a cos C. So the result becomes c2= a2+ b2! 2 a b cos C.Dr. Neal, WKU Example 1. Find the remaining side of the triangle: 100º1512 Solution. Because we have Side-Angle-Side, we can use the Law of Cosines to find the third side which we will call c. Then c2= 152+ 122! 2 " 15 "12 " cos100º, so c = (152+ 122! 2 " 15 "12 " cos100º) ≈ 20.7729 Side-Side-Side If we have the three sides a triangle, then we can find any of the angles using the Law of Cosines. To solve for angle C, we have c2= a2+ b2! 2 a b cos C ! 2 ab cosC = a2+ b2! c2 ! cosC =a2+ b2! c22a b ! C = cos!1a2+ b2! c22a b" # $ $ % & ' ' Similarly, A = cos!1b2+ c2! a22 bc" # $ $ % & ' ' and B = cos!1a2+ c2! b22a c" # $ $ % & ' ' . Recall that cos!1" is always between 0º and 180º (or between 0 an π in radians), so we will always have one unique solution for any angle. Example 2. (i) Find angles A and B in the triangle below. (ii) Find the height of the triangle. (iii) Compute the area with 12! base ! height and with Heron’s Formula. 101520ABDr. Neal, WKU Solution. (i) We know that 202= 102+ 152! 2 " 10 " 15 " cos A. So then cos A =102+ 152! 2022 " 10 " 15 = !75300 and A = cos!1!75300" # $ % & ' ≈ 104.4775º Likewise, 102= 152+ 202! 2 " 15 " 20 " cos B. Thus, cos B =152+ 202! 1022 " 15 " 20=525600 and B = cos!1525600" # $ % & ' ≈ 28.955º (ii) Once we have angle A, then we have its supplement != 180º " A. But angle A and angle != 180º " A will have the same sine value (one is in the 1st Quadrant, the other is in the 2nd Quadrant). In the drawing below, we see that sin!=h10, so h = 10sin!= 10sin(180º "A) = 10sin A # 10 sin(104.4775º). 101520ABh! (iii) The area of the original triangle is then 12! 15 ! h "12! 15 ! 10 sin(104.4775º ) ≈ 72.61844173 sq. units. Using Heron’s Formula, we let s =10 + 15 + 202= 22. 5. Then the area is 22.5(22.5 ! 20)(22.5 ! 15)(22.5 ! 10) = 22.5 ! 2. 5 ! 7.5! 12.5 ≈ 72.61843774 The slight difference comes from rounding off angle A = cos!1!75300" # $ % & ' to 104.4775º. The answers would be identical if we use 12!15 !10 sin cos"1"75300# $ % & ' ( # $ % & ' ( .Dr. Neal, WKU Example 3. (i) Find the third side of the triangle below. (ii) Find angle B in the triangle. (iii) Find the area. 40º1622B Solution. (i) Call the third side c. Then c = a2+ b2! 2a b cosC = (162+ 222! 2 " 16 " 22 " cos40º ) ≈ 14.167 (ii) We now have 222! 162+ 14.1672" 2 # 16 #14.167 # cos B, which gives cos B !162+ 14.1672" 2222 # 16 #14.167 and B ! cos"1162+ 14.1672" 2222 #16 #14.167$ % & & ' ( ) ) ≈ 93.4519º (iii) We can find the height by sin(40º) =h16 which gives h = 16sin(40º). The area is then 12! 22 ! 16sin(40º) ≈ 113.13 square units. 40º1622h Using Heron’s Formula, we have s !16 + 22 + 14.1672= 26.0835 and Area ≈ 26.0835(26.0835 ! 16)(26.0835 ! 22)(26.0835 ! 14.167) ≈ 113.13 (no noticeable difference)Dr. Neal, WKU Example 4. Find the direct distance from Point A to Point C in the following cases: (i) From Point A, move 20 feet in a direction 25º South of West to Point B. Then from Point B, move 25 feet in a direction 30º East of South to Point C. (ii) From Point A, move 10 feet in a direction 40º East of North to Point B. Then from Point B, move 12 feet in a direction 20º South of East to Point C. Solution. (i) We have two sides 20' and 25', but we must find the angle in between and the apply the Law of Cosines to find the distance b from Point A to Point B. From the second drawing below, we see that this angle is 85º. Then the desired distance is b = (202+ 252! 2 " 20 " 25 " cos(85º)) ≈ 30.624 ft WS25ºS30º20'25'ABC AS30º20'25'60ºBC25º25º65ºb (ii) Similarly, we find the angle in between to be 180º – 50º – 20º = 110º. Then the distance is b = (102+ 122! 2 " 10 " 12 " cos(110º )) ≈ 18.0578 ft 40º20º10'12'ABC 40º20º10'