Download The Law of Sines: Determining Missing Angles and Sides in Triangles - Prof. David K. Neal and more Study notes Trigonometry in PDF only on Docsity! Dr. Neal, Spring 2009 MATH 117 The Law of Sines Besides Side-Angle-Side (SAS) and Side-Side-Side (SSS), there are two other congruence forms that completely determine a triangle. These are Angle-Side-Angle (ASA) and Side-Angle-Angle (SAA). In each case, the third angle can be determined because all the angles must sum to 180º. So with ASA and SAA, we have all three angles, but we must find the other two sides. The Law of Cosines will not apply because it requires two sides to find the third side. And we may not have a right triangle, so we usually can’t use right-triangle trig. In this case, we use the Law of Sines. We again label the sides of the triangle as a , b , c and the angles as A , B , and C . As usual, side a is opposite angle A , side b is opposite angle B , and side c is opposite angle C . Then, a sin A = b sin B = c sin C Law of Sines a b c C A B Example 1. Find the remaining sides in the following triangles: 10 85º 55º a c 100º 30º 40 w v (i) (ii) Solution. (i) First, let A = 85º and C = 55º. Then B = 180º – 85º – 55º = 40º. The given side is b = 10. To find sides a and c , we have a sin85º = 10 sin40º → a = 10 sin85º sin40º ≈ 15.498 c sin55º = 10 sin40º → c = 10 sin55º sin40º ≈ 12.7437 (ii) Let W = 30º and V = 180º – 100º – 30º = 50º. Then w sin30º = 40 sin100º = v sin50º → w = 40 sin30º sin100º ≈ 20.3 and v = 40 sin50º sin100º ≈ 31.114 Dr. Neal, Spring 2009 When we have Side-Angle-Side, then we can use the Law of Cosines to find the third side. But it may be easier to use the Law of Sines to find the other two angles. But now we use the “reciprocal” form: sin A a = sin B b = sin C c Example 2. Find the remaining side and angles in the triangle below: 30º 50 40 A B Solution. To find side c , we use the Law of Cosines: c2 = 402 + 502 − 2(40)(50)cos30º ; c = (402 + 502 − 2 × 40 × 50 × cos30º) ≈ 25.217 Now let c = 25.217, C = 30º, a = 50, and b = 40. Then sin A 50 = sin30º 25.217 = sin B 40 Then, sin A = 50 sin30º 25.217 and sin B = 40 sin30º 25.217 . But both of these equations have two solutions, a first quadrant angle and a second quadrant angle. However, the two shortest sides in a triangle must have opposite angles that are acute (less than 90º). Thus, angle B must be less than 90º . So B = sin−1 40 sin30º 25.217 ≈ 52.4776º. Finally, A ≈ 180º – 30º – 52.4776º = 97.5224º. Note: Using the inverse sine to solve for A , we have sin−1 50 sin30º 25.217 ≈ 82.478º which is not the correct value of A . We then need A = 180º – 82.478º to give an obtuse (second quadrant) angle.
