9.4 The Law of Cosines Objective To use the law of cosines to find unknown parts of a triangle.

Data Required for Solving Oblique Triangles Case 1One side and two angles known: »SAA or ASA Case 2Two sides and one angle not included between the sides known: »SSA  This case may lead to more than one solution. Ambiguous Case Case 3Two sides and one angle included between the sides known: »SAS Case 4Three sides are known: »SSS

Solving an SAS Triangle The Law of Sines was good for –ASA- two angles and the included side –SAA- two angles and any side –SSA- two sides and an opposite angle (being aware of possible ambiguity) Why would the Law of Sines not work for an SAS triangle? ° No side opposite from any angle to get the ratio

Deriving the Law of Cosines Write an equation using Pythagorean theorem for shaded triangle. b h a k c - k A B C c

Deriving the Law of Cosines Write an equation using Pythagorean theorem for shaded triangle. h b a k c A B C c + k  - A c + (– bcosA) c – bcosA

Law of Cosines Similarly Note the pattern Include Opposite S A S Use these to find missing sides

Law of Cosines When  C = 90 o, cosC = 0, then the Law of Cosines reduces to c 2 = a 2 + b 2, which indicates a right triangle. When  C 0, then c 2 < a 2 + b 2 by the amount 2abcosC. When  C > 90 o, cosC a 2 + b 2 by the amount –2abcosC.

If we solve the law of cosines for cos A, cos B, cos C, we obtain Law of Cosines S S S Use these to find missing angles

Choose the Method Solving an Oblique Triangle GivenBegin by Using Two angles and any side (SAA or ASA) Law of Sines Two sides and an angle opposite one of them (SSA) ( Ambiguous Case ) Law of Sines Two sides and their included angle (SAS) Law of Cosines Three sides (SSS) Law of Cosines

Using the law of cosines, we can easily identify acute and obtuse angles. The law of sines does not distinguish between obtuse and acute angles, however, because both types of angle have positive sine values. The Law of Sines and Cosines

Applying the Cosine Law Now use it to solve the triangle we started with Label sides and angles –Side c first ° A B C c Case 3Two sides and one angle included between the sides known: »SAS

Now calculate the angles –use b 2 = a 2 + c 2 – 2ac cos B and solve for B ° A B C c = 6.65 Applying the Cosine Law  o The remaining angle determined by subtraction  A = 180 o – o – 26 o = o o

Example 1: The leading edge of each wing of the B-2 Stealth Bomber measures feet in length. The angle between the wing's leading edges is o. What is the wing span (the distance from A to C)? Applying the Cosine Law: Wing Span A C [Solution 1] Not use the Law of Cosine  172 ft [Solution 2] Use the Law of Cosine  172 ft

Try It Out Determine the area of these triangles 127° ° 42.8° 17.9 a) b)  115 u 2 a A 60.9 o  97.9 u 2

Cost of a Lot Example 2: An industrial piece of real estate is priced at $4.15 per square foot. Find the cost of a triangular lot measuring 324 feet by 516 feet by 412 feet [Solution] Use the Law of Cosines to find the angle opposite to the shortest side. A Case 4Three sides are known: »SSS

Example 3: The lengths of the sides of a triangle are 5, 10, and 12. Solve the triangle. [Solution] Make a sketch Case 4Three sides are known: SSS Applying the Cosine Law

Example 4: In the diagram at the right, AB = 5, BD = 2, DC = 4, and CA = 7. Find AD. [Solution] First we apply the law of cosines to  ABC to find out the cosB: Then we apply the law of sines to  ABD to find out AD: Thus AD = 5.

19 20 o 53 y 2 = x 2 + z 2 – 2·x·z·cosY 19 2 = x – 2·x·53·cos20 o Law of Cosines - SSA x x 2 – 106·cos 20 o x + (53 2 – 19 2 ) = 0 Y Z X Example 5: In  XYZ,  Y = 20 o, z = 53, y = 19. Find YZ or x. Z 19 [Solution] This situation is SSA, which may have ambiguous cases. It seems we have to apply the law of sines. However, the question is only asks to find a side, YZ, or x. We then use the law of cosines

19 20 o 53 x Y Z X Z Law of Cosines - SSA Example 5: In  XYZ,  Y = 20 o, z = 53, y = 19. Find YZ or x.

Example 6: Find the three angles of the triangle. C BA Find the angle opposite the longest side first. Law of Sines: 36.3   26.4  [Solution]

Example 7: Solve the triangle  Law of Sines: 37.2  C BA  Law of Cosines: [Solution]

Example 8: Application Two ships leave a port at 9 A.M. One travels at a bearing of N 53  W at 12 mph, and the other travels at a bearing of S 67  W at 16 mph. How far apart will the ships be at noon? 53  67  c 36 mi 48 mi C At noon, the ships have traveled for 3 hours. Angle C = 180  – 53  – 67  = 60  The ships will be approximately 43 miles apart. 43 mi 60  N

Assignment P. 352 #1 – 18, 22