789 8.2 The Law of Cosines NOTE If we let C = 90° in the third form of the law of cosines, then cos C = cos 90° = 0, and the formula becomes c2 = a2 + b2. Pythagorean theorem The Pythagorean theorem is a special case of the law of cosines. Using the Law of Cosines b = 12.9 m a c = 15.4 m A B C 42.38 Figure 15 EXAMPLE 2 Applying the Law of Cosines (SAS) Solve triangle ABC if A = 42.3°, b = 12.9 m, and c = 15.4 m. SOLUTION See Figure 15. We start by finding side a with the law of cosines. a2 = b2 + c2 - 2bc cos A Law of cosines a2 = 12.92 + 15.42 - 2112.92115.42 cos 42.3° Substitute. a2 ≈109.7 Use a calculator. a ≈10.47 m Of the two remaining angles B and C, B must be the smaller because it is opposite the shorter of the two sides b and c. Therefore, B cannot be obtuse. sin A a = sin B b Law of sines (alternative form) sin 42.3° 10.47 = sin B 12.9 Substitute. sin B = 12.9 sin 42.3° 10.47 Multiply by 12.9 and rewrite. B ≈56.0° Use the inverse sine function. The easiest way to find C is to subtract the measures of A and B from 180°. C = 180° - A - B Angle sum formula, solved for C C≈180° - 42.3° - 56.0° Substitute. C≈81.7° Subtract. Take square roots and choose the positive root. EXAMPLE 1 Applying the Law of Cosines (SAS) A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that b = 259 m, a = 423 m, and angle ACB measures 132° 40′. Find the distance c. See Figure 14. SOLUTION We can use the law of cosines here because we know the lengths of two sides of the triangle and the measure of the included angle. c2 = a2 + b2 - 2ab cos C Law of cosines c2 = 4232 + 2592 - 21423212592 cos 132° 40′ Substitute. c2 ≈394,510.6 Use a calculator. c ≈628 The distance between the points is approximately 628 m. C b = 259 m a = 423 m A B c 1328 409 Figure 14 Take the square root of each side. Choose the positive root. S Now Try Exercise 39. S Now Try Exercise 19.
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