THE LAW OF SINES COSINES LAW OF SINES

THE LAW OF SINES & COSINES

LAW OF SINES The Law of Sines applies to AAS, ASA, SSA(special case). The Law of Sines In any triangle ABC, B a c A Copyright © 2009 Pearson Education, Inc. b C

EXAMPLE In , e = 4. 56, E = 43º, and G = 57º. Solve the triangle. Solution: Draw the triangle. We have AAS. Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution continued Find F: F = 180º – (43º + 57º) = 80º Use law of sines to find the other two sides. Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution continued We have solved the triangle. Copyright © 2009 Pearson Education, Inc.

Check for Understanding. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.

Check for Understanding. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.

Determine if the data supports 1 unique triangle, 2 triangles That are not congruent or 0 triangles. Copyright © 2009 Pearson Education, Inc.

25 Copyright © 2009 Pearson Education, Inc.

Determine if the data supports 1 unique triangle, 2 triangles That are not congruent or 0 triangles. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.

Determine if the data supports 1 unique triangle, 2 triangles That are not congruent or 0 triangles. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.

LAW OF COSINES The Law of Cosines In any triangle ABC, B a c A b C Thus, in any triangle, the square of a side is the sum of the squares of the other two sides, minus twice the product of those sides and the cosine of the included angle. When the included angle is 90º, the law of cosines reduces to the Pythagorean theorem. Copyright © 2009 Pearson Education, Inc.

WHEN TO USE THE LAW OF COSINES The Law of Cosines is used to solve triangles given two sides and the included angle (SAS) or given three sides (SSS). Copyright © 2009 Pearson Education, Inc.

EXAMPLE In !ABC, a = 32, c = 48, and B = 125. 2º. Solve the triangle. Solution: Draw and label a triangle. Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution continued Use the law of cosines to find the third side, b. We need to find the other two angle measures. We can use either the law of sines or law of cosines. Using the law of cosines avoids the possibility of the ambiguous case. So use the law of cosines. Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution continued Find angle A. Now find angle C. C ≈ 180º – (125. 2º + 22º) C ≈ 32. 8º Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solve !RST, r = 3. 5, s = 4. 7, and t = 2. 8. Solution: Draw and label a triangle. Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution continued Similarly, find angle R. Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution continued Now find angle T. T ≈ 180º – (95. 86º + 47. 80º) ≈ 36. 34º Copyright © 2009 Pearson Education, Inc.

THE AREA OF A TRIANGLE The area of any is one half the product of he lengths of two sides and the sine of the included angle: Copyright © 2009 Pearson Education, Inc.

EXAMPLE A university landscaping architecture department is designing a garden for a triangular area in a dormitory complex. Two sides of the garden, formed by the sidewalks in front of buildings A and B, measure 172 ft and 186 ft, respectively, and together form a 53º angle. The third side of the garden, formed by the sidewalk along Crossroads Avenue, measures 160 ft. What is the area of the garden to the nearest square foot? Copyright © 2009 Pearson Education, Inc.

EXAMPLE Solution: Use the area formula. The area of the garden is approximately 12, 775 ft 2. Copyright © 2009 Pearson Education, Inc.

Note: s is the “semi-perimeter. ”