EXAMPLE 1 Solve a triangle for the AAS

EXAMPLE 1 Solve a triangle for the AAS or ASA case ABC with C = 107°, B = 25°, and b = 15. SOLUTION First find the angle: A = 180° – 107° – 25° = 48°. By the law of sines, you can write a 15 c = = sin 48° sin 25° sin 107° a c 15 Write two equations, each 15 = = sin 48° sin 25° with one variable. sin 107° sin 25°

EXAMPLE 1 Solve a triangle for the AAS or ASA case 15 sin 48° 15 sin 107° a = sin 25° Solve for each variable. c = sin 25° a 26. 4 c Use a calculator. ANSWER In ABC, A = 48°, a 26. 4, and c 33. 9

EXAMPLE 2 Solve the SSA case with one solution ABC with A = 115°, a = 20, and b = 11. SOLUTION First make a sketch. Because A is obtuse and the side opposite A is longer than the given adjacent side, you know that only one triangle can be formed. Use the law of sines to find B.

EXAMPLE 2 Solve the SSA case with one solution sin B sin 115° = 20 11 11 sin 115° sin B = 20 B = 29. 9° Law of sines 0. 4985 Multiply each side by 11. Use inverse sine function. You then know that C 180° – 115° – 29. 9° = 35. 1°. Use the law of sines again to find the remaining side length c of the triangle.

EXAMPLE 2 Solve the SSA case with one solution c 20 Law of sines sin 35. 1° = sin 115° 20 sin 35. 1° c = Multiply each side by sin 35. 1°. sin 115° c 12. 7 Use a calculator. ANSWER In ABC, B 29. 9°, C 35. 1°, and c 12. 7.

EXAMPLE 3 Solve Examine the SSA case with no solution ABC with A = 51°, a = 3. 5, and b = 5. SOLUTION Begin by drawing a horizontal line. On one end form a 51° angle (A) and draw a segment 5 units long (AC , or b). At vertex C, draw a segment 3. 5 units long (a). You can see that a needs to be at least 5 sin 51° 3. 9 units long to reach the horizontal side and form a triangle. So, it is not possible to draw the indicated triangle.

EXAMPLE 4 Solve the SSA case with two solutions ABC with A = 40°, a = 13, and b = 16. SOLUTION First make a sketch. Because b sin A = 16 sin 40° 10. 3, and 10. 3 < 16 (h < a < b), two triangles can be formed. Triangle 1 Triangle 2

EXAMPLE 4 Solve the SSA case with two solutions Use the law of sines to find the possible measures of B. sin B sin 40° = 16 13 16 sin 40° sin B = 13 Law of sines 0. 7911 Use a calculator. There are two angles B between 0° and 180° for which sin B 0. 7911. One is acute and the other is obtuse. Use your calculator to find the acute angle: sin– 1 0. 7911 52. 3°. The obtuse angle has 52. 3° as a reference angle, so its measure is 180° – 52. 3° = 127. 7°. Therefore, B 52. 3° or B 127. 7°.

EXAMPLE 4 Solve the SSA case with two solutions Now find the remaining angle C and side length c for each triangle. Triangle 1 C Triangle 2 180° – 40° – 52. 3° = 87. 7° 13 c = sin 87. 7° sin 40° 13 sin 87. 7° c = sin 40° ANSWER In Triangle 1, B 52. 3°, C 87. 7°, and c 20. 2. C 180° – 40° – 127. 7° = 12. 3° 13 c = sin 12. 3° sin 40° 20. 2 13 sin 12. 3° c = sin 40° 4. 3 ANSWER In Triangle 2, B 127. 7°, C 12. 3°, and c 4. 3.

EXAMPLE 5 Find the area of a triangle Biology Black-necked stilts are birds that live throughout Florida and surrounding areas but breed mostly in the triangular region shown on the map. Find the area of this region.

EXAMPLE 5 Find the area of a triangle SOLUTION The area of the region is: Area = 1 bc sin A 2 Write area formula. 1 (125) (223) sin 54. 2° Substitute. = 2 11, 300 Use a calculator. ANSWER The area of the region is about 11, 300 square miles.

EXAMPLE 1 Solve a triangle for the SAS case ABC with a = 11, c = 14, and B = 34°. SOLUTION Use the law of cosines to find side length b. b 2 = a 2 + c 2 – 2 ac cos B Law of cosines b 2 = 112 + 142 – 2(11)(14) cos 34° Substitute for a, c, and B. b 2 61. 7 Simplify. b 2 61. 7 7. 85 Take positive square root.

EXAMPLE 1 Solve a triangle for the SAS case Use the law of sines to find the measure of angle A. sin A sin B Law of sines = a b sin 34° sin A = 7. 85 11 11 sin 34° sin A = 7. 85 A sin – 1 0. 7836 Substitute for a, b, and B. 0. 7836 Multiply each side by 11 and Simplify. 51. 6° Use inverse sine. The third angle C of the triangle is C 94. 4°. ANSWER In ABC, b 7. 85, A 180° – 34° – 51. 6° = 51. 68, and C 94. 48.

EXAMPLE 2 Solve a triangle for the SSS case ABC with a = 12, b = 27, and c = 20. SOLUTION First find the angle opposite the longest side, AC. Use the law of cosines to solve for B. b 2 = a 2 + c 2 – 2 ac cos B Law of cosines 272 = 122 + 202 – 2(12)(20) cos B Substitute. 272 = 122 + 202 = cos B – 2(12)(20) – 0. 3854 cos B B cos – 1 (– 0. 3854) Solve for cos B. Simplify. 112. 7° Use inverse cosine.

EXAMPLE 2 Solve a triangle for the SSS case Now use the law of sines to find A. sin A sin B = a b Law of sines sin A sin 112. 7° = 12 27 Substitute for a, b, and B. 12 sin 112. 7° sin A = 27 A sin– 1 0. 4100 24. 2° Multiply each side by 12 and simplify. Use inverse sine. The third angle C of the triangle is C 43. 1°. ANSWER In ABC, A 24. 2, B 180° – 24. 2° – 112. 7° = 112. 7, and C 43. 1.

EXAMPLE 3 Use the law of cosines in real life Science Scientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to 180°, the more efficiently the organism walked. The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B.

EXAMPLE 3 Use the law of cosines in real life SOLUTION Law of cosines b 2 = a 2 + c 2 – 2 ac cos B 3162 = 1552 + 1972 – 2(155)(197) cos B Substitute. 3162 = 1552 + 1972 = cos B – 2(155)(197) B Solve for cos B. – 0. 6062 cos B Simplify. cos – 1 (– 0. 6062) 127. 3° Use inverse cosine. ANSWER The step angle B is about 127. 3°.

EXAMPLE 4 Solve a multi-step problem Urban Planning The intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown. SOLUTION STEP 1 Find the semiperimeter s. s = 1 (a + b + c ) = 1 (170 + 240 + 350) = 380 2 2

EXAMPLE 4 Solve a multi-step problem STEP 2 Use Heron’s formula to find the area of Area = ABC. s (s – a) (s – b) (s – c) = 380 (380 – 170) (380 – 240) (380 – 350) 18, 300 The area of the traffic triangle is about 18, 300 square yards.