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Solve Triangle with Given Angle and Side Lengths

This text provides step-by-step solutions for solving triangle problems using the laws of sines and cosines. It includes examples for the AAS, ASA, SSA, and SSS cases.

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Solve Triangle with Given Angle and Side Lengths

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

  2. 15 sin 107° a = c = sin 25° a c 26.4 33.9 InABC,A = 48°,a 26.4,andc 33.9. ANSWER 15 sin 48° sin 25° EXAMPLE 1 Solve a triangle for the AAS or ASA case Solve for each variable. Use a calculator.

  3. Solve ABCwithA = 115°,a = 20, andb = 11. EXAMPLE 2 Solve the SSA case with one solution SOLUTION First make a sketch. Because Ais 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.

  4. = 11 sin 115° 0.4985 sin B = 20 29.9° B = You then know that C 180° – 115° – 29.9° = 35.1°. Use the law of sines again to find the remaining side length cof the triangle. sinB sin 115° 20 11 EXAMPLE 2 Solve the SSA case with one solution Law of sines Multiply each side by 11. Use inverse sine function.

  5. = c = c c sin 35.1° 12.7 ANSWER 20 InABC,B 29.9°,C 35.1°,andc 12.7. sin 115° 20 sin 35.1° sin 115° EXAMPLE 2 Solve the SSA case with one solution Law of sines Multiply each side by sin 35.1°. Use a calculator.

  6. Solve ABCwithA = 51°,a = 3.5, and b = 5. 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 aneeds to be at least 5sin51°3.9 units long to reach the horizontal side and form a triangle. So, it is not possible to draw the indicated triangle. EXAMPLE 3 Examine the SSA case with no solution SOLUTION

  7. Solve ABCwithA = 40°,a = 13, and b = 16. First make a sketch. Because bsinA = 16sin 40°10.3, and 10.3 < 13 < 16(h < a < b), two triangles can be formed. Triangle 1 Triangle 2 EXAMPLE 4 Solve the SSA case with two solutions SOLUTION

  8. sinB = 16 16 sin 40° sin 40° 0.7911 = sinB 13 13 There are two angles Bbetween 0° and 180° for which sinB 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 Use the law of sines to find the possible measures of B. Law of sines Use a calculator.

  9. Triangle 1 Triangle 2 C 180° – 40° – 52.3° = 87.7° C 180° – 40° – 127.7° = 12.3° c c = = sin 87.7° sin 12.3° 13 sin 12.3° 13 sin 87.7° c 20.2 c 4.3 = = sin 40° sin 40° 13 13 sin 40° sin 40° In Triangle 1,B 52.3°, C 87.7°, In Triangle 2,B 127.7°, C 12.3°, ANSWER ANSWER andc 20.2. andc 4.3. EXAMPLE 4 Solve the SSA case with two solutions Now find the remaining angle Cand side length c for each triangle.

  10. 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 Biology

  11. Area = bc sin A (125) (223) sin 54.2° = 11,300 1 1 2 2 ANSWER The area of the region is about 11,300 square miles. EXAMPLE 5 Find the area of a triangle SOLUTION The area of the region is: Write area formula. Substitute. Use a calculator.

  12. Solve ABCwith a = 11,c = 14, and B = 34°. b2 61.7 b2 61.7 7.85 EXAMPLE 1 Solve a triangle for the SAS case SOLUTION Use the law of cosines to find side length b. b2 = a2 + c2 – 2ac cosB Law of cosines b2 = 112 + 142 – 2(11)(14) cos 34° Substitute for a, c, and B. Simplify. Take positive square root.

  13. = sin 34° sinA = 7.85 11 sin B sin A 0.7836 sin A = a b A sin–1 0.7836 51.6° 11 sin 34° The third angle Cof the triangle isC180° – 34° – 51.6° = 94.4°. 7.85 InABC,b 7.85,A 51.68,andC 94.48. ANSWER EXAMPLE 1 Solve a triangle for the SAS case Use the law of sines to find the measure of angle A. Law of sines Substitute for a, b, and B. Multiply each side by 11 and Simplify. Use inverse sine.

  14. SolveABC witha = 12,b = 27, and c = 20. First find the angle opposite the longest side, AC. Use the law of cosines to solve for B. 272 = 122 + 202 cosB = – 2(12)(20) – 0.3854cosB B cos –1 (– 0.3854) 112.7° EXAMPLE 2 Solve a triangle for the SSS case SOLUTION b2 = a2 + c2 – 2ac cosB Law of cosines 272 = 122 + 202 – 2(12)(20) cosB Substitute. Solve for cosB. Simplify. Use inverse cosine.

  15. sin B sin A = b a sin A = 12 0.4100 sin A = sin 112.7° 12 sin 112.7° 27 The third angle Cof the triangle isC180° – 24.2° – 112.7° = 43.1°. 27 A sin–1 0.4100 24.2° InABC,A 24.2,B 112.7,andC 43.1. ANSWER EXAMPLE 2 Solve a triangle for the SSS case Now use the law of sines to find A. Law of sines Substitute for a, b, and B. Multiply each side by 12 and simplify. Use inverse sine.

  16. 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 to180°, the more efficiently the organism walked. EXAMPLE 3 Use the law of cosines in real life Science The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B.

  17. 3162 = 1552 + 1972 cosB = – 2(155)(197) – 0.6062 cos B B cos –1 (– 0.6062) 127.3° ANSWER The step angle Bis about 127.3°. EXAMPLE 3 Use the law of cosines in real life SOLUTION b2 = a2 + c2 – 2ac cosB Law of cosines 3162 = 1552 + 1972 – 2(155)(197) cosB Substitute. Solve for cos B. Simplify. Use inverse cosine.

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

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

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