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The law of sines cannot be used to solve a triangle where two sides and the included angle (SAS) are given or three sides (SSS) are given. In such cases we use the cosine rule which is defined as follows: A B C If b and c are two sides of a triangle and A, the included angle of these sides, then: a a 2 = b 2 + c 2 – 2 bc cos A b c The Law of Cosines If a and b are two sides and C the included angle: c 2 = a 2 + b 2 – 2 ab cos C If a and c are two sides and B the included angle: b 2 = a 2 + c 2 – 2 ac cos B

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Summary of methods for solving triangles. 1.One side and two angles known. (SAA or ASA) 1. Find the third angle (A + B + C = 180 o ) 2. Find the remaining sides using sine rule 2.Two sides and any one angle known (SSA) 1. Use sine law to find an angle. 2. Find the third angle (A + B + C = 180 o 3. Find the third side using sine rule. 3.Two sides and included angle known (SAS) 1. Find the third side using cosine rule. 2. Find the smaller of the remaining angles using sine rule. 3. Find the third angle (A + B + C). 4.Three sides are known (SSS) 1. Find the largest angle using cosine rule. 2.Use sine rule to find remaining two angles.

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Solve the triangle with C = 28.3 o, b = 5.71 cm, a = 4.21 cm to two decimal places. Two sides and the included angle are given. C B A We use the cosine rule: c c 2 = a 2 + b 2 – 2 ab cos C b a = – 2 × 4.21 × 5.71 cos 28.3= = 2.83 cm Now we use the sine rule to find angles A and B = A = sin -1 (0.7053) = o in the 1 st quadrant We dont need to solve for the second quadrant angle. B = 180 – ( o ) = o

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Solve the triangle with A = 67.3 o, b = 37.9 m, c = 40.8 m. Two sides and the included angle are given We use the cosine rule for the third side. a 2 = b 2 + c 2 – 2 bc cos A = – 2 × 37.9 × 40.8 cos 67.3= = 43.7 m Now we use the sine rule to find angles B and C = C = sin -1 (0.8613) = 59.5 o B = 180 – ( ) = 53.2 o

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Solve the triangle with B = o, a = 15.1 m, c = 19.2 m. Two sides and the included angle are given We use the cosine rule for the third side. b 2 = a 2 + c 2 – 2 ac cos B = – 2 × 15.1 × 19.2 cos 168.2= = 34.1 m Now we use the sine rule to find angles A and C = A = sin -1 (0.0906) = 5.2 o C = 180 – ( o ) = 6.6 o

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Solve the triangle with a = 3.0 m, b = 5.0 m, c = 6.0 m. Three sides are known We find C, the largest angle opposite to the largest side c = 6.0 m c 2 = a 2 + b 2 – 2 ab cos C2 ab cos C = a 2 + b 2 – c 2 = C = cos -1 ( ) = 94 o Use sine rule to find one of the remaining angles. = A= sin -1 (0.4989)= 30 o B = 180 – (93.8 o o )= 56 o

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Solve the triangle with a = 187 m, b = 214 m, c = 325 m. Three sides are known We find C, the largest angle opposite to the largest side c = 325 m c 2 = a 2 + b 2 – 2 ab cos C2 ab cos C = a 2 + b 2 – c 2 = C = cos -1 ( )= o Use sine rule to find one of the remaining angles. = A= sin -1 (0.5469)= 33.2 o B = 180 – (108.1 o o )= 38.7 o

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Two ships leave a harbor together, traveling on courses that make an angle 135 o 40. How far apart are they when they each travel 402 km? 135 o 40 BC a b c A Harbor A is the harbor AC = AB = 402 km We find side BC = a A = 135 o 40 Since b = c, B = C (isosceles triangle) B + C = 180 – 135 o 40= 44 o 20 B = (44 o 20)/2= 22 o 10 = 745 km

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The sides of a parallelogram are 4.0 cm and 6.0 cm. One angle is 58 o while another is 122 o. Find the lengths of its diagonals. ABCD is a parallelogram. A B C D A = C = 58 o B = D = 122 o 58 o 122 o DB = a is a diagonal. a AB = 6.0 cm d AD = 4.0 cm b Using cosine law in ABD BD 2 = AD 2 + AB 2 – 2 × AD × AB cos A = – 2 × 4 × 6 cos 58 o = = 5.2 cm AC = is a diagonal. d Using cosine law in ADB AC 2 = AD 2 + DC 2 – 2 AD × DC cos D = – 2 × 4 × 6 cos 122 = 77.4 = 8.8 cm

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Solve the triangle with C = 58.4 o, b = 7.23 cm, c = 6.54 cm. This is a case of SSA A B C Use sine rule to find one of the missing angles. a b c = B 1 = sin -1 (0.9416) = 70.3 o in the 1 st quadrant B 2 = 180 – 70.3 = in the 2 nd quadrant. A 1 = 180 – (58.4 o o ) = 51.3 o A 2 = 180 – (58.4 o o ) = 11.9 o = 5.99 cm = 1.58 cm

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