Solution of Triangles COSINE RULE

Cosine Rule  2 sides and one included angle given. e.g. b = 10cm, c = 7 cm and  A = 55° or, a = 14cm, b = 10 cm and  C = 48° B A C a b c a 2 = b 2 + c 2 – 2bc cos A b 2 = c 2 + a 2 – 2ca cos B c 2 = a 2 + b 2 - 2ab cos C

Example ABC is a triangle with  A = 50º, AB = 9cm dan AC = 6 cm. Find the length BC A B C 6cm 9 cm 50º Using cosine rule, a 2 = b 2 + c 2 – 2bc cos A = – 2(6)(9)cos 50º = a = sqrt (47.58) = cm

Cosine Rule  3 sides given. (solve for angles) hence Cos A = (b 2 + c 2 – a 2 ) ÷ 2bc or Cos B = (c 2 + a 2 – b 2 ) ÷ 2ca or Cos C = (a 2 + b 2 – c 2 )  2ab B A C a b c a 2 = b 2 + c 2 – 2bc cos A b 2 = c 2 + a 2 – 2ca cos B c 2 = a 2 + b 2 - 2ab cos C

EXAMPLE PQR is a triangle with PQ = 10cm, PR = 5cm, and QR = 13 cm. Find the largest angle in the triangle. P Q R 5cm 10cm 13cm Using cosine rule, p 2 =q 2 + r 2 – 2qr cos P cos P = ( – 13 2 ) ÷ 2(5)(10) = −0.44  P = 116º 6’

Solve the triangle using sine rule and cosine rule. ABC is a straight line. Find  ABD and  BCD. AB C D 3cm 7cm 4cm 6cm Using cosine rule, cos  ABD = ( – 6 2 ) ÷ 2(3)(4) = −  ABD = 117º 17’ Hence  DBC = 62  43’ Using sine rule, sin  BCD =  BCD = 30  31’

Area of Triangle Area of triangle = B A C a b c h b sin C = h

Example 16.5 cm 8 cm 120  B C A Area of triangle = 0.5  16.5  8  sin 120º = cm 2