Solution of Triangles COSINE RULE Cosine Rule A

  • Slides: 8
Download presentation
Solution of Triangles COSINE RULE

Solution of Triangles COSINE RULE

Cosine Rule A b c B a C a 2 = b 2 +

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

Example ABC is a triangle with A = 50º, AB = 9 cm dan

Example ABC is a triangle with A = 50º, AB = 9 cm dan AC = 6 cm. Find the length BC B 9 cm 50º A 6 cm Using cosine rule, a 2 = b 2 + c 2 – 2 bc cos A = 62 + 92 – 2(6)(9)cos 50º = 47. 58 C a = sqrt (47. 58) = 6. 896 cm

Cosine Rule A b c B a C a 2 = b 2 +

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

EXAMPLE PQR is a triangle with PQ = 10 cm, PR = 5 cm,

EXAMPLE PQR is a triangle with PQ = 10 cm, PR = 5 cm, and QR = 13 cm. Find the largest angle in the triangle. Q 13 cm 10 cm P 5 cm Using cosine rule, p 2 =q 2 + r 2 – 2 qr cos P = (52 + 102 – 132) ÷ 2(5)(10) = − 0. 44 P = 116º 6’ R

Solve the triangle using sine rule and cosine rule. ABC is a straight line.

Solve the triangle using sine rule and cosine rule. ABC is a straight line. Find ABD and BCD. Using cosine rule, D 7 cm 6 cm A 3 cm cos ABD = (42 + 32 – 62) ÷ 2(3)(4) = − 0. 45833 ABD = 117º 17’ C Hence DBC = 62 43’ 4 cm B Using sine rule, sin BCD = 0. 5079 BCD = 30 31’

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

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

Example C 16. 5 cm 120 ° A 8 cm B Area of triangle

Example C 16. 5 cm 120 ° A 8 cm B Area of triangle = 0. 5 16. 5 8 sin 120º = 57. 16 cm 2