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Solution of Triangles

Solution of Triangles. COSINE RULE. A. b. c. B. C. a. 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 °. a 2 = b 2 + c 2 – 2bc cos A b 2 = c 2 + a 2 – 2ca cos B

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Solution of Triangles

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  1. Solution of Triangles COSINE RULE

  2. A b c B C a 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° a2 = b2 + c2 – 2bc cos A b2 = c2 + a2 – 2ca cos B c2 = a2 + b2 - 2ab cos C

  3. Example ABC is a triangle with A = 50º, AB = 9cm dan AC = 6 cm. Find the length BC B Using cosine rule, a2 = b2 + c2 – 2bc cos A = 62 + 92 – 2(6)(9)cos 50º = 47.58 a = sqrt (47.58) = 6.896 cm 9 cm 50º C A 6cm

  4. A b c B C a Cosine Rule 3 sides given. (solve for angles) hence Cos A = (b2 + c2 – a2 ) ÷ 2bc or Cos B = (c2 + a2 – b2) ÷ 2ca or Cos C = (a2 + b2 – c2)  2ab a2 = b2 + c2 – 2bc cos A b2 = c2 + a2 – 2ca cos B c2 = a2 + b2 - 2ab cos C

  5. EXAMPLE PQR is a triangle with PQ = 10cm, PR = 5cm, and QR = 13 cm. Find the largest angle in the triangle. Using cosine rule, p2 =q2 + r2 – 2qr cos P cos P = (52 + 102 – 132) ÷ 2(5)(10) = −0.44 P = 116º 6’ Q 13cm 10cm R P 5cm

  6. Solve the triangle using sine rule and cosine rule. ABC is a straight line. Find ABD and BCD. D Using cosine rule, cos ABD = (42 + 32 – 62) ÷ 2(3)(4) = −0.45833 ABD = 117º 17’ Hence DBC = 6243’ 7cm 6cm 4cm C A 3cm B Using sine rule, sin BCD = 0.5079 BCD = 3031’

  7. A b c B C a Area of Triangle Area of triangle = h bsinC = h

  8. Area of triangle = 0.5  16.5  8  sin 120º = 57.16 cm2 C 16.5 cm ° 120 A B 8 cm Example

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