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Solving Triangles: Trigonometric Concepts and Applications

This guide explores core concepts of trigonometry focused on solving triangles using trigonometric ratios, including the sine, cosine, and tangent functions. It covers practical applications such as the Cosine Rule and Sine Rule, providing step-by-step examples to find unknown sides and angles in various triangles. Understanding angles in both degrees and radians is included, alongside utilizing Pythagorean theorem applications where relevant. This resource is designed for students looking to deepen their comprehension of triangles in trigonometry.

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Solving Triangles: Trigonometric Concepts and Applications

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  1. Trigonometry Solving Triangles

  2. Two old angels Skipped over heaven Carrying a harp Solving Triangles HYP OPP  ADJ

  3. 3 1 1 3 tan 45 º = 1 tan 60 º = tan 30 º = 1 2 1 2 3 2 sin 45 º = 3 sin 30 º = sin 60 º = 1 2 1 2 3 2 cos 45 º = cos 60 º = cos 30 º = Angles are given in radians π radians = 180º 2 π 3 π 6 π 4 π 2 = 45º = 60º = 30º = 90º Trigonometric ratios in surd form 30º 2 60º 1 Page 9 of tables 45º 1 45º 1

  4. Cosine Rule c Page 9 of tables C b a A B a b c a2=b2+c2– 2bccosA b2=a2+c2– 2accosB c2=a2+b2– 2abcosC

  5. Cosine Rule c By Pythagoras’ Theorem a2 = (c – x)2 + h2 a2 = c2 – 2cx + x2 + h2 a2 = b2 + c2 – 2cx b a h A a b x c c – x b2=x2+h2 a2 = b2+c2– 2bccosA

  6. w 10 65o 6·2 89o 6 l 13.8 147o 11 8 m Cosine Rule The Cosine Rule can be used to find a third side of a triangle if you have the other two sides and the angle between them. Included angle

  7. Examples  Find the unknown side in the triangle below: l 5 m Identify sides a,b,c and angle Ao 43o Write down the Cosine Rule 12 m a = lb = 5 c = 12 A = 43º Substitute values and find a2 a2 = b2+c2– 2bccosA Take square root of both sides a2 = 52 + 122 – 2  5  12cos43o a2 = 25+ 144 – 120(0·731) a2 = 81·28 a =9·02 m

  8. Examples  Find the unknown side in the triangle below: a = ? b = 12·2 c = 17·5 A = 137º 17·5 cm 137o 12·2 cm a2 = b2+c2– 2bccosA a2 = 12·22+ 17·52– ( 2  12·2  17·5  cos 137o ) a2 = 148·84 + 306·25 – ( 427 – 0·731 ) a2 = 455·09 + 312·137 a2 = 767·227 a = 27·7 cm

  9. Examples  Find the two possible values for the unknown side. a = 6 b = 10 c = x A = 20º x 6 20o 10 a2 = b2+c2– 2bccosA 62= 102+x2– (2  10 x cos 20o) 36 = 100 +x2– 20x( 0·9397) 0 =x2– 18·79x+ 64

  10. Examples  Find the two possible values for the unknown side. a = 6 b = 10 c = x A = 20º a= 1 b=–18·79 c= 64 x 6 0 =x2– 18·79x+ 64 20o 10

  11. 43 cm (1) 78o 31 cm L 6·3 cm (3) 110o G (2) 8·7 cm M 5·2 m 38o 8 m  Find the length of the unknown side in the triangles below: G = 12.4cm L = 47.5cm M = 5·05 m

  12. Sine Rule c c C A B a a b b Page 9 of tables

  13. _____ _____ _____ a sin A b sin B c sin C == a 50º 7 82º b c 5 d In the triangle abc, d is a point on [bc]. bd = 5 cm, ac = 7cm, bca = 82º and cad = 50º . (i) Find dc, correct to the nearest cm. Sine Rule Angle sum of ∆ is 180º dc ______ ______ __ 7 = sin 50º sin 48º 48º Multiply both sides by sin 50º = 7·215… 7 = 7 cm

  14. In the triangle abc, d is a point on [bc]. bd = 5 cm, ac = 7cm, bca = 82º and cad = 50º . a (ii)Find ab, correct to the nearest cm. Cosine Rule: a2=b2+c2– 2bc cosA 50º 7 ab2= 122+ 72– 2(12)(7)cos82º = 144 + 49 – 168cos82º 82º b c = 169·6189… 5 d 7 ab = 169·6189 + = 12 12 = 13·02… = 13 cm

  15. Must be the included angle Area of triangle c c C A B a a b b Page 6 of tables

  16. 1 2 Area = (3)(4)sin55 1 2 Area of triangle =absinC Calculate the area of the triangle shown. Give your answer correct to one decimal place. 3 cm = 4·9149… 55º 4 cm = 4·9 cm2 C must be the included angle

  17. Area = (14)(18·4)sin70 1 2 1 2 Area of triangle =absinC Find the area of triangle abc, correct to the nearest whole number. c 66º 14 44º 70º a b = 121·0324… 18·4 C must be the included angle = 121units2 |abc| =180 – 44 – 66 = 70

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