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Trigonometry

Trigonometry. Solving Triangles. T wo o ld a ngels S kipped o ver h eaven C arrying a harp. Solving Triangles. HYP. OPP. . ADJ. 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

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Trigonometry

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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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