2. Trigonometry Solving Triangles

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

4. Trigonometric ratios in surd form 1 1 1 2 30º 45º 60º Angles are given in radians π radians = 180º π3π3 = 60º π2π2 = 90º π4π4 = 45º π6π6 = 30º 3 1 3 2 3 1 tan 60 º = sin 60 º = 1212 cos 60 º = tan 30 º = 1212 sin 30 º = 3 2 cos 30 º = 2 tan 45 º = 1 1 2 sin 45 º = 1 2 cos 45 º = Page 9 of tables

5. c ab Cosine Rule C AB a 2 = b 2 + c 2 – 2bccosA ba c b 2 = a 2 + c 2 – 2accosB c 2 = a 2 + b 2 – 2abcosC Page 9 of tables

6. By Pythagoras’ Theorem a 2 = (c – x) 2 + h 2 a 2 = c 2 – 2cx + x 2 + h 2 a 2 = b 2 + c 2 – 2cx Cosine Rule c ab a 2 = b 2 + c 2 – 2bccosA c b a xc – x b2 = x2 + h2b2 = x2 + h2 A h

7. 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. 89 o 13.8 6·26·2 w 147 o 8 11 m Cosine Rule 6 10 65 o l Included angle

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

9. 137 o 17·5 cm 12·2 cm a 2 = 12·2 2 + 17·5 2 – ( 2  12·2  17·5  cos 137 o ) a 2 = 148·84 + 306·25 – ( 427  – 0·731 ) a 2 = 455·09 + 312·137 a 2 = 767·227a = 27·7 cm  Find the unknown side in the triangle below: Examples a 2 = b 2 + c 2 – 2bccosA a = ? b = 12·2 c = 17·5 A = 137º

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

11. 20 o x 6 0 = x 2 – 18·79x + 64  Find the two possible values for the unknown side. Examples 10 a = 6 b = 10 c = x A = 20º a  1 b  –18·79 c  64

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

13. c ab Sine Rule C AB ba c Page 9 of tables

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

15. (ii) Find  ab , correct to the nearest cm.  In the triangle abc, d is a point on [bc].  bd  = 5 cm,  ac  = 7cm,  bca  = 82º and  cad  = 50º. 5d 50º 7 += 1212 Cosine Rule: a 2 = b 2 + c 2 – 2bc cosA  ab  2 = 12 2 + 7 2 – 2(12)(7)cos82º = 144 + 49 – 168cos82º = 169·6189… = 13·02… = 13 cm  ab  = 169·6189 a cb 7 82º

16. c ab Area of triangle C AB ba c Page 6 of tables Must be the included angle

17. Calculate the area of the triangle shown. Give your answer correct to one decimal place. Area of triangle = absin C 1212 Area = (3)(4) sin 55  1212 = 4·9149… 4 cm 3 cm C must be the included angle = 4·9 cm 2 55º

18. Find the area of triangle abc, correct to the nearest whole number. Area of triangle = absin C 1212 Area = (14)(18·4) sin 70  1212 = 121·0324… 18·4 14 c ab 44º 66º C must be the included angle |  abc | = 180  – 44  – 66  = 70  70º = 121units 2