This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1. The area of the triangle shown below is 2.21 cm 2 . The length of the shortest side is x cm and the other two sides are 3 x cm and ( x + 3) cm. x 3 x x + 3 (a) Using the formula for the area of the triangle, write down an expression for sin in terms of x . (2) (b) Using the cosine rule, write down and simplify an expression for cos in terms of x. (2) (c) (i) Using your answers to parts (a) and (b), show that, 2 2 2 2 2 3 42 . 4 1 2 3 2 3  =  x x x x (1) (ii) Hence find (a) the possible values of x; (2) (b) the corresponding values of , in radians , using your answer to part (b) above. (3) (Total 10 marks) 2. (a) Using the formula for the area of a triangle gives A = 2 1 x 3 x sin (M1) sin = 2 3 42 . 4 x (A1) 2 (b) Using the cosine rule gives cos = x x x x x 3 2 ) 3 ( ) 3 ( 2 2 2 + + (M1) = 2 2 2 3 2 3 x x x (A1) 2 1 (c) (i) Substituting the answers from (a) and (b) into the identity cos 2 = 1 sin 2 gives (M1) 2 2 2 2 2 3 42 . 4 1 2 3 2 3  =  x x x x (AG) (ii) (a) x = 1.24, 2.94 (G1)(G1) (b) = arccos  2 2 2 3 2 3 x x x = 1.86 radians or = 0.171 (accept 0.172) radians (3 sf) (G1)(G1) 6 Notes: Some calculators may not produce answers that are as accurate as required, especially if they use zoom and trace to find the answers. Allow 0.02 difference in the value of x, with appropriate ft for . Award (M1)(G1)(G0) for correct answers given in degrees (106 or 9.84). Award (M1)(G1)(G0) if the answers are not given to 3 sf Award (M0)(G2) for correct answers without working. [10] 3. In a triangle ABC, C B A = 30, AB = 6 cm and AC = 2 3 cm. Find the possible lengths of [BC]. Working: Answer: .................................................................. (Total 3 marks) 2 4. Note: Award full marks for exact answers or answers given to three significant figures. Method 1 : Using the sine rule: 2 3 30 sin 6 C sin = sin C = 2 1 C = 45, 135. (M1) Again, = sin15 BC or sin105 BC 30 sin 2 3 Thus, BC = 2 6 sin 105 or 2 6 sin 15 BC = 8.20 cm or BC = 2.20 cm. (A1)(A1) (C3) Method 2: Using the cosine rule: AC 2 = 6 2 + BC 2 2(6)(BC)cos 30 18 = 36 + BC 2 3 6 BC (M1) Therefore, BC 2 ( 3 6 )BC + 18 = 0 Therefore, (BC 3 3 ) 2 = 27 18 = 9 Therefore, BC = 3 3 3, ie BC = 8.20 cm or BC = 2.20 cm. (A1)(A1) (C3) Method 3: B A 6 30 3 2 3 2 D C 1 C 2 In ABD, AD = 3 cm, and BD = 3 3 27 = cm. (A1) In AC 1 D, C 1 D = 3 Also, C 2 D = 3. (A1) Therefore BC = ( 3 3 3) cm, ie BC = 8.20 cm or BC = 2.20 cm. (A1) (C3) Note: If only one answer is given, award a maximum of (M1) (A1)....
View
Full
Document
 Spring '10
 wilson
 Math, Angles

Click to edit the document details