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triangles assignment HL

# triangles assignment HL - 1 The area of the triangle shown...

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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 θ in terms of x . (2) (b) Using the cosine rule, write down and simplify an expression for cos θ 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 θ, 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 θ (M1) sin θ = 2 3 42 . 4 x (A1) 2 (b) Using the cosine rule gives cos θ = 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

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(c) (i) Substituting the answers from (a) and (b) into the identity cos 2 θ = 1 – sin 2 θ 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) θ = arccos - - 2 2 2 3 2 3 x x x θ = 1.86 radians or θ = 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 θ. 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 ˆ 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 3 0 º 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).

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