{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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
Background image of page 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).
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}