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Additional Mathematics Trigonometry Topic assessment 1 The figure shows the cross section of a roof of a building. The length of the roof AC is 7 metres and the angles at B and C are 52° and 38° respectively. A 7 C h 38° x 52° B Find the height, h, above the base and the length of the roof, AB. [5] 2 Find the value of x in the range 0° x 360° for which sin x 53 and tan x 34 . [2] 3 You are given that sin y 54 and cos y 53 . Find the exact value of tan y. [2] 4 Solve the equation sin2 x = 0.9 in the range 0° x 360°. [5] 5 (i) Show that cos x 2sin2 x = 2cos2 x + cos x 2. [2] (ii) Hence, or otherwise, solve the equation cos x 2 sin2 x + 2 = 0 in the range 0° x 360°. [5] 6 Solve the following equations in the range 180° x 180°. (i) sin x = 5cos x [2] (ii) 3tan2 x + 2tan x 5 = 0 [5] (iii) 4cos2 x + sin x 3 = 0 [5] 1 of 7 05/06/13 © MEI AM Trigonometry Assessment solutions 7 In the triangle ABC shown, AC = 8, angle C = 38° and angle B = 52°. A 8 38° C 8 9 52° B (i) Find the length of the side AB. [3] (ii) Calculate the area of the triangle. [3] In the triangle ABC, AB = 6, BC = 8 and CA = 9. Calculate the size of the largest angle in this triangle. [4] In the quadrilateral ABCD shown below, AB = 8, BC = 5, angle B = 70°, angle DAC = 40° and angle DCA = 50°. B 70° 5 8 C 50° A 45° D (i) Calculate the length of the diagonal AC. [4] (ii) Calculate the length of the side AD. [4] 2 of 7 05/06/13 © MEI AM Trigonometry Assessment solutions 10 A coast line runs East-West and on the coast there are two harbours, A and B. A boat sails from A on a bearing of 040° for 5 miles to point C before altering course to 135° to sail directly to B. C 135° N 5 040° B A Find (i) the distance AB, [5] (ii) the distance sailed. [4] Total: 60 3 of 7 05/06/13 © MEI AM Trigonometry Assessment solutions Solutions to topic assessment 1. Using left-hand triangle: Using right-hand triangle: sin 38 h 7 h 7 sin 38 4.31 metres (3 s.f.) sin 52 x h x h sin 52 M1 A2 5.47 metres (3 s.f.) 2. Since sin x is positive and tan x is negative, x lies in the 2nd quadrant. The acute angle for which sin x 53 is 36.9° The corresponding angle in the second quadrant is 180° - 36.9° = 143.1°. 3. tan y sin y cos y 54 4 53 3 M1 A1 B1 B1 M1 A1 4. sin 2 x 0.9 sin x 0.94868... , so there is a value of x in all four quadrants. B2 The acute angle for which sin x 0.94868... is x = 71.6° The equivalent angle in the second quadrant is 180° - 71.6° = 108.4°. The equivalent angle in the second quadrant is 180° + 71.6° = 251.6°. The equivalent angle in the second quadrant is 360° - 71.6° = 288.4°. The solution is x = 71.6°, 108.4°, 251.6° and 288.4°. 5. (i) Using the identity sin 2 x cos2 x 1 cos x 2 sin 2 x cos x 2(1 cos 2 x ) cos x 2 2 cos 2 x B3 M1 A1 2 cos 2 x cos x 2 (ii) cos x 2 sin 2 x 2 0 2 cos 2 x cos x 2 2 0 2 cos 2 x cos x 0 cos x(2 cos x 1) B1 cos x 0 or cos x 21 M1 A1 cos x 0 x 90 or 270 If cos x 21 , x is in the 2nd or 3rd quadrant 4 of 7 M1 A1 05/06/13 © MEI AM Trigonometry Assessment solutions The acute angle with cosine of 21 is 60°, so the equivalent angle in the 2nd quadrant is 180° - 60° = 120°, and the equivalent angle in the 3rd quadrant is 180° + 60° = 240°. The solution is x = 90°, 120°, 240°, 270°. 6. (i) (ii) sin x 5 cos x sin x 5 cos x tan x 5 tan is positive in the 1st and 3rd quadrants. The acute angle for which tan x 5 is 78.7°. The equivalent angle in the 3rd quadrant for -180° x 180° is 78.7° - 180° = -101.3° M1 A1 The solution is x = -101.3°, 78.7°. 3 tan 2 x 2 tan x 5 0 (3 tan x 5 )(tan x 1) 0 M1 A1 M1 tan x 53 or 1 If tan x 53 , x is in the 2nd or 4th quadrants, so x = 180° - 59.0° = 121.0° or x = -59.0°. If tan x 1 , x is in the 1st or 3rd quadrants, so x = 45° or x = 45° - 180° = 135° A2 The solution is x = -135°, -59.0°, 45°, 121.0°. (iii) 4cos 2 x sin x 3 0 4(1 sin 2 x ) sin x 3 0 4 4 sin 2 x sin x 3 0 4 sin 2 x sin x 1 0 1 1 4 4 1 1 17 8 8 1 17 If sin x , x is in the 1st or 2nd quadrant, so x = 39.8° 8 or x = 180° - 39.8° = 140.2° sin x 1 17 , x is in the 3rd or 4th quadrant, 8 so x = 23.0° - 180° = -157.0° or x = -23.0° If sin x The solution is x = -157.0°, -23.0°, 39.8°, 140.2°. 5 of 7 05/06/13 © MEI A2 AM Trigonometry Assessment solutions 7. (i) (ii) Using the sine rule: AB 8 sin 38 sin 52 8 sin 38 AB 6.25 (3 s.f.) sin 52 M1 A2 A 180 38 52 90 , so the triangle is right-angled Taking AB as the base means AC is the height Area 21 8 6.250 25.0 square units. B1 M1 A1 B1 8. The largest angle is opposite the longest side, so the largest angle is B. 6 2 8 2 92 Using the cosine rule: cosB 2 68 M1 A2 B 78.6 (3 s.f.) 9. (i) Using the cosine rule on triangle ABC: b 2 a 2 c 2 2ac cosB B1 b 2 5 2 8 2 2 5 8 cos 70 b 7.85 (3 s.f.) M1 A2 (ii) Using triangle ACD: D = 180° - 45° - 50° = 85° AD Using the sine rule: sin 50 AC B1 B1 sin 85 AC sin 50 AD 6.04 (3 s.f.) sin 85 M1 A1 10. C 1350 400 450 N 5 0400 A 450 500 B B1 for angle C M1 A1 for angle B (i) Using the sine rule: 5 AB sin 45 sin 85 5 sin 85 AB 7.04 miles (3 s.f.) sin 45 6 of 7 M1 A1 05/06/13 © MEI AM Trigonometry Assessment solutions (ii) 5 BC sin 45 sin 50 5 sin 50 BC 5.42 (3 s.f.) sin 45 Distance sailed = AC + CB = 5 + 5.42 = 10.42 miles. Using the sine rule: 7 of 7 M1 A1 M1 A1 05/06/13 © MEI