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The figure below shows a circular clock, indicating 5 o’clock. The hours hand is 5 cm long and the minutes hand is 7 cm long. Calculate the angle between the two hands at this time. Calculate the shortest distance between the tips of the two hands at this time, correct to 3 significant figures. 30 ° 30° ° 360 ÷ 12 30 1. 2. © T Madas

The figure below shows a circular clock, indicating 5 o’clock. The hours hand is 5 cm long and the minutes hand is 7 cm long. Calculate the angle between the two hands at this time. Calculate the shortest distance between the tips of the two hands at this time, correct to 3 significant figures. x 2 = 72 + 52 – 2 x 7 x 5 x cos 150° x 2 = 49 + 25 – 70 cos 150° x x 2 ≈ 134. 462 c c c By the cosine rule : 7 cm 1. 2. x ≈ 11. 6 cm 5 cm © T Madas

In the triangle ABC, AC = 8 cm, AB = 9 cm and RCAB = 75° P is a point on BC such that RAPB = 90°. 1. Calculate the length of BC [3 s. f. ] 2. Calculate the length AP [3 s. f. ] C [BC ]2 = 82 + 92 – 2 x 8 x 9 x cos 75° 8 cm P [BC ]2 = 64 + 81 – 144 cos 75° [BC ]2 ≈ 107. 7300575 BC A 75° c c c By the cosine rule on ABC : ≈ 10. 37930910 B 9 cm BC ≈ 10. 37930910 BC ≈ 10. 4 cm [3 s. f. ] © T Madas

In the triangle ABC, AC = 8 cm, AB = 9 cm and RCAB = 75° P is a point on BC such that RAPB = 90°. 1. Calculate the length of BC [3 s. f. ] 2. Calculate the length AP [3 s. f. ] C θ 75° B 9 cm BC ≈ 10. 37930910 BC ≈ 10. 4 cm θ ≈ 48. 11° [3 s. f. ] sin 75° = sinθ x 8 10. 379 8 c sinθ = 8 sin 75° 10. 379 c 8 x sinθ ≈ 0. 7445 θ ≈ sin-1 (0. 7445) c c P 8 cm A By the sine rule on ABC : θ ≈ 48. 11° © T Madas

In the triangle ABC, AC = 8 cm, AB = 9 cm and RCAB = 75° P is a point on BC such that RAPB = 90°. 1. Calculate the length of BC [3 s. f. ] 2. Calculate the length AP [3 s. f. ] C A B 9 cm BC ≈ 10. 37930910 BC ≈ 10. 4 cm θ ≈ 48. 11° [3 s. f. ] 9 = Sin(48. 11°) x 9 AP = 9 Sin(48. 11°) θ 75° AP AP ≈ 6. 6998528 c 8 cm 9 x c c AP = sinθ AB P c By trigonometry on APB : AP ≈ 6. 70 cm [3 s. f. ] © T Madas

The triangle ABC has AB = 30 m, RABC = 30° and has an area of 300 m 2. Calculate the perimeter of the triangle, giving your answer correct to 3 significant figures. The area of the triangle ABC 300 A y c 30 m m 40 x x 30 x x x sin 30° 300 = 1 2 x x 30 x x x 1 x 4 4 x 300 = 1 2 2 m 2 C 1200 = 30 x c 30° 30 x x x sin 30° c A = 1 2 c B x = 40 © T Madas

The triangle ABC has AB = 30 m, RABC = 30° and has an area of 300 m 2. Calculate the perimeter of the triangle, giving your answer correct to 3 significant figures. B y 2 = 302 + 402 – 2 x 30 x 40 x cos 30° y 2 = 900 + 1600 – 2400 cos 30° 40 y 2 ≈ 421. 539 30 m m 30° 300 m 2 A 20. 53 y m c c c By the cosine rule on ABC y ≈ 20. 53 m C Therefore the perimeter of VABC to 3 s. f. is 90. 5 m © T Madas

In the triangle below AB = x cm, BC = (x + 2) cm, RC = 30° and sin. A = ¾. Calculate the value of x. By the sine rule on ABC : A sin. C = ¾ 2 30° diagram not to scale sin 30° C x = = 1 2 4 x 3 4 x = x+2 3 4 1 2 (x + 2) x 4 3 x = 2 (x + 2) 3 x = 2 x + 4 c x x c c c x+ c B x = 4 © T Madas

In the triangle ABC, AB = 8 cm, BC = 6 cm and AC = 12 cm. Point D is a point on AC such that BD = 4 cm. Calculate to the nearest degree: 1. RACB 2. RBDC B cm by the cosine rule 4 c 8 cm m 6 θ A D C 82 64 144 cosθ = = 62 + 122 – 2 x 6 x 12 x cosθ 36 + 144 – 144 cosθ 36 + 144 – 64 116 c c c 12 cm © T Madas

In the triangle ABC, AB = 8 cm, BC = 6 cm and AC = 12 cm. Point D is a point on AC such that BD = 4 cm. Calculate to the nearest degree: 1. RACB 2. RBDC B 6 m by the cosine rule 4 c 8 cm cm θ A D C 12 cm 144 cosθ = 116 © T Madas

In the triangle ABC, AB = 8 cm, BC = 6 cm and AC = 12 cm. Point D is a point on AC such that BD = 4 cm. Calculate to the nearest degree: 1. RACB 2. RBDC B by the cosine rule 36° θ D C 12 cm 144 cosθ = 116 cosθ = 0. 8056 θ = cos-1[ 0. 8056] θ ≈ 36° c c c A cm 4 c 8 cm m 6 © T Madas

In the triangle ABC, AB = 8 cm, BC = 6 cm and AC = 12 cm. Point D is a point on AC such that BD = 4 cm. Calculate to the nearest degree: 1. RACB 2. RBDC B α cm by the sine rule 36° D C sinα ≈ 0. 882 α = 62° α = sin-1[ 0. 882 ] c c sinα = sin 36° 6 4 sinα = 6 x sin 36° 4 c 12 cm c A 4 c 8 cm m 6 © T Madas

In the triangle below AB = 7 cm, BC = x cm, AC = (x + 2) cm and RBAC = 60°. Calculate the value of x. By the cosine rule on ABC : BC 2 = AB 2 + AC 2 – 2 x AB x AC x cos. A x 2 = 72 + (x + 2)2 – 2 x 7 x (x + 2) x cos 60° x+2 x B x 2 = 49+ x 2 + 2 x 2 x x – 14(x + 2) x 12 x 2 = 49+ x 2 + 4 x – 7(x + 2) x 2 = 49+ x 2 + 4 x – 7 x – 14 7 0 = 39 – 3 x 60° A diagram not to scale 3 x = 39 c c c c C x = 13 all lengths in cm © T Madas

In a triangle ABC, AB = x cm, BC = 13 cm, AC = (x + 8) cm and RBAC = 60°. Calculate the value of x. By the cosine rule on ABC : BC 2 = AB 2 + AC 2 – 2 x AB x AC x cos. A 132 = x 2 + (x + 8)2 – 2 x x x (x + 8) x cos 60° 13 x+8 1 B 169 = x 2 + 82 + 2 x 8 x x – 2 x (x + 8) x 2 169 = 2 x 2 + 64 + 16 x – x (x + 8) 169 = 2 x 2 + 64 + 16 x – x 2 – 8 x x 169 = x 2 + 8 x + 64 60° A diagram not to scale 0 = x 2 + 8 x + 64 – 169 c c c c C x 2 + 8 x – 105 = 0 all lengths in cm © T Madas

In a triangle ABC, AB = x cm, BC = 13 cm, AC = (x + 8) cm and RBAC = 60°. Calculate the value of x. x 2 + 8 x – 105 = 0 ( x – 7 ) ( x + 15) = 0 c c C x+8 13 B x 1 3 5 7 105 35 21 15 60° A diagram not to scale all lengths in cm © T Madas

In a triangle ABC, AB = x cm, BC = 13 cm, AC = (x + 8) cm and RBAC = 60°. Calculate the value of x. x 2 + 8 x – 105 = 0 ( x – 7 ) ( x + 15) = 0 13 x+8 7 cm B x c c C x= -15 60° A diagram not to scale all lengths in cm © T Madas

A regular octagon is inscribed in a circle of radius 5 cm. Calculate the perimeter of the octagon, giving your answer correct to 3 significant figures. [AB ]2 = 52 + 52 – 2 x 5 x cos 45° A [AB ]2 = 25 + 25 – 50 cos 45° cm O [AB ]2 ≈ 14. 6467 5 45° B AB c c c By the cosine rule on OAB ≈ 3. 8268 cm The perimeter of the octagon 8 x 3. 8268 = 30. 6 cm [ 3 s. f. ] © T Madas

The square ABCD has side length of 2 m. M and N are the midpoints of AB and BC respectively. By using the cosine rule, or otherwise, prove that cosine of 4 5 RMDN is A M 1 2 5 1 N 5 θ D B 2 m 1 C as required © T Madas

ABCDEF is a regular hexagon with side length of 8 cm and MNOPQR is also a regular hexagon produced by joining the midpoints of the sides of ABCDEF. Work out the size of RBAF Calculate the length MN, correct to d. p. 1. 2. C P D O Q B E 60° R N 120° 60° A M 60° F © T Madas

ABCDEF is a regular hexagon with side length of 8 cm and MNOPQR is also a regular hexagon produced by joining the midpoints of the sides of ABCDEF. Work out the size of RBAF Calculate the length MN, correct to 2 d. p. 1. 2. C P N D O m 4 c Q 120° B E A 4 cm M R N 120° A M F © T Madas

ABCDEF is a regular hexagon with side length of 8 cm and MNOPQR is also a regular hexagon produced by joining the midpoints of the sides of ABCDEF. 1. 2. Work out the size of RBAF Calculate the length MN, correct to 2 d. p. N m 4 c x 2 x 2 x 42 + 42 – 2 x 4 x cos 120° 16 + 16 – 32 x (-0. 5) 32 + 16 48 = 6. 93 [2 d. p. ] = = c c by the cosine rule 120° A 4 cm M © T Madas

Boat A is 7. 2 km from a lighthouse H on a bearing of 195°. Boat B is 11. 4 km from H on a bearing of 293°. 1. Calculate the distance between A and B 14. 3 km 2. Calculate the bearing of boat A from boat B N By the cosine rule on AHB d 2 = 7. 22 + 11. 42 – 2 x 7. 2 x 11. 4 x cos 98° d 2 = 51. 84 + 129. 96 – 164. 16 cos 98° B 11. 4 98° 5 . 30 7. 2 14 d d 2 ≈ 204. 65 H c c c N d ≈ 14. 305 km 195° A © T Madas

Boat A is 7. 2 km from a lighthouse H on a bearing of 195°. Boat B is 11. 4 km from H on a bearing of 293°. 1. Calculate the distance between A and B 14. 3 km 2. Calculate the bearing of boat A from boat B 143° 14 5 . 30 d 98° H 195° c 67° sinθ = sin 98° x 7. 2 14. 3 sinθ = 7. 2 sin 98° 14. 3 c 113° 3 θ 0 11. 4 ° By the sine rule on AHB : 7. 2 x 7. 2 B N sinθ ≈ 0. 498 θ ≈ sin-1 (0. 498) θ ≈ c c N 30° A © T Madas

D C 128 A 13 5 m 2° m B A statue is mounted on the top of a building. In the diagram, AC represents the height of the building and CD represents the statue. The points A, C and D lie on the same line. Other known measurements are marked on the diagram. Calculate to 3 significant figures: 1. The height of the statue 2. The height of the building © T Madas

D x C 128 13 5 m 2° m A B By the cosine rule: x 2 = 1282 + 1352 – 2 x 128 x 135 x cos 2° x 2 = 16384 +18225 – 34560 cos 2° x 2 ≈ 70. 053 x ≈ 8. 37 m [3 s. f. ] © T Madas