 Download Presentation INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

# INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

Download Presentation ## INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Trigonometry Statistics UNIT 2 : Graphs, Charts & Tables Simultaneous Equations EXIT

2. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Trigonometry UNIT 2 : Please choose a question to attempt from the following: 1 2 3 4 5 6 Back to Unit 2 Menu EXIT

3. B 40cm A 25° C 50cm TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

4. B 40cm A 25° C 50cm 1. For area of a triangle questions with an angle present use: TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. What would you like to do now? 2. Always remember to round your answer if the questions asks you to. Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

5. B 40cm A 25° C 50cm Area of  = 423cm2 TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. What would you like to do now? Try another like this Go to full solution Go to Comments Go to Trigonometry Menu EXIT

6. Question 1 B 40cm A 25° C 50cm 1. For area of triangles questions where an angle is present use: Find the area of the following triangle to the nearest cm2. Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 = 422.61… 2. Remember to round if asked to. = 423 to nearest unit Area of  = 423cm2 Continue Solution Try another like this Comments Trigonometry Menu Back to Home

7. Markers Comments B 1 2 40cm c 25° b A C 50cm Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1. For area of triangles questions where an angle is present use: Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 = 422.61… 2. Remember to round if asked to. = 423 to nearest unit Area of  = 423cm2 Next Comment Trigonometry Menu Back to Home

8. K 6.5m 150° L M 8m TRIGONOMETRY : Question 1B Find the area of the following triangle. Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

9. K 6.5m 150° L M 8m 1. For area of a triangle questions with an angle present use: TRIGONOMETRY : Question 1B Find the area of the following triangle. What would you like to do now? 2. Always remember to round your answer if the questions asks you to. Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

10. K 6.5m 150° L M 8m Area of  = 13m2 TRIGONOMETRY : Question 1B Find the area of the following triangle. What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT

11. Question 1B K 6.5m 150° L M 8m 1. For area of triangles questions where an angle is present use: Find the area of the following triangle. Area of  = ½ kmsinL° = 8 x 6.5 x sin150°  2 = 13 Area of  = 13m2 Begin Solution Continue Solution Comments Trigonometry Menu Back to Home

12. Markers Comments K 6.5m m 150° k L M 8m Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1. For area of triangles questions where an angle is present use: Area of  = ½ kmsinL° = 8 x 6.5 x sin150°  2 = 13 Area of  = 13m2 Next Comment Trigonometry Menu Back to Home

13. N TRIGONOMETRY : Question 2 Two helicopters leave an air base. The first flies on a bearing of 340° at 160km/hr. The second flies due east at 200km/hr. How far apart will they be after 21/2 hours? Answer to the nearest 10km. Get hint Reveal answer only 340° Go to full solution Go to Comments Go to Trigonometry Menu EXIT

14. N TRIGONOMETRY : Question 2 Two helicopters leave an air base. The first flies on a bearing of 340° at 160km/hr. The second flies due east at 200km/hr. How far apart will they be after 21/2 hours? Answer to the nearest 10km. What would you like to do now? 1. Identify what you need to find and the information you have to help you. 4. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 2. Calculate as many of the missing angles as possible. 3. Make a sketch to clarify matters. 5. Substitute known values, remembering to use brackets as appropriate. Reveal answer only 340° Go to full solution Go to Comments Go to Trigonometry Menu EXIT

15. N TRIGONOMETRY : Question 2 Two helicopters leave an air base. The first flies on a bearing of 340° at 160km/hr. The second flies due east at 200km/hr. How far apart will they be after 21/2 hours? Answer to the nearest 10km. What would you like to do now? Try another like this 340° Go to full solution Go to Comments Go to Trigonometry Menu Distance is 740km EXIT

16. Question 2 b a 400km N 110° A c 500km 1. Identify what needs to be found. 160km/hr. 2. Need distances travelled and angle between flight paths. 340° d1 = speed1 x time = 160 x 2.5 = 400km How far apart will they be after 21/2 hours? d2 = speed2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise 20° Full angle = 20° + 90° = 110° 90° 3. Sketch triangle. 200km/hr Continue Solution Try another like this Comments Trigonometry Menu Back to Home

17. Question 2 N 4. Apply Cosine rule. 160km/hr. a2 = b2 + c2 – (2bccosA°) 340° 5. Substitute known values and remember to use brackets. How far apart will they be after 21/2 hours? = 4002 + 5002 – (2 x 400 x 500 x cos110°) 20° = 546808.05.. 90° a = 546808.05.. = 739.46.... 200km/hr 6. Remember to round answer if asked to. Continue Solution = 740 Try another like this Distance is 740km Comments Trigonometry Menu Back to Home

18. Markers Comments 400km 110° 500km 1. Identify what needs to be found. Note: Bearings are measured clockwise from N. 2. Need distances travelled and angle between flight paths. d1 = speed1 x time = 160 x 2.5 = 400km d2 = speed2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise Full angle = 20° + 90° = 110° 3. Sketch triangle. Next Comment Trigonometry Menu Back to Home

19. Markers Comments Check formulae list for the cosine rule: a2 = b2 + c2 – 2bc cosA (2 sides and the included angle) Relate to variables used a2 = b2 + c2 – 2bc cosA 4. Apply Cosine rule. a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. = 4002 + 5002 – (2 x 400 x 500 x cos110°) = 546808.05.. a = 546808.05.. = 739.46.... 6. Remember to round answer if asked to. Next Comment = 740 Trigonometry Menu Distance is 740km Back to Home

20. N 195° SE TRIGONOMETRY : Question 2B Two ships sail from a port. The first travels for two hours on a bearing of 195° at a speed of 18mph. The second travels south-east for three hours at a speed of 15mph. How far apart will they be? Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

21. N 195° SE TRIGONOMETRY : Question 2B Two ships sail from a port. The first travels for two hours on a bearing of 195° at a speed of 18mph. The second travels south-east for three hours at a speed of 15mph. How far apart will they be? 2. Calculate as many of the missing angles as possible. What would you like to do now? 1. Identify what you need to find and the information you have to help you. 4. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 3. Make a sketch to clarify matters. Reveal answer only 5. Substitute known values, remembering to use brackets as appropriate. Go to full solution Go to Comments Go to Trigonometry Menu EXIT

22. N 195° SE Distance is 41.2 miles TRIGONOMETRY : Question 2B Two ships sail from a port. The first travels for two hours on a bearing of 195° at a speed of 18mph. The second travels south-east for three hours at a speed of 15mph. How far apart will they be? What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT

23. Question 2B A 45miles 60° c 36miles N b a 195° SE 1. Identify what needs to be found. How far apart will they be? 2. Need distances travelled and angle between paths. 1350 d1 = speed1 x time = 18 x 2 = 36miles d2 = speed2 x time = 15 x 3 = 45miles 3hr@15mph 2hr@18mph NB: SE = 135° Angle = 195° - 135° = 60° 600 3. Sketch triangle. Begin Solution Continue Solution Comments Trigonometry Menu Back to Home

24. Question 2B N 195° SE 4. Apply Cosine rule. How far apart will they be? a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. 1350 3hr@15mph = 362 + 452 – (2 x 36 x 45 x cos60°) 2hr@18mph = 1701 600 a = 1701 = 41.243.... = 41.2 Begin Solution Continue Solution Distance is 41.2 miles Comments Trigonometry Menu Back to Home

25. Comments A 45miles 60° c 36miles b a 1. Identify what needs to be found. Note: Bearings are measured clockwise from N. 2. Need distances travelled and angle between flight paths. d1 = speed1 x time = 18 x 2 = 36miles d2 = speed2 x time = 15 x 3 = 45miles NB: SE = 135° Angle = 195° - 135° = 60° 3. Sketch triangle. Next Comment Trigonometry Menu Back to Home

26. Comments Check formulae list for the cosine rule: a2 = b2 + c2 – 2bc cosA (2 sides and the included angle) Relate to variables used a2 = b2 + c2 – 2bc cosA 4. Apply Cosine rule. a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. = 362 + 452 – (2 x 36 x 45 x cos60°) = 1701 a = 1701 = 41.243.... = 41.2 Next Comment Distance is 41.2 miles Trigonometry Menu Back to Home

27. P 10cm Q S 22cm 15cm R TRIGONOMETRY : Question 3 In the kite shown below PQ = 10cm, QR = 15cm & diagonal PR = 22cm. (a) Find the size of angle QPR. (b) Hence find the area of the kite to the nearest square unit. Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

28. P 3. For area of a triangle with an angle present use: 10cm Q S 22cm 15cm R TRIGONOMETRY : Question 3 In the kite shown below PQ = 10cm, QR = 15cm & diagonal PR = 22cm. (a) Find the size of angle QPR. (b) Hence find the area of the kite to the nearest square unit. 1. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 4. Always remember to round your answer if the questions asks you to. What would you like to do now? 2. Substitute known values, remembering to use brackets as appropriate. Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

29. P 10cm Q S 22cm angleQPR = 35.3° = 127cm2 15cm R TRIGONOMETRY : Question 3 In the kite shown below PQ = 10cm, QR = 15cm & diagonal PR = 22cm. (a) Find the size of angle QPR. (b) Hence find the area of the kite to the nearest square unit. What would you like to do now? Try another like this Go to full solution Go to Comments Go to Trigonometry Menu EXIT

30. Question 3 r2 + q2 - p2 cosP = 2rq = 102 + 222 - 152 2 x 10 x 22 P 10cm Q S 22cm angleQPR = cos-1(0.8159..) = 35.3° 15cm R 1. To find angle when you have 3 sides use 2nd version of cosine rule: = (102 + 222 - 152)  (2 x 10 x 22) = 0.8159… 2. Remember to use inverse function to find angle. (a) Find the size of angle QPR Continue Solution Try another like this Comments Trigonometry Menu Back to Home

31. Question 3 P 10cm Q S 22cm = 127cm2 15cm R 1. Kite = 2 identical triangles. For area of triangles where an angle is present use: (b) Area QPR = ½ qrsinP° = 10 x 22 x sin35.3°  2 = 63.56..cm2 2. Remember to double this and round answer. Area of kite = 2 x 63.56.. (b) Hence find the area of the kite to the nearest square unit = 127.12..cm2 Continue Solution Try another like this Comments Trigonometry Menu Back to Home

32. Markers Comments r2 + q2 - p2 cosP = 2rq = 102 + 222 - 152 2 x 10 x 22 b2 + c2 – a2 2bc angleQPR = cos-1(0.8159..) = 35.3° Check the formulae list for the second form of the cosine rule: 1. To find angle when you have 3 sides use 2nd version of cosine rule: ( 3 sides) cosA = = (102 + 222 - 152)  (2 x 10 x 22) = 0.8159… 2. Remember to use inverse function to find angle. Next Comment Trigonometry Menu Back to Home

33. Markers Comments r2 + q2 - p2 cosP = 2rq = 102 + 222 - 152 2 x 10 x 22 P 10cm Relate to variables used: cosP = = r Q q2 + r2 – p2 2qr q 22cm p 222 + 102 – 152 2x22x10 15cm angleQPR = cos-1(0.8159..) = 35.3° R 1. To find angle when you have 3 sides use 2nd version of cosine rule: = (102 + 222 - 152)  (2 x 10 x 22) = 0.8159… 2. Remember to use inverse function to find angle. Next Comment Trigonometry Menu Back to Home

34. Markers Comments r2 + q2 - p2 cosP = 2rq = 102 + 222 - 152 2 x 10 x 22 angleQPR = cos-1(0.8159..) = 35.3° Note: When keying in to calculator work out the top line and the bottom line before dividing or use brackets. 1. To find angle when you have 3 sides use 2nd version of cosine rule: = (102 + 222 - 152)  (2 x 10 x 22) = 0.8159… 2. Remember to use inverse function to find angle. Next Comment Trigonometry Menu Back to Home

35. E 15cm 15cm 25cm H F 15cm 15cm G TRIGONOMETRY : Question 3B The sides of a rhombus are each 15cm while the main diagonal is 25cm Find the size of angle EFH and hence find the area of the rhombus to the nearest square unit. Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

36. E 15cm 15cm 3. For area of a triangle with an angle present use: 25cm H F 15cm 15cm G TRIGONOMETRY : Question 3B The sides of a rhombus are each 15cm while the main diagonal is 25cm Find the size of angle EFH and hence find the area of the rhombus to the nearest square unit. What would you like to do now? 1. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule Reveal answer only 4. Always remember to round your answer if the questions asks you to. Go to full solution 2. Substitute known values, remembering to use brackets as appropriate. Go to Comments Go to Trigonometry Menu EXIT

37. angleEFH = 33.6° E = 415cm2 15cm 15cm 25cm H F 15cm 15cm G TRIGONOMETRY : Question 3B The sides of a rhombus are each 15cm while the main diagonal is 25cm Find the size of angle EFH and hence find the area of the rhombus to the nearest square unit. What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT

38. Question 3B e2 + h2 - f2 (a) cosF = 2eh = 152 + 252 - 152 2 x 25 x 15 E 15 15 25 H F 15 15 G 1. To find angle when you have 3 sides use 2nd version of cosine rule: = (152 + 252 - 152)  (2 x 25 x 15) = 0.8333… 2. Remember to use inverse function to find angle. (a) Find the size of angle EFH angleEFH = cos-1(0.8333..) = 33.6° Continue Solution Comments Trigonometry Menu Back to Home

39. Question 3B E 15 15 25 H F 15 15 G 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: 33.60 (b) Area EFH = ½ ehsinF° = 25 x 15 x sin33.6°  2 = 207.52....cm2 2. Remember to double this and round answer. Area of kite = 2 x 207.52...... (b) Hence find the area to the nearest square unit = 415.04..cm2 = 415cm2 Continue Solution Comments Trigonometry Menu Back to Home

40. Markers Comments e2 + h2 - f2 (a) cosF = 2eh = 152 + 252 - 152 2 x 25 x 15 b2 + c2 – a2 2bc Check the formulae list for the second form of the cosine rule: 1. To find angle when you have 3 sides use 2nd version of cosine rule: ( 3 sides) cosA = = (152 + 252 - 152)  (2 x 25 x 15) = 0.8333… 2. Remember to use inverse function to find angle. angleEFH = cos-1(0.8333..) = 33.6° Next Comment Trigonometry Menu Back to Home

41. Markers Comments e2 + h2 - f2 (a) cosF = 2eh = 152 + 252 - 152 2 x 25 x 15 Relate to variables used: cosF = = E e2 + h2 – f2 2eh 15cm 15cm h f 222 + 102 – 152 2x22x10 e H 25cm F 1. To find angle when you have 3 sides use 2nd version of cosine rule: = (152 + 252 - 152)  (2 x 25 x 15) = 0.8333… 2. Remember to use inverse function to find angle. angleEFH = cos-1(0.8333..) = 33.6° Next Comment Trigonometry Menu Back to Home

42. Markers Comments E 1 2 15cm 15cm h f e H 25cm F Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: (b) Area EFH = ½ ehsinF° = 25 x 15 x sin33.6°  2 = 207.52....cm2 2. Remember to double this and round answer. Area of kite = 2 x 207.52...... = 415.04..cm2 = 415cm2 Next Comment Trigonometry Menu Back to Home

43. T 105° 5.9cm 35° V U 10cm TRIGONOMETRY : Question 4 In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

44. T 105° 5.9cm 35° V U 10cm TRIGONOMETRY : Question 4 In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. 3. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 5. Always remember to round your answer if the questions asks you to. What would you like to do now? 1. For perimeter need all three sides. So must find TU. 2. Calculate unknown angles. Reveal answer only Go to full solution 4. Substitute known values, remembering to use brackets as appropriate. Go to Comments Go to Trigonometry Menu EXIT

45. T 105° 5.9cm 35° V U 10cm TRIGONOMETRY : Question 4 In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. = 22.6cm What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT

46. Question 4 T v t = v 10 105° = sinV° sinT° 5.9cm sin40° sin105° 35° V U 10cm • Perimeter requires all three sides. • So we need to find TU . 2. Whether you use Sine or Cosine rule need angle V. 400 Angle V = 180° - 35° - 105° = 40° 3. If we use Sine rule: Find the perimeter to one decimal place. 4. Substitute known values: Continue Solution Comments Trigonometry Menu Back to Home

47. Question 4 T v 10 105° = 5.9cm sin40° sin105° 35° V U 10cm 4. Substitute known values: 5. Cross multiply: 400 v x sin105° = 10 x sin40° v = 10 x sin40°  sin105° Find the perimeter to one decimal place. = 6.654… = 6.7cm 6. Answer the question: Continue Solution Perim of  = (6.7 + 10 + 5.9)cm Comments = 22.6cm Trigonometry Menu Back to Home

48. Markers Comments T v t b Sine B c Sine C = v 10 105° v = sinV° sinT° 5.9cm sin40° sin105° t a Sine A 35° V U 10cm Since we can pair off two angles with the opposite sides Sine Rule • Perimeter requires all three sides. • So we need to find TU . 2. Whether you use Sine or Cosine rule need angle V. Refer to the Formulae List : = = Angle V = 180° - 35° - 105° = 40° 3. If we use Sine rule: 4. Substitute known values: Next Comment Trigonometry Menu Back to Home

49. Markers Comments 10 Sine 105˚ v Sine 40˚ v 10 = sin40° sin105° 4. Substitute known values: Go straight to values : = 5. Cross multiply: v x sin105° v = 10 x sin40°  sin105° = 6.654… = 6.7cm 6. Answer the question: Next Comment Perim of  = (6.7 + 10 + 5.9)cm Trigonometry Menu = 22.6cm Back to Home

50. C N A 50° B 12km S TRIGONOMETRY : Question 5 Lighthouse C is on a bearing of 050° from lighthouse A and northwest of lighthouse B. Lighthouse B is 12km due east of lighthouse A. A ship(S) is sailing directly from lighthouse B to lighthouse A. How close does it come to lighthouse C? Hints Answer only Full solution Comments Trig Menu EXIT