I.G.C.S.E. Trigonometry. You can access the solutions from the end of each question

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1 I.G..S.E. Trigonometry Index: Please click on the question number you want Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 You can access the solutions from the end of each question

2 Question 1 1. For the triangles below calculate the missing lengths. a. b. y 3 cm x 12 m 5 cm 9 m lick here to read the solution to this question

3 Solution to question 1 a. Using Pythagoras theorem we have 3 cm x 5 cm x = x = = = 34 = 5.83cm b. y 12 m Using Pythagoras theorem, remembering that 12 m is the otenuse = y y = m y = = 63 = 7.94m lick here to read the question again

4 Question 2 For the triangles below find the missing letters. a. b cm a 7.5 cm 22 b c. d. 4.6 cm 7.3 cm c cm d lick here to read the solution to this question

5 Solution to question 2 First label the sides of each triangle, acent, osite and otenuse. a cm a The missing side is the osite side and we have the acent and angle hence a tanθ = tan34 = 1.2 a = 1.2 tan34 = 0.809cm b. 7.5 cm 22 b The missing side is the acent side and we have the otenuse and angle hence b cosθ = cos22 = 7.5 b = 7.5 cos22 = 6.95cm c. 7.3 cm c 4.6 cm The angle is and we know the osite side and otenuse hence 4. 6 sinθ = sinc = = = c sin 39.1 d cm d The missing side is the otenuse side and we have the acent side and angle hence 3.45 cosθ = cos17.5 = d 3.45 d = = 3.62cm cos17.5 lick here to read the question again

6 Question 3 In the diagram below find the following lengths a. b. c cm lick here to read the solution to this question

7 Solution to question cm a. First draw out triangle separately cosθ = cos27 = 5.6 = cos27 = 6.29cm cm 27 b. First draw out triangle separately cosθ = cos32 = 5.6 cos cos32 = = 5.33cm cos27 c. onsidering triangle. 5.6 cos27 32 sinθ = sin32 = 5.6 cos sin32 = = 3.33cm cos27 lick here to read the question again

8 Question 4 ship sails from Porthampton on a bearing of 120 towards a buoy marker for 60 km and then changes course to a bearing of 150 for another 80 km until it reaches Littlehampton. a. raw a scale diagram using a scale of 1 cm to 10 km and find the distance and bearing of Littlehampton from Porthampton. b. Using trigonometry, calculate how far east and how far south the ship has travelled. c. alculate the distance and bearing of Littlehampton from Porthampton. lick here to read the solution to this question

9 Solution to question 4 a. Scale drawing scale 1 cm to 10 km. N P 120 earing to be measured istance to be measured 60 km N 150 istance = 135 km, measured with a ruler. 80 km earing is137 a protractor. measured with b. rawing out triangles P and L separately we have P L 60 km 80 km L P cosθ = cos30 = 60 P = 60 cos30 sinθ = sin30 (East) = 60 = 60 sin30 (Souh) t P cosθ = cos60 = 80 P = 80 cos60 sinθ = sin60 (East) = 80 = 80 sin60 (Souh) t lick here to continue with solution or go to next page

10 The total distance east is = 60cos cos60 = = 92.0km The total distance south is = 60sin sin60 = = 99.3km c. The distance PL is found by Pythagoras theorem ( 60cos30 80cos60 ) ( 60sin30 80sin60 ) = = 135km 2 2 PL = We need to find angle N ˆ PL so we use triangle PL P 60cos sin60 60sin sin60 L tanθ = 1 60sin sin60 PL ˆ = tan 60cos cos60 = ut we need to find the bearing of Littlehampton from Portshampton, which is given by = lick here to read the question again

11 Question 5 The height of an eye of a man is 175 cm. He is standing 20 m from a building, which has a flagpole on it. He looks up at an angle of elevation 23 and sees the top of the building. He then looks up at the top of the flagpole, which has an angle of elevation of 28. a. alculate the height of the building. b. alculate the height of the flagpole. lick here to read the solution to this question

12 Solution to question 5 First draw a diagram m m a. onsider triangle m b. onsider triangle m tanθ = tan23 = 20 = 20 tan23 = 8.49m The height of the wall is = = = 10.2m tanθ = tan28 = 20 = 20 tan28 = 10.63m The height of the flagpole from the ground is = = m Height of flagpole is = = 2.14m lick here to read the question again

13 Question 6 rectangular box EFGH has = 5cm, = 6cm and E = 3cm alculate H G a., b. G, E F c. The angle ˆ G. 3 cm 5cm 6cm lick here to read the solution to this question

14 Solution to question 6 H G E F 3 cm 5cm 6cm a. onsider triangle y Pythagoras theorem 6 cm = = 61 = 7.81cm 5 cm b. onsider triangle G. G y Pythagoras theorem G = = 70 = 8.37cm ( ) Note 61 = m 3 cm c. onsider triangle G. ˆ 1 3 tanθ = G = tan 21.0 = 61 lick here to read the question again