I.G..S.E. Trigonometry 01 Index: Please click on the question number you want Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 You can access the solutions from the end of each question
Question 1 y using Pythagoras Theorem, find x in the diagram below. 12 cm x 16 cm lick here to read the solution to this question
Solution to question 1 12 cm x 16 cm y using Pythagoras theorem x = 400 = 20cm x = 12 + 16 2 2 2 = 144 + 256 = 400 lick here to read the question again
Question 2 Find the missing sides marked with letters a. 7cm b. c. a 25 13.2 cm 17 b 5.6 cm 34 c d. Find the height of the isosceles triangle given = 12 cmand ˆ = 49. 49 lick here to read the solution to this question
Solution to question 2 a. a 25 7cm tanθ = a tan25 = 7 a = 7 tan25 = 3.26cm b. c. d. 5.6 cm 13.2 cm 34 17 17 c b sinθ = b sin17 = 13.2 b = 13.2sin17 = 3.86cm cosθ = 5.6 cos34 = c 5.6 c = = 6.75cm cos34 49 N 49 N 6 cm 12 cm onsidering N N tanθ = tan49 = cn = 6tan49 = 6.90cm 6 lick here to read the question again
Question 3 Find the missing angles a. b. 7 cm θ 15 cm 6.5 cm θ 3.3 cm α lick here to read the solution to this question
Solution to question 3 a. 7 cm θ 15 cm cosθ = 7 1 7 cosθ = θ = cos = 62. 2 15 15 b. 3.3 cm 6.5 cm θ tanθ = 3.3 1 3.3 tanθ = θ = tan = 26.9 6. 5 6.5 3.3 cm Either α = 180 90 26.9 = 63.1 6.5 cm 26.9 α or tanθ = 6.5 1 6.5 tanθ = θ = tan 63.1 3. 3 = 3.3 lick here to read the question again
Question 4 n aeroplane flies on a bearing of 132 for 150 km as shown in the diagram in the diagram below. Find how far east and south the aeroplane has flown. N N 132 W E 150 km S lick here to read the solution to this question
Solution to question 4 n aeroplane flies on a bearing of the diagram below. 132 for 150 km as shown in the diagram in N N 132 East θ W E 150 km South S onsider θ = 132 90 = 42 42 150 km The distance travelled east is given by. cosθ = cos 42 = = 150cos 42 = 111km 150 The distance travelled south is given by. sinθ = sin42 = = 150sin42 = 100 km 150 lick here to read the question again
Question 5 The angle of depression of a ship from the top of a cliff is 44. If the cliff is 260 m high find how far the boat is form the base of the cliff. liff 44 260m lick here to read the solution to this question
Solution to question 5 liff 260m θ 44 onsider. θ = 90 44 = 46 260m 46 tanθ = tan46 = 260 = 260tan46 = 269m lick here to read the question again
Question 6 man is 105 m from the base of a tower. He measures the angle of elevation of the top of the tower to be 43. If the man is 1.5 m tall, find the height of the tower. Tower 43 1.5 m lick here to read the solution to this question 105 m
Solution to question 6 Tower 1.5 m 43 D 105 m The height of the tower is given by D + tanθ = tan43 = 105 = 105tan43 = 97. 9m The height of the tower is D + = 1.5 + 97.9 = 99.4m lick here to read the question again
Question 7 vertical pole P stands at one corner of a horizontal field measuring 8 m by 6 m, as shown in the diagram below. P D 32 8 m 6 m a. y considering the triangle DP find the height of the pole P. b. y considering the triangle P calculate the angle of elevation of P from. c. alculate the length of the diagonal of the rectangle D. d. alculate the angle of elevation of P from. lick here to read the solution to this question
Solution to question 7 P D 32 6 m 8 m a. onsidering the triangle DP P D 32 8 m The height of the pole is given by P P tanθ = tan32 = P = 8 tan32 = 5. 00m 8 b. onsidering P P 8tan32 m θ 6 m The angle of elevation of P from is given by angle ˆ P. 8tan32 8tan32 tanθ = tan P ˆ = P ˆ = tan = 39.8 6 6 1 lick here to continue with solution or go to next page
c. D 6 m 8 m y Pythagoras theorem we have that = 8 + 6 2 2 2 = 64 + 36 = 100 = 100 = 10m d. onsidering P P 8tan32 m θ 10 m The angle of elevation of P from is given by angle ˆ P. ˆ 8tan32 tanθ = tan P = P ˆ = tan = 26.6 10 10 lick here to read the question again 1 8tan32