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Trigonometry

Trigonometry. 3D Trigonometry. s. p , q and r are points on level ground, [ sr ] is a vertical flagpole of height h . The angles of elevation of the top of the flagpole from p and q are α and β , respectively. h. β. 30 º. q. r. α. 60 º.

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Trigonometry

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  1. Trigonometry 3D Trigonometry

  2. s p, q and r are points on level ground, [sr] is a vertical flagpole of height h. The angles of elevation of the top of the flagpole from p and q are α and β, respectively. h β 30º q r α 60º (i)If |α|= 60º and |β|= 30º, express |pr| and |qr| in terms of h. p

  3. s h p r s OPP h 60º ADJ 30º q r 60º p

  4. s h OPP 30º q r ADJ 60º p

  5. (ii)Find|pq| in terms of h, if tan qrp= A s Pythagoras’ Theorem h 30º q r 60º p a2 = b2+c2– 2bccosA

  6. q r p a2 = b2+c2– 2bccosA

  7. slanted edge Sommets 5 1 90 72 -17.00787 2 270 72 -17.00787 3 270 252 -17.00787 4 90 252 -17.00787 5 180 162 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 70.4549365234253 18.5097949350555 -90.9217083697797 2 245.53709547791 30.1395555679914 -131.05880197977 3 204.102094663231 100.878766349906 -291.305813700551 4 29.0199357087471 89.2490057169698 -251.168720090559 5 139.83922903418 137.698138286049 -157.341972681488 The great pyramid at Giza in Egypt has a square base and four triangular faces. The base of the pyramid is of side 230 metres and the pyramid is 146 metres high. The top of the pyramid is directly above the centre of the base. (i) Calculate the length of one of the slanted edges, correct to the nearest metre. Pythagoras’ theorem x 230 m 2 = 105800 x 162·6 146 x= 325·269.. = 162·6 162·6 230 m 2 Sommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 2006 Paper 2 Q5 (b)

  8. slanted edge Sommets 5 1 90 72 -17.00787 2 270 72 -17.00787 3 270 252 -17.00787 4 90 252 -17.00787 5 180 162 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 70.4549365234253 18.5097949350555 -90.9217083697797 2 245.53709547791 30.1395555679914 -131.05880197977 3 204.102094663231 100.878766349906 -291.305813700551 4 29.0199357087471 89.2490057169698 -251.168720090559 5 139.83922903418 137.698138286049 -157.341972681488 The great pyramid at Giza in Egypt has a square base and four triangular faces. The base of the pyramid is of side 230 metres and the pyramid is 146 metres high. The top of the pyramid is directly above the centre of the base. (i) Calculate the length of one of the slanted edges, correct to the nearest metre. Pythagoras’ theorem l 146 m 2 = 47754·76 l 146 l= 218·528.. 162·6 162·6 m = 219 m Sommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 2006 Paper 2 Q5 (b)

  9. slanted edge 1 2 1 2 Area of triangle = base × height = (230)(186·4) Sommets 5 1 90 72 -17.00787 2 270 72 -17.00787 3 270 252 -17.00787 4 90 252 -17.00787 5 180 162 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 70.4549365234253 18.5097949350555 -90.9217083697797 2 245.53709547791 30.1395555679914 -131.05880197977 3 204.102094663231 100.878766349906 -291.305813700551 4 29.0199357087471 89.2490057169698 -251.168720090559 5 139.83922903418 137.698138286049 -157.341972681488 (ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces) Pythagoras’ theorem 219 m h =34736 = 186·375.. = 186·4 m 115 m 230 m 2 h = 21436 m2 Sommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 2006 Paper 2 Q5 (b)

  10. slanted edge Sommets 5 1 90 72 -17.00787 2 270 72 -17.00787 3 270 252 -17.00787 4 90 252 -17.00787 5 180 162 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 70.4549365234253 18.5097949350555 -90.9217083697797 2 245.53709547791 30.1395555679914 -131.05880197977 3 204.102094663231 100.878766349906 -291.305813700551 4 29.0199357087471 89.2490057169698 -251.168720090559 5 139.83922903418 137.698138286049 -157.341972681488 (ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces) Pythagoras’ theorem 219 m h =34736 = 186·375.. = 186·4 m 115 m 2 h Total area = 21436  4 = 85744 m2 = 86000 m2 Sommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 Faces 5 1 4 4 3 2 1 2 3 1 2 5 3 3 2 3 5 4 3 3 4 5 5 3 4 1 5 BSommets 5 1 99 63 -17.00787 2 225 63 -17.00787 3 225 189 -17.00787 4 99 189 -17.00787 5 162 126 68.03149 2006 Paper 2 Q5 (b)

  11. r h θ p q 3x s r t 2θ x h θ p q 3x pqrs is a vertical wall of height h on level ground. p is a point on the ground in front of the wall. The angles of elevation of r from p is θ and the angle of elevation of s from p is 2θ. |pq| = 3|pt|. Find θ. 2005 Paper 2 Q5 (c)

  12. s h 2θ p t x s r t 2θ x h θ p q 3x pqrs is a vertical wall of height h on level ground. p is a point on the ground in front of the wall. The angles of elevation of r from p is θ and the angle of elevation of s from p is 2θ. |pq| = 3|pt|. Find θ. 2005 Paper 2 Q5 (c)

  13. s r t 2θ x h θ 3 x tanθ x tan2θ = p q 3x Let t = tan θ 2005 Paper 2 Q5 (c)

  14. d abc is an isosceles triangle on a horizontal plane, such that |ab|=|ac|= 5 and |bc|= 4. m is the midpoint of [bc]. 2 5 b a A (i)Find |bac| to the nearest degree. 5 m 4 c 4.D Question 4

  15. d 21 2 Ðamd = tan 21 abc is an isosceles triangle on a horizontal plane, such that |ab|=|ac|= 5 and |bc|= 4. m is the midpoint of [bc]. 2 5 b a (ii)A vertical pole [ad] is erected at a such that |ad| =2, find |amd| to the nearest degree. 2 5 m c 2 = am 21 amd 4.D Question 4

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