1. Trigonometry 3D Trigonometry

2. r s h p q β α 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. (i) If | α | = 60º and | β | = 30º, express | pr | and | qr | in terms of h. 30º 60º

3. s h rp ADJ OPP s h r p q 30º 60º

4. s h r p q OPP ADJ 30º 60º

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

6. r p q a 2 = b 2 + c 2 – 2bccosA

7. slanted edge 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. 230 m Pythagoras’ theorem x  105800 2 325·269.. x 2 x x  162·6  146 2006 Paper 2 Q5 (b)

8. slanted edge 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. 146 m 162·6 m Pythagoras’ theorem l  47754·76 2 l  218·528.. l  219 m 162·6 146 2006 Paper 2 Q5 (b)

9. (ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces) slanted edge 219 m 230 m  34736 2  186·375.. h  186·4 m h 115 m Pythagoras’ theorem Area of triangle  base × height 1212  (230)(186·4) 1212  21436 m 2 2006 Paper 2 Q5 (b)

10. slanted edge (ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces) 219 m  34736 2  186·375.. h  186·4 m h 115 m Pythagoras’ theorem Total area  21436  4  85744 m 2  86000 m 2 2006 Paper 2 Q5 (b)

11. θ 2θ2θ 3x3x x q 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. p s r h t θ 3x3x q p r h 2005 Paper 2 Q5 (c)

12. θ 2θ2θ 3x3x x q 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. p s r h t 2θ2θ x t p s h 2005 Paper 2 Q5 (c)

13. tan θx θ 2θ2θ 3x3x x q p s r h t  tan 2θx3 Let t = tan θ 2005 Paper 2 Q5 (c)

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

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