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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. q h β 30º (i) If | α | = 60º and | β | = 30º, express | pr | and | qr | in terms of h. 60º α p r s s h OPP 60º p ADJ h tan 60 pr h 3 pr r q h 30º 60º h pr 3 p r s h tan 30 qr 1 h 3 qr q 30º ADJ qr 3h p 60º h OPP r (ii) Find | pq | in terms of h, if tan qrp = 8. s Pythagoras’ Theorem x 2 12 ( 8 )2 1 8 9 q 30º 3h 3x 8 A 1 p 1h cos A 3 r 60º h 3 a2 = b2 + c2 – 2bccosA 2 h h 1 qp 3h 2 3h 3 3 3 2 2 2 2 9 h h 2 h h 2 3h 2 h2 3 3 3 1 2 cos A 8h 3 3 q r 2 qp 2 2 8h 3 8 h 3 p h 3 a2 = b2 + c2 – 2bccosA 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. slanted edge 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 2 2302 2302 x 105800 x 2 146 162·6 x 325·269.. 162·6 2 2006 Paper 2 Q5 (b) 162·6 230 m 230 m 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. slanted edge 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 2 1462 162·62 l l 47754·76 l 218·528.. 2 146 162·6 219 m 2006 Paper 2 Q5 (b) 162·6 m 146 m (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 h 115 2 2 2 slanted edge 219 m h h2 2192 1152 115 m230 m h 2 34736 186·375.. 186·4 m 1 Area of triangle base × height 2 1 (230)(186·4) 2 2 21436 m 2006 Paper 2 Q5 (b) (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 h 115 2 2 2 slanted edge 219 m h2 2192 1152 115 m h 2 34736 186·375.. 186·4 m Total area 21436 4 85744 m2 86000 m2 2006 Paper 2 Q5 (b) h s r h θ p q 3x 2θ p h tan 3x r t x h θ 3x h 3x tan 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) q s s h p 2θ x 2θ t p h tan 2 x r t x h θ 3x h x tan 2 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) q s 2 tan tan 2 1 tan 2 2 tan 3xtan θ xtan 2θ 1 tan 2 2θ p 3t 1 t t 3t 0 t 1 3t 1 t 3 2 x h θ q 3x Let t = tan θ t 0 2t 3t 2 1 t 3 r t 2 2 2t 3t 3t 2t 0 1 3t 0 1 t tan 3 2005 Paper 2 Q5 (c) 3 2 6 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 5 (i) Find | bac | to the nearest degree. a 2 b2 c 2 2bc cos A m4 42 52 52 2(5)(5)cos A 16 25 25 50cos A 16 50 50cos A c 50cos A 50 16 34 cos A 50 A 47·156.... A 47 a d abc is an isosceles triangle on a horizontal plane, such that |ab| |ac| 5 and |bc| 4. m is the midpoint of [bc]. (ii) A vertical pole [ad] is erected at a such that |ad | 2, find |amd | to the nearest degree. am 2 5 2 2 5 b 2 m am 25 4 am 21 amd 2 21 a 21 5 2 2 2 tan amd 2 c amd 23·578.. 24